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1938 | 1938_1 | A solid is bounded by two bases in the horizontal planes $z = h/2$ and $z = -h/2$, and by such a surface that the area of every section in a horizontal plane is given by a formula of the sort \[ \text{Area} = a_0 z^3 + a_1 z^2 + a_2 z + a_3 \] (where as special cases some of the coefficients may be $0$). Find the volume formula where $B_1$ and $B_2$ are the areas of the bases, and $M$ is the area of the middle horizontal section. Also fine the value of $a_0$ for which the formulas for the volume of a cone and of a sphere can be included in this formula and add it to the expression of the volume you find for the final answer. | The volume in question is given by \[ V = \int_{-h/2}^{h/2} (a_0 z^3 + a_1 z^2 + a_2 z + a_3) dz \] \[ = \frac{a_1 h^3}{12} + a_3 h. \] On the other hand, the base areas and $M$ are given by \[ B_1 = \frac{a_0 h^3}{8} + \frac{a_1 h^2}{4} + \frac{a_2 h}{2} + a_3, \] \[ B_2 = -\frac{a_0 h^3}{8} + \frac{a_1 h^2}{4} - \frac{a_2 h}{2} + a_3, \] \[ M = a_3, \] so that the suggested expression \( \frac{1}{6} h[B_1 + B_2 + 4M] \) works out to be \[ \frac{1}{6} h \left( \frac{a_1 h^2}{2} + 6a_3 \right) = \frac{a_1 h^3}{12} + a_3 h = V \] as required.\n\nThe formula \( V = \boxed{\frac{1}{6}h(B_1 + B_2 + 4M)} \) is known in solid geometry as the prismoidal formula. It is closely related to Simpson's rule in numerical integration. | algebraic | putnam (modified boxing) | Geometry Calculus | A solid is bounded by two bases in the horizontal planes $z = h/2$ and $z = -h/2$, and by such a surface that the area of every section in a horizontal plane is given by a formula of the sort \[ \text{Area} = a_0 z^3 + a_1 z^2 + a_2 z + a_3 \] (where as special cases some of the coefficients may be $0$). Show that the volume is given by the formula \[ V = \frac{1}{6} h[B_1 + B_2 + 4M], \] where $B_1$ and $B_2$ are the areas of the bases, and $M$ is the area of the middle horizontal section. Show that the formulas for the volume of a cone and of a sphere can be included in this formula when $a_0 = 0$. | The volume in question is given by \[ V = \int_{-h/2}^{h/2} (a_0 z^3 + a_1 z^2 + a_2 z + a_3) dz \] \[ = \frac{a_1 h^3}{12} + a_3 h. \] On the other hand, the base areas and $M$ are given by \[ B_1 = \frac{a_0 h^3}{8} + \frac{a_1 h^2}{4} + \frac{a_2 h}{2} + a_3, \] \[ B_2 = -\frac{a_0 h^3}{8} + \frac{a_1 h^2}{4} - \frac{a_2 h}{2} + a_3, \] \[ M = a_3, \] so that the suggested expression \( \frac{1}{6} h[B_1 + B_2 + 4M] \) works out to be \[ \frac{1}{6} h \left( \frac{a_1 h^2}{2} + 6a_3 \right) = \frac{a_1 h^3}{12} + a_3 h = V \] as required.\n\nThe formula \( V = \frac{1}{6} h[B_1 + B_2 + 4M] \) is known in solid geometry as the prismoidal formula. It is closely related to Simpson's rule in numerical integration. | 0 |
1938 | 1938_2 | A can buoy is to be made of three pieces, namely, a cylinder and two equal cones, the altitude of each cone being equal to the altitude of the cylinder. For a given area of surface, find the ratio of $h_0/r_0$ where $h_0$ is the hight at which the maximum volume is achieved and $r_0$ is the corresponding radius at maximum volume? | Let $r$ be the radius of the cylinder, and $h$ its altitude. The given condition is \[ S = 2\pi rh + 2(\pi r\sqrt{h^2 + r^2}) = \text{constant}, \] and the volume of the buoy is \[ V = \pi r^2 h + \frac{2\pi r^2 h}{3} = \frac{5\pi r^2 h}{3}. \] The required problem is to find the maximum value of $V$ subject to condition $(1)$. This can be done by the method of Lagrange multipliers, but in this particular problem it is easier to solve $(1)$ for $h$ and express $V$ as a function of $r$. We have \[ (S - 2\pi rh)^2 = 4\pi^2r^2(h^2 + r^2), \] whence \[ h = \frac{S^2 - 4\pi^2r^4}{4\pi S}, \] and the expression for $V$ becomes \[ V = \frac{5r}{12S}(S^2 - 4\pi^2r^4). \] Since $r$ and $V$ must be positive, the domain of interest is given by \[ 0 < r < \sqrt[4]{\frac{S^2}{4\pi^2}}. \] We compute the derivative and equate it to zero to get \[ \frac{dV}{dr} = \frac{5S}{12} - \frac{100\pi^2r^4}{12S} = 0. \] The only critical value is \[ r_0 = \sqrt[4]{\frac{S^2}{20\pi^2}}. \] Since $V \to 0$ as $r \to 0$ or as $r \to \sqrt[4]{S^2/4\pi^2}$, and is positive in between, the critical value $r_0$ yields a maximum for $V$. The corresponding value of $h$ is found from $(3)$ to be $h_0 = \frac{2}{5}\sqrt{5} r_0$. The shape of the buoy is completely determined by the ratio \[ \frac{h_0}{r_0} = \boxed{\frac{2}{5}\sqrt{5}}. \] | numerical | putnam (modified boxing) | Geometry Calculus | A can buoy is to be made of three pieces, namely, a cylinder and two equal cones, the altitude of each cone being equal to the altitude of the cylinder. For a given area of surface, what shape will have the greatest volume? | Let $r$ be the radius of the cylinder, and $h$ its altitude. The given condition is \[ S = 2\pi rh + 2(\pi r\sqrt{h^2 + r^2}) = \text{constant}, \] and the volume of the buoy is \[ V = \pi r^2 h + \frac{2\pi r^2 h}{3} = \frac{5\pi r^2 h}{3}. \] The required problem is to find the maximum value of $V$ subject to condition $(1)$. This can be done by the method of Lagrange multipliers, but in this particular problem it is easier to solve $(1)$ for $h$ and express $V$ as a function of $r$. We have \[ (S - 2\pi rh)^2 = 4\pi^2r^2(h^2 + r^2), \] whence \[ h = \frac{S^2 - 4\pi^2r^4}{4\pi S}, \] and the expression for $V$ becomes \[ V = \frac{5r}{12S}(S^2 - 4\pi^2r^4). \] Since $r$ and $V$ must be positive, the domain of interest is given by \[ 0 < r < \sqrt[4]{\frac{S^2}{4\pi^2}}. \] We compute the derivative and equate it to zero to get \[ \frac{dV}{dr} = \frac{5S}{12} - \frac{100\pi^2r^4}{12S} = 0. \] The only critical value is \[ r_0 = \sqrt[4]{\frac{S^2}{20\pi^2}}. \] Since $V \to 0$ as $r \to 0$ or as $r \to \sqrt[4]{S^2/4\pi^2}$, and is positive in between, the critical value $r_0$ yields a maximum for $V$. The corresponding value of $h$ is found from $(3)$ to be $h_0 = \frac{2}{5}\sqrt{5} r_0$. The shape of the buoy is completely determined by the ratio \[ \frac{h_0}{r_0} = \frac{2}{5}\sqrt{5}. \] | 0 |
1938 | 1938_3 | If a particle moves in the plane, we may express its coordinates $x$ and $y$ as functions of the time $t$. If $x = t^3 - t$ and $y = t^4 + t$, show that the curve has a point of inflection at $t = 0$ and find the maximum speed/absolute value of velocity it achieves durings its entire movement.$. | If the velocity vector at time $t$ is of length $v$ and has direction $\theta$, then $\dot{x} = v \cos \theta$, $\dot{y} = v \sin \theta$, and $\ddot{x} = \dot{v} \cos \theta - v \dot{\theta} \sin \theta$, $\ddot{y} = \dot{v} \sin \theta + v \dot{\theta} \cos \theta$. Thus $\dot{x} \ddot{y} - \ddot{x} \dot{y} = v^2 \dot{\theta}$ and $v^2 = \dot{x}^2 + \dot{y}^2$. From the given parametric data,\n\n$v^2 \dot{\theta} = (3t^2 - 1)12t^2 - (4t^3 + 1)6t = 6t(2t^3 - 2t - 1)$\n\n$v^2 = 16t^6 + 9t^4 + 8t^3 - 6t^2 + 2.$\n\nFrom the $v^2 \dot{\theta}$ relation, $\dot{\theta}$ changes sign as $t$ passes through 0. To rule out the possibility of a cusp point, one notes that $v^2 \neq 0$ at $t = 0$. Hence the curve has an inflection point at $t = 0$. Also\n\n$\frac{d(v^2)}{dt} = 96t^5 + 36t^4 + 24t^2 - 12t$\n\n$\left.\frac{d(v^2)}{dt}\right|_{t=0} = 0$\n\n$\left.\frac{d^2(v^2)}{dt^2}\right|_{t=0} = -\boxed{12}.$\n\nSo $v^2$ has a (local) maximum at $t = 0$. Hence the speed $v$ also has a local maximum. | numerical | putnam | Calculus Analysis | If a particle moves in the plane, we may express its coordinates $x$ and $y$ as functions of the time $t$. If $x = t^3 - t$ and $y = t^4 + t$, show that the curve has a point of inflection at $t = 0$ and that the velocity of the moving particle has a maximum at $t = 0$. | First Solution. If the velocity vector at time $t$ is of length $v$ and has direction $\theta$, then $\dot{x} = v \cos \theta$, $\dot{y} = v \sin \theta$, and $\ddot{x} = \dot{v} \cos \theta - v \dot{\theta} \sin \theta$, $\ddot{y} = \dot{v} \sin \theta + v \dot{\theta} \cos \theta$. Thus $\dot{x} \ddot{y} - \ddot{x} \dot{y} = v^2 \dot{\theta}$ and $v^2 = \dot{x}^2 + \dot{y}^2$. From the given parametric data,\n\n$v^2 \dot{\theta} = (3t^2 - 1)12t^2 - (4t^3 + 1)6t = 6t(2t^3 - 2t - 1)$\n\n$v^2 = 16t^6 + 9t^4 + 8t^3 - 6t^2 + 2.$\n\nFrom the $v^2 \dot{\theta}$ relation, $\dot{\theta}$ changes sign as $t$ passes through 0. To rule out the possibility of a cusp point, one notes that $v^2 \neq 0$ at $t = 0$. Hence the curve has an inflection point at $t = 0$. Also\n\n$\frac{d(v^2)}{dt} = 96t^5 + 36t^4 + 24t^2 - 12t$\n\n$\left.\frac{d(v^2)}{dt}\right|_{t=0} = 0$\n\n$\left.\frac{d^2(v^2)}{dt^2}\right|_{t=0} = -12.$\n\nSo $v^2$ has a (local) maximum at $t = 0$. Hence the speed $v$ also has a local maximum. | 0 |
1938 | 1938_5 | Evaluate the following limits:\n\n(i) $\lim_{n \to \infty} \frac{n^2}{e^n}$.\n\n(ii) $\lim_{x \to 0} \frac{1}{x} \int_0^x (1 + \sin 2t)^{1/t} \, dt and return the sume as the final answer.$ | \textbf{(i)} It follows from L'Hospital's rule that\n\[\lim_{x \to \infty} \frac{x^2}{e^x} = \lim_{x \to \infty} \frac{2x}{e^x} = \lim_{x \to \infty} \frac{2}{e^x} = 0,\] whence the desired limit is zero.\n\nAlternatively, one could use the fact that for $x > 0$,\n\[ e^x = \sum_{n=0}^\infty \frac{x^n}{n!} > \frac{x^3}{6}, \] whence \[ 0 < \frac{x^2}{e^x} < \frac{6}{x}, \] and, as $x \to \infty$, $\lim_{x \to \infty} 6/x = 0$, whence the desired limit is zero.\n\n\textbf{(ii)} By L'Hospital's rule,\n\[ \lim_{x \to 0} \frac{1}{x} \int_0^x (1 + \sin 2t)^{1/t} \, dt = \lim_{x \to 0} (1 + \sin 2x)^{1/x}, \] provided the latter limit exists. Let $\phi(x) = (1 + \sin 2x)^{1/x}$. Then\n\[ \log \phi(x) = \frac{\log (1 + \sin 2x)}{x}. \] Again using L'Hospital's rule,\n\[ \lim_{x \to 0} \log \phi(x) = \lim_{x \to 0} \frac{2 \cos 2x}{1 + \sin 2x} = 2. \] Since the exponential function is continuous,\n\[ \lim_{x \to 0} \phi(x) = e^2. \] Therefore,\n\[ \lim_{x \to 0} \frac{1}{x} \int_0^x (1 + \sin 2t)^{1/t} \, dt = e^2. \] This makes the final answer \boxed{e^2}. | numerical | putnam (modified boxing) | Calculus | Evaluate the following limits:\n\n(i) $\lim_{n \to \infty} \frac{n^2}{e^n}$.\n\n(ii) $\lim_{x \to 0} \frac{1}{x} \int_0^x (1 + \sin 2t)^{1/t} \, dt.$ | \textbf{Solution.}\n\n\textbf{(i)} It follows from L'Hospital's rule that\n\[\lim_{x \to \infty} \frac{x^2}{e^x} = \lim_{x \to \infty} \frac{2x}{e^x} = \lim_{x \to \infty} \frac{2}{e^x} = 0,\] whence the desired limit is zero.\n\nAlternatively, one could use the fact that for $x > 0$,\n\[ e^x = \sum_{n=0}^\infty \frac{x^n}{n!} > \frac{x^3}{6}, \] whence \[ 0 < \frac{x^2}{e^x} < \frac{6}{x}, \] and, as $x \to \infty$, $\lim_{x \to \infty} 6/x = 0$, whence the desired limit is zero.\n\n\textbf{(ii)} By L'Hospital's rule,\n\[ \lim_{x \to 0} \frac{1}{x} \int_0^x (1 + \sin 2t)^{1/t} \, dt = \lim_{x \to 0} (1 + \sin 2x)^{1/x}, \] provided the latter limit exists. Let $\phi(x) = (1 + \sin 2x)^{1/x}$. Then\n\[ \log \phi(x) = \frac{\log (1 + \sin 2x)}{x}. \] Again using L'Hospital's rule,\n\[ \lim_{x \to 0} \log \phi(x) = \lim_{x \to 0} \frac{2 \cos 2x}{1 + \sin 2x} = 2. \] Since the exponential function is continuous,\n\[ \lim_{x \to 0} \phi(x) = e^2. \] Therefore,\n\[ \lim_{x \to 0} \frac{1}{x} \int_0^x (1 + \sin 2t)^{1/t} \, dt = e^2. \] | 0 |
1938 | 1938_8_i | Let $A_{ik}$ be the cofactor of $a_{ik}$ in the determinant\n\[ d = \begin{vmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{vmatrix}. \] Let $D$ be the corresponding determinant with $a_{ik}$ replaced by $A_{ik}$. Find the value of $D in terms of $d$. | Let $\alpha$ be the matrix of the given determinant with elements $a_{ik}$ and let $\beta$ be the matrix of the cofactors $A_{ik}$, and let $\gamma$ be the transpose of $\beta$. Then the product matrix $\alpha \gamma$ is a diagonal matrix with all entries on the main diagonal equal to $d$.\n\nThus $\det(\alpha \gamma) = d^4 = (\det \alpha)(\det \gamma) = (\det \alpha)(\det \beta) = dD$. The equation\n\n\[ dD = d^4. \]\n\n is an identity between polynomials in the 16 matrix entries regarded as independent indeterminates. Since there certainly exists a $4 \times 4$ matrix whose determinant is not zero, $d$ is non-zero in the polynomial ring. Since the polynomial ring is an integral domain, the result\n\n\[ D = \boxed{d^3} \]\n\n follows from (1). | algebraic | putnam (modified boxing) | Linear Algebra | Let $A_{ik}$ be the cofactor of $a_{ik}$ in the determinant\n\[ d = \begin{vmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{vmatrix}. \] Let $D$ be the corresponding determinant with $a_{ik}$ replaced by $A_{ik}$. Prove $D = d^3$. | Let $\alpha$ be the matrix of the given determinant with elements $a_{ik}$ and let $\beta$ be the matrix of the cofactors $A_{ik}$, and let $\gamma$ be the transpose of $\beta$. Then the product matrix $\alpha \gamma$ is a diagonal matrix with all entries on the main diagonal equal to $d$.\n\nThus $\det(\alpha \gamma) = d^4 = (\det \alpha)(\det \gamma) = (\det \alpha)(\det \beta) = dD$. The equation\n\n\[ dD = d^4. \]\n\n is an identity between polynomials in the 16 matrix entries regarded as independent indeterminates. Since there certainly exists a $4 \times 4$ matrix whose determinant is not zero, $d$ is non-zero in the polynomial ring. Since the polynomial ring is an integral domain, the result\n\n\[ D = d^3 \]\n\n follows from (1). | 0 |
1938 | 1938_8_ii | Let $P(y) = Ay^2 + By + C$ be a quadratic polynomial in $y$. If the roots of the quadratic equation $P(y) - y = 0$ are $a$ and $b$ ($a \neq b$), show that $a$ and $b$ are roots of the biquadratic equation $P(P(y)) - y = 0$. Hence write down a quadratic equation which will give the other two roots, $c$ and $d$, of the biquadratic. Apply this result to solving the following biquadratic equation:\n\n\[ (y^2 - 3y + 2)^2 - 3(y^2 - 3y + 2) + 2 - y = 0 and give the absolute value of the difference of all its roots. \] | Since $a$ is a root of $P(y) - y = 0$, we have $P(a) = a$. Then $P(P(a)) = P(a) = a$, so $a$ is a root of $P(P(y)) - y = 0$. Similarly, $b$ is a root of this biquadratic.\n\nLet $Q(y) = P(P(y)) - y$. To find the other zeros of $Q$, note that $P(y) - y = Ay^2 + (B - 1)y + C = A(y - a)(y - b)$, whence $A(a + b) = 1 - B$. Then\n\n\[ Q(y) = P(P(y)) - P(y) + P(y) - y \]\n\[ = A\{P(y) - a\}\{P(y) - b\} + A(y - a)(y - b) \]\n\[ = A\{A(y - a)(y - b) + y - a\}\{A(y - a)(y - b) + y - b\} + A(y - a)(y - b) \]\n\[ = A(y - a)(y - b)R(y), \]\n\nwhere\n\[ R(y) = \{A(y - b) + 1\}\{A(y - a) + 1\} + 1 \]\n\[ = AP(y) + Ay - A(a + b) + 2 \]\n\[ = A^2y^2 + A(B + 1)y + AC + B + 1. \]\n\nThe roots $c$ and $d$ are the zeros of $R$, so the required quadratic equation for $c$ and $d$ is\n\[ A^2y^2 + A(B + 1)y + AC + B + 1 = 0. \]\n\nIn the special case given, $A = 1$, $B = -3$, $C = 2$, and $R(y) = y^2 - 2y$. The zeros of $P(y) - y$ are $2 \pm \sqrt{2}$, so the zeros of $Q$ are $2 \pm \sqrt{2}, 0$, and $2$ with the final answer being \boxed{2\sqrt{2} - 2}. | numerical | putnam (modified boxing) | Algebra | Let $P(y) = Ay^2 + By + C$ be a quadratic polynomial in $y$. If the roots of the quadratic equation $P(y) - y = 0$ are $a$ and $b$ ($a \neq b$), show that $a$ and $b$ are roots of the biquadratic equation $P(P(y)) - y = 0$. Hence write down a quadratic equation which will give the other two roots, $c$ and $d$, of the biquadratic. Apply this result to solving the following biquadratic equation:\n\n\[ (y^2 - 3y + 2)^2 - 3(y^2 - 3y + 2) + 2 - y = 0. \] | Since $a$ is a root of $P(y) - y = 0$, we have $P(a) = a$. Then $P(P(a)) = P(a) = a$, so $a$ is a root of $P(P(y)) - y = 0$. Similarly, $b$ is a root of this biquadratic.\n\nLet $Q(y) = P(P(y)) - y$. To find the other zeros of $Q$, note that $P(y) - y = Ay^2 + (B - 1)y + C = A(y - a)(y - b)$, whence $A(a + b) = 1 - B$. Then\n\n\[ Q(y) = P(P(y)) - P(y) + P(y) - y \]\n\[ = A\{P(y) - a\}\{P(y) - b\} + A(y - a)(y - b) \]\n\[ = A\{A(y - a)(y - b) + y - a\}\{A(y - a)(y - b) + y - b\} + A(y - a)(y - b) \]\n\[ = A(y - a)(y - b)R(y), \]\n\nwhere\n\[ R(y) = \{A(y - b) + 1\}\{A(y - a) + 1\} + 1 \]\n\[ = AP(y) + Ay - A(a + b) + 2 \]\n\[ = A^2y^2 + A(B + 1)y + AC + B + 1. \]\n\nThe roots $c$ and $d$ are the zeros of $R$, so the required quadratic equation for $c$ and $d$ is\n\[ A^2y^2 + A(B + 1)y + AC + B + 1 = 0. \]\n\nIn the special case given, $A = 1$, $B = -3$, $C = 2$, and $R(y) = y^2 - 2y$. The zeros of $P(y) - y$ are $2 \pm \sqrt{2}$, so the zeros of $Q$ are \boxed{$2 \pm \sqrt{2}, 0$, and $2$}. | 0 |
1938 | 1938_9 | Find all the ranges of the domain of the equation \[ y y'' - 2(y')^2 = 0 \] which pass through the point $x = 1, y = 1$ and return them in list format, i.e., inside [] being comma separated. | $1 / y^3$ is an integrating factor since\n\[ \frac{d}{dx} \left( \frac{y'}{y^2} \right) = \frac{y y'' - 2(y')^2}{y^3} = 0. \]\nTherefore $y' / y^2 = C$ and $-1 / y = Cx + D$ for appropriate constants $C$ and $D$. In order that the solution pass through $(1, 1)$, we require that $C + D = -1$. Hence\n\[ y = \frac{1}{1 + C(1 - x)}. \]\nConversely, any function of this form satisfies the equation and the initial conditions. If $C = 0$, this is a constant function and its domain may be taken as $(-\infty, +\infty)$. If $C \neq 0$, the right member of (1) becomes infinite for $x = (1 + C) / C$, so the domain of (1) must be restricted to\n\[ (-\infty, \frac{1 + C}{C}) \text{ if } C > 0, \]\n\[ (\frac{1 + C}{C}, \infty) \text{ if } C < 0 with the final answer being \boxed{[(-\infty, y^2/y' + 1), (y^2/y' + 1, \infty)]}. \] | algebraic | putnam (modified boxing) | Differential Equations | Find all the solutions of the equation \[ y y'' - 2(y')^2 = 0 \] which pass through the point $x = 1, y = 1$. | First Solution. $1 / y^3$ is an integrating factor since\n\[ \frac{d}{dx} \left( \frac{y'}{y^2} \right) = \frac{y y'' - 2(y')^2}{y^3} = 0. \]\nTherefore $y' / y^2 = C$ and $-1 / y = Cx + D$ for appropriate constants $C$ and $D$. In order that the solution pass through $(1, 1)$, we require that $C + D = -1$. Hence\n\[ y = \frac{1}{1 + C(1 - x)}. \]\nConversely, any function of this form satisfies the equation and the initial conditions. If $C = 0$, this is a constant function and its domain may be taken as $(-\infty, +\infty)$. If $C \neq 0$, the right member of (1) becomes infinite for $x = (1 + C) / C$, so the domain of (1) must be restricted to\n\[ (-\infty, \frac{1 + C}{C}) \text{ if } C > 0, \]\n\[ (\frac{1 + C}{C}, \infty) \text{ if } C < 0. \] | 0 |
1938 | 1938_10 | A horizontal disc of diameter 3 inches is rotating at 4 revolutions per minute. A light is shining at a distant point in the plane of the disc. An insect is placed at the edge of the disc furthest from the light, facing the light. It at once starts crawling, and crawls so as always to face the light, at 1 inch per second. Set up the differential equation of motion, and find at what point the insect again reaches the edge of the disc. | Choose both rectangular and polar coordinate systems so that the origin is at the center of the disc, the insect is initially at $(3/2, 0)$, the distant light at $(-\infty, 0)$, and the disc rotates counterclockwise. Suppose that at time $t$ the insect’s position is $(x, y)$ in cartesian coordinates, and $(r, \theta)$ in polar coordinates.\n\nThen as long as the insect is on the disc, the horizontal and vertical components of its velocity are respectively\n\[ V_x = \frac{dx}{dt} = -1 - \frac{2\pi r}{15} \sin \theta = -1 - \frac{2\pi}{15} y, \] \n\[ V_y = \frac{dy}{dt} = \frac{2\pi r}{15} \cos \theta = \frac{2\pi}{15} x. \]\nDifferentiating (1) and using (2) we get\n\[ \frac{d^2x}{dt^2} = -\frac{2\pi}{15} \frac{dy}{dt} = -\left( \frac{2\pi}{15} \right)^2 x, \]\nwhence the differential equation governing $x$ is\n\[ \frac{d^2x}{dt^2} + \left( \frac{2\pi}{15} \right)^2 x = 0. \]\nThe solution to (3) is\n\[ x = A \cos \left( \frac{2\pi}{15} t - \phi \right), \]\nand from (1)\n\[ y = A \sin \left( \frac{2\pi}{15} t - \phi \right) - \frac{15}{2\pi}. \]\nTherefore the motion is uniform circular motion along the circle\n\[ x^2 + \left( y + \frac{15}{2\pi} \right)^2 = A^2 \]\nwhich has center at $(0, -15/2\pi)$ and radius $A$. Here $A$ can be evaluated from the initial conditions\n\[ x = \frac{3}{2}, y = 0 \text{ when } t = 0, \]\ngiving\n\[ A^2 = (3/2)^2 + (15/2\pi)^2. \]\nBy symmetry this circle cuts the boundary of the disc again at $\boxed{(-3/2, 0)}$, so the insect will leave the disc at that point. | numerical | putnam | Calculus Geometry Differential Equations | A horizontal disc of diameter 3 inches is rotating at 4 revolutions per minute. A light is shining at a distant point in the plane of the disc. An insect is placed at the edge of the disc furthest from the light, facing the light. It at once starts crawling, and crawls so as always to face the light, at 1 inch per second. Set up the differential equation of motion, and find at what point the insect again reaches the edge of the disc. | Choose both rectangular and polar coordinate systems so that the origin is at the center of the disc, the insect is initially at $(3/2, 0)$, the distant light at $(-\infty, 0)$, and the disc rotates counterclockwise. Suppose that at time $t$ the insect’s position is $(x, y)$ in cartesian coordinates, and $(r, \theta)$ in polar coordinates.\n\nThen as long as the insect is on the disc, the horizontal and vertical components of its velocity are respectively\n\[ V_x = \frac{dx}{dt} = -1 - \frac{2\pi r}{15} \sin \theta = -1 - \frac{2\pi}{15} y, \] \n\[ V_y = \frac{dy}{dt} = \frac{2\pi r}{15} \cos \theta = \frac{2\pi}{15} x. \]\nDifferentiating (1) and using (2) we get\n\[ \frac{d^2x}{dt^2} = -\frac{2\pi}{15} \frac{dy}{dt} = -\left( \frac{2\pi}{15} \right)^2 x, \]\nwhence the differential equation governing $x$ is\n\[ \frac{d^2x}{dt^2} + \left( \frac{2\pi}{15} \right)^2 x = 0. \]\nThe solution to (3) is\n\[ x = A \cos \left( \frac{2\pi}{15} t - \phi \right), \]\nand from (1)\n\[ y = A \sin \left( \frac{2\pi}{15} t - \phi \right) - \frac{15}{2\pi}. \]\nTherefore the motion is uniform circular motion along the circle\n\[ x^2 + \left( y + \frac{15}{2\pi} \right)^2 = A^2 \]\nwhich has center at $(0, -15/2\pi)$ and radius $A$. Here $A$ can be evaluated from the initial conditions\n\[ x = \frac{3}{2}, y = 0 \text{ when } t = 0, \]\ngiving\n\[ A^2 = (3/2)^2 + (15/2\pi)^2. \]\nBy symmetry this circle cuts the boundary of the disc again at $(-3/2, 0)$, so the insect will leave the disc at that point. | 0 |
1938 | 1938_11 | Given the parabola $y^2 = 2mx$, what is the length of the shortest chord that is normal to the curve at one end? | Any point on the parabola has coordinates of the form $(2mt^2, 2mt)$. Let $AB$ be a chord normal to the parabola at $A$. Say $A = (2mt^2, 2mt)$ and $B = (2ms^2, 2ms)$. The slope of $AB$ is $1/(s + t)$, and the slope of the tangent at $A$ is $1/(2t)$. Hence $s + t = -1/(2t)$.\n\nThe length $L$ of $AB$ is given by\n\[ L^2 = 4m^2[(s^2 - t^2)^2 + (s - t)^2] = 4m^2(s - t)^2[(s + t)^2 + 1]. \]\nSubstituting $s = -t - 1/(2t)$ we have\n\[ L^2 = 4m^2 \left( \frac{4t^2 + 1}{2t} \right)^2 \frac{1 + 4t^2}{4t^2} = \frac{m^2}{4} \frac{(4t^2 + 1)^3}{t^4}. \]\nWe seek the value of $t$ which minimizes $L$, so we may just as well choose $t$ to minimize\n\[ \frac{4t^2 + 1}{t^{4/3}} = 4t^{2/3} + t^{-4/3}. \]\nSetting the derivative equal to zero, we find two critical points, $t = \pm \sqrt{2}/2$. Since $L \to \infty$ as $t \to 0$, $\pm \infty$, these two critical values both give minima. Either of the two shortest chords is of length $\boxed{3\sqrt{3} |m|}$, from (1). | algebraic | putnam | Geometry Calculus | Given the parabola $y^2 = 2mx$, what is the length of the shortest chord that is normal to the curve at one end? | Any point on the parabola has coordinates of the form $(2mt^2, 2mt)$. Let $AB$ be a chord normal to the parabola at $A$. Say $A = (2mt^2, 2mt)$ and $B = (2ms^2, 2ms)$. The slope of $AB$ is $1/(s + t)$, and the slope of the tangent at $A$ is $1/(2t)$. Hence $s + t = -1/(2t)$.\n\nThe length $L$ of $AB$ is given by\n\[ L^2 = 4m^2[(s^2 - t^2)^2 + (s - t)^2] = 4m^2(s - t)^2[(s + t)^2 + 1]. \]\nSubstituting $s = -t - 1/(2t)$ we have\n\[ L^2 = 4m^2 \left( \frac{4t^2 + 1}{2t} \right)^2 \frac{1 + 4t^2}{4t^2} = \frac{m^2}{4} \frac{(4t^2 + 1)^3}{t^4}. \]\nWe seek the value of $t$ which minimizes $L$, so we may just as well choose $t$ to minimize\n\[ \frac{4t^2 + 1}{t^{4/3}} = 4t^{2/3} + t^{-4/3}. \]\nSetting the derivative equal to zero, we find two critical points, $t = \pm \sqrt{2}/2$. Since $L \to \infty$ as $t \to 0$, $\pm \infty$, these two critical values both give minima. Either of the two shortest chords is of length $\boxed{3\sqrt{3} |m|}$, from (1). | 0 |
1938 | 1938_12 | 12. From the center of a rectangular hyperbola a perpendicular is dropped upon a variable tangent. Find the locus of the foot of the perpendicular. Obtain the equation of the locus in polar coordinates, and sketch the curve. | Let the axes be the asymptotes, so that $xy = a^2$ is the equation of the given hyperbola. Let the point $(h, k)$ be on the hyperbola. Then $hk = a^2$ and the equation of the tangent line at $(h, k)$ is $hy + kx - 2hk = 0$.\n\nThe $x$ and $y$ intercepts of this tangent line are $2h$ and $2k$ respectively. Let $(r, \theta)$ be the polar coordinates of the foot of the perpendicular from the origin to the tangent line. Then $2h \cos \theta = r$ and $2k \sin \theta = r$, and hence\n\[ r^2 = 4hk \sin \theta \cos \theta \quad \text{or} \quad r^2 = 2a^2 \sin 2\theta. \tag{1} \]\nThis is the polar equation of the desired locus. We have shown that the foot of every perpendicular lies on (1).\n\nConversely, every point satisfying (1), except the pole, is the foot of some perpendicular: Given such a point, $P = (r, \theta)$, which must be in either the first or third quadrant, the equations $2h \cos \theta = r$ and $2k \sin \theta = r$ determine a point $(h, k)$ on the hyperbola and $P$ is the foot of the perpendicular on the tangent at $(h, k)$.\n\nThe locus is the well-known lemniscate of Bernoulli. The final equation is \boxed{r^2 = 2a^2 \sin 2\theta}. | algebraic | putnam | Geometry Algebra | 12. From the center of a rectangular hyperbola a perpendicular is dropped upon a variable tangent. Find the locus of the foot of the perpendicular. Obtain the equation of the locus in polar coordinates, and sketch the curve. | Let the axes be the asymptotes, so that $xy = a^2$ is the equation of the given hyperbola. Let the point $(h, k)$ be on the hyperbola. Then $hk = a^2$ and the equation of the tangent line at $(h, k)$ is $hy + kx - 2hk = 0$.\n\nThe $x$ and $y$ intercepts of this tangent line are $2h$ and $2k$ respectively. Let $(r, \theta)$ be the polar coordinates of the foot of the perpendicular from the origin to the tangent line. Then $2h \cos \theta = r$ and $2k \sin \theta = r$, and hence\n\[ r^2 = 4hk \sin \theta \cos \theta \quad \text{or} \quad \boxed{r^2 = 2a^2 \sin 2\theta}. \tag{1} \]\nThis is the polar equation of the desired locus. We have shown that the foot of every perpendicular lies on (1).\n\nConversely, every point satisfying (1), except the pole, is the foot of some perpendicular: Given such a point, $P = (r, \theta)$, which must be in either the first or third quadrant, the equations $2h \cos \theta = r$ and $2k \sin \theta = r$ determine a point $(h, k)$ on the hyperbola and $P$ is the foot of the perpendicular on the tangent at $(h, k)$.\n\nThe locus is the well-known lemniscate of Bernoulli. | 0 |
1938 | 1938_13 | Find the shortest distance between the plane $Ax + By + Cz + 1 = 0$ and the ellipsoid $x^2/a^2 + y^2/b^2 + z^2/c^2 = 1$. (For brevity, let\n\[ h = \frac{1}{\sqrt{A^2 + B^2 + C^2}} \quad \text{and} \quad m = \sqrt{a^2A^2 + b^2B^2 + c^2C^2}. \]\nState algebraically the condition that the plane shall lie outside the ellipsoid. | If the given plane intersects the ellipsoid, then the minimum distance is zero. If the plane fails to intersect the ellipsoid, then the shortest distance is the distance between the given plane and the nearer of the two tangent planes to the ellipsoid that are parallel to the given plane.\n\nThe tangent plane to the ellipsoid at the point $(x_0, y_0, z_0)$ is\n\[ \frac{x_0x}{a^2} + \frac{y_0y}{b^2} + \frac{z_0z}{c^2} = 1. \]\nIf this plane is parallel to $Ax + By + Cz + 1 = 0$, then\n\[ \frac{x_0}{a^2} = kA, \quad \frac{y_0}{b^2} = kB, \quad \text{and} \quad \frac{z_0}{c^2} = kC, \]\nwhere $k$ is a constant. Since\n\[ 1 = \frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} + \frac{z_0^2}{c^2} = k^2[a^2A^2 + b^2B^2 + c^2C^2], \]\nwe get $|k| = 1/m$.\n\nThe distance from the origin to the given plane is\n\[ \frac{1}{\sqrt{A^2 + B^2 + C^2}} = h. \]\nSince the parallel tangent plane can be written in the form\n\[ k(Ax + By + Cz) = 1, \]\nthe distance from the origin to either parallel tangent plane is\n\[ \frac{1}{|k|\sqrt{A^2 + B^2 + C^2}} = hm. \]\nHence if $m < 1$, the given plane lies farther from the origin than the tangent planes, and it does not cut the ellipsoid. The distance from the ellipsoid to the given plane in this case is $h(1 - m)$. But if $m \geq 1$, the given plane either lies between the tangent planes or coincides with one of them, so it cuts the ellipsoid and the distance is zero. Thus the final answer is \boxed{hm}. | algebraic | putnam | Geometry Analysis | Find the shortest distance between the plane $Ax + By + Cz + 1 = 0$ and the ellipsoid $x^2/a^2 + y^2/b^2 + z^2/c^2 = 1$. (For brevity, let\n\[ h = \frac{1}{\sqrt{A^2 + B^2 + C^2}} \quad \text{and} \quad m = \sqrt{a^2A^2 + b^2B^2 + c^2C^2}. \]\nState algebraically the condition that the plane shall lie outside the ellipsoid. | If the given plane intersects the ellipsoid, then the minimum distance is zero. If the plane fails to intersect the ellipsoid, then the shortest distance is the distance between the given plane and the nearer of the two tangent planes to the ellipsoid that are parallel to the given plane.\n\nThe tangent plane to the ellipsoid at the point $(x_0, y_0, z_0)$ is\n\[ \frac{x_0x}{a^2} + \frac{y_0y}{b^2} + \frac{z_0z}{c^2} = 1. \]\nIf this plane is parallel to $Ax + By + Cz + 1 = 0$, then\n\[ \frac{x_0}{a^2} = kA, \quad \frac{y_0}{b^2} = kB, \quad \text{and} \quad \frac{z_0}{c^2} = kC, \]\nwhere $k$ is a constant. Since\n\[ 1 = \frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} + \frac{z_0^2}{c^2} = k^2[a^2A^2 + b^2B^2 + c^2C^2], \]\nwe get $|k| = 1/m$.\n\nThe distance from the origin to the given plane is\n\[ \frac{1}{\sqrt{A^2 + B^2 + C^2}} = h. \]\nSince the parallel tangent plane can be written in the form\n\[ k(Ax + By + Cz) = 1, \]\nthe distance from the origin to either parallel tangent plane is\n\[ \frac{1}{|k|\sqrt{A^2 + B^2 + C^2}} = \boxed{hm}. \]\nHence if $m < 1$, the given plane lies farther from the origin than the tangent planes, and it does not cut the ellipsoid. The distance from the ellipsoid to the given plane in this case is $h(1 - m)$. But if $m \geq 1$, the given plane either lies between the tangent planes or coincides with one of them, so it cuts the ellipsoid and the distance is zero. | 0 |
1939 | 1939_1 | Find the length of the curve $y^2 = x^3$ from the origin to the point where the tangent makes an angle of $45^\circ$ with the x-axis. | The arc in the first quadrant is represented by the equation $y = x^{3/2}$, and its slope is $\frac{3}{2}x^{1/2}$. The point $P(x_0, y_0)$ where the tangent makes an angle of $45^\circ$ is determined from the relation $\frac{3}{2}x_0^{1/2} = 1$, whence $x_0 = \frac{4}{9}$. The desired length is therefore\n\[ \int_0^{4/9} \sqrt{1 + \frac{9x}{4}} \, dx = \frac{8}{27} \left( 1 + \frac{9x}{4} \right)^{3/2} \Bigg|_0^{4/9} = \boxed{\frac{8}{27}(2\sqrt{2} - 1)}. \] | numerical | putnam | Calculus Geometry | Find the length of the curve $y^2 = x^3$ from the origin to the point where the tangent makes an angle of $45^\circ$ with the x-axis. | The arc in the first quadrant is represented by the equation $y = x^{3/2}$, and its slope is $\frac{3}{2}x^{1/2}$. The point $P(x_0, y_0)$ where the tangent makes an angle of $45^\circ$ is determined from the relation $\frac{3}{2}x_0^{1/2} = 1$, whence $x_0 = \frac{4}{9}$. The desired length is therefore\n\[ \int_0^{4/9} \sqrt{1 + \frac{9x}{4}} \, dx = \frac{8}{27} \left( 1 + \frac{9x}{4} \right)^{3/2} \Bigg|_0^{4/9} = \boxed{\frac{8}{27}(2\sqrt{2} - 1)}. \] | 0 |
1939 | 1939_2 | A point $P$ is taken on the curve $y = x^3$. The tangent at $P$ meets the curve again at $Q$. Given the slope of the curve at $Q$ is $n$ times the slope at $P$ find the value of $n$. | Let $P$ have coordinates $(x_0, y_0)$; then the slope at $P$ is $3x_0^2$. The equation of the tangent at $P$ is $y = 3x_0^2(x - x_0) + x_0^3$. The points of intersection of the tangent and the original curve are determined by the relation\n\[ x^3 = 3x_0^2(x - x_0) + x_0^3, \]\nwhich is equivalent to\n\[ (x - x_0)^2(x + 2x_0) = 0. \]\nHence the second point of intersection is $(-2x_0, -8x_0^3)$. The slope at this point is $12x_0^2$, which is four times the slope at $P$, as was to be proved.\n\nIf $x_0 = 0$, the tangent does not really meet the curve again. However, since the tangent in this case has a triple point of intersection with the curve, instead of the usual double point of intersection, it is reasonable to say that it meets the curve "again" at $(0,0)$ making $n = \boxed{4}$. | numerical | putnam (modified boxing) | Geometry Calculus | A point $P$ is taken on the curve $y = x^3$. The tangent at $P$ meets the curve again at $Q$. Prove that the slope of the curve at $Q$ is four times the slope at $P$. | Let $P$ have coordinates $(x_0, y_0)$; then the slope at $P$ is $3x_0^2$. The equation of the tangent at $P$ is $y = 3x_0^2(x - x_0) + x_0^3$. The points of intersection of the tangent and the original curve are determined by the relation\n\[ x^3 = 3x_0^2(x - x_0) + x_0^3, \]\nwhich is equivalent to\n\[ (x - x_0)^2(x + 2x_0) = 0. \]\nHence the second point of intersection is $(-2x_0, -8x_0^3)$. The slope at this point is $12x_0^2$, which is four times the slope at $P$, as was to be proved.\n\nIf $x_0 = 0$, the tangent does not really meet the curve again. However, since the tangent in this case has a triple point of intersection with the curve, instead of the usual double point of intersection, it is reasonable to say that it meets the curve "again" at $(0,0)$. | 0 |
1939 | 1939_3 | Find the cubic equation whose roots are the cubes of the roots of
\[ x^3 + ax^2 + bx + c = 0. \] | First Solution. Let the roots of the given cubic equation be $x_1, x_2, x_3$. Then the roots of the desired equation are $x_1^3, x_2^3, x_3^3$. From
\[ x^3 + ax^2 + bx + c = (x - x_1)(x - x_2)(x - x_3), \]
it follows that
\[ x_1 + x_2 + x_3 = -a, \quad x_1x_2 + x_2x_3 + x_3x_1 = b, \quad x_1x_2x_3 = -c. \]
Let the desired cubic equation be
\[ x^3 + Ax^2 + Bx + C = (x - x_1^3)(x - x_2^3)(x - x_3^3) = 0. \]
Then we have
\[ (x_1 + x_2 + x_3)^3 = x_1^3 + x_2^3 + x_3^3 + 3(x_1 + x_2 + x_3)(x_1x_2 + x_2x_3 + x_3x_1) - 3x_1x_2x_3, \]
whence
\[ A = -(x_1^3 + x_2^3 + x_3^3) = a^3 - 3ab + 3c. \]
Also,
\[ (x_1x_2 + x_2x_3 + x_3x_1)^3 = x_1^3x_2^3 + x_2^3x_3^3 + x_3^3x_1^3 + 3abc - 3c^2 \]
and hence
\[ B = x_1^3x_2^3 + x_2^3x_3^3 + x_3^3x_1^3 = b^3 - 3abc + 3c^2. \]
Finally
\[ C = -x_1^3x_2^3x_3^3 = c^3. \]
Thus the desired cubic equation is
\[ \boxed{x^3 + (a^3 - 3ab + 3c)x^2 + (b^3 - 3abc + 3c^2)x + c^3 = 0}. \] | algebraic | putnam | Algebra | Find the cubic equation whose roots are the cubes of the roots of
\[ x^3 + ax^2 + bx + c = 0. \] | First Solution. Let the roots of the given cubic equation be $x_1, x_2, x_3$. Then the roots of the desired equation are $x_1^3, x_2^3, x_3^3$. From
\[ x^3 + ax^2 + bx + c = (x - x_1)(x - x_2)(x - x_3), \]
it follows that
\[ x_1 + x_2 + x_3 = -a, \quad x_1x_2 + x_2x_3 + x_3x_1 = b, \quad x_1x_2x_3 = -c. \]
Let the desired cubic equation be
\[ x^3 + Ax^2 + Bx + C = (x - x_1^3)(x - x_2^3)(x - x_3^3) = 0. \]
Then we have
\[ (x_1 + x_2 + x_3)^3 = x_1^3 + x_2^3 + x_3^3 + 3(x_1 + x_2 + x_3)(x_1x_2 + x_2x_3 + x_3x_1) - 3x_1x_2x_3, \]
whence
\[ A = -(x_1^3 + x_2^3 + x_3^3) = a^3 - 3ab + 3c. \]
Also,
\[ (x_1x_2 + x_2x_3 + x_3x_1)^3 = x_1^3x_2^3 + x_2^3x_3^3 + x_3^3x_1^3 + 3abc - 3c^2 \]
and hence
\[ B = x_1^3x_2^3 + x_2^3x_3^3 + x_3^3x_1^3 = b^3 - 3abc + 3c^2. \]
Finally
\[ C = -x_1^3x_2^3x_3^3 = c^3. \]
Thus the desired cubic equation is
\[ \boxed{x^3 + (a^3 - 3ab + 3c)x^2 + (b^3 - 3abc + 3c^2)x + c^3 = 0}. \] | 0 |
1939 | 1939_4 | Find the equations of the two straight lines each of which cuts all the four straight lines
\[
x = 1, y = 1, z = 0; \quad z = 1, x = 0; \quad y = 1, z = 0; \quad x = y = -6z and return the sum of the equations of the 2 lines.
\] | Suppose the required line $L$ meets the given lines in the points $A$, $B$, $C$, and $D$ respectively. Then
\[ A = (1, 0, a), \quad B = (b, 1, 0), \quad C = (0, c, 1), \quad D = (6d, 6d, -d) \]
for some numbers $a$, $b$, $c$, and $d$. Treat $A$, $B$, $C$, and $D$ as vectors. The condition that they be collinear is that the vectors
\[ B - A = (b - 1, 1, -a), \quad C - A = (-1, c, 1 - a), \quad D - A = (6d - 1, 6d, -d - a) \]
be proportional.
The proportionality of the first two tells us that
\[ c = \frac{1}{1 - b} = \frac{a - 1}{a}, \]
while the first and third give
\[ 6d = \frac{1 - 6d}{1 - b} = \frac{a + d}{a}. \]
Rewrite the middle member here using $(1)$:
\[ 6d = (1 - 6d) \frac{a - 1}{a} = \frac{a + d}{a}. \]
Clearing fractions:
\[ 6ad = a + d. \]
Adding these equations, we find $4d = a + 1$, so
\[ 6a(a + 1) = 24ad = 4(a + d) = 5a + 1. \]
The quadratic equation $6a(a + 1) = 5a + 1$ has roots
\[ a = \frac{1}{3}, -\frac{1}{2}, \]
and the corresponding values of the other unknowns are
\[ b = \frac{3}{2}, \frac{2}{3}, \quad c = -2, 3, \quad d = \frac{1}{3}, \frac{1}{8}. \]
The direction vectors of the lines (proportional to $B - A$, $C - A$, and $D - A$) in the two cases are $(3, 6, -2)$ and $(-2, 6, 3)$, respectively. The two lines are given parametrically by
\[ L_1: s \mapsto \left(1, 0, \frac{1}{3} \right) + s(3, 6, -2), \]
and
\[ L_2: t \mapsto \left(1, 0, -\frac{1}{2} \right) + t(-2, 6, 3). \]
These lines cross the given lines (in order) for
\[ s = 0, \frac{1}{6}, -\frac{1}{3}, \frac{1}{3} \quad \text{and} \quad t = 0, \frac{1}{6}, \frac{1}{2}, \frac{1}{8}. \]
In non-parametric form $L_1$ is given by
\[ y = 2(x - 1) = 1 - 3z \]
and $L_2$ is given by
\[ y = 3(1 - x) = 2z + 1. \] This makes the final answer \boxed{2 - z}. | algebraic | putnam (modified boxing) | Geometry | Find the equations of the two straight lines each of which cuts all the four straight lines
\[
x = 1, y = 1, z = 0; \quad z = 1, x = 0; \quad y = 1, z = 0; \quad x = y = -6z.
\] | Suppose the required line $L$ meets the given lines in the points $A$, $B$, $C$, and $D$ respectively. Then
\[ A = (1, 0, a), \quad B = (b, 1, 0), \quad C = (0, c, 1), \quad D = (6d, 6d, -d) \]
for some numbers $a$, $b$, $c$, and $d$. Treat $A$, $B$, $C$, and $D$ as vectors. The condition that they be collinear is that the vectors
\[ B - A = (b - 1, 1, -a), \quad C - A = (-1, c, 1 - a), \quad D - A = (6d - 1, 6d, -d - a) \]
be proportional.
The proportionality of the first two tells us that
\[ c = \frac{1}{1 - b} = \frac{a - 1}{a}, \]
while the first and third give
\[ 6d = \frac{1 - 6d}{1 - b} = \frac{a + d}{a}. \]
Rewrite the middle member here using $(1)$:
\[ 6d = (1 - 6d) \frac{a - 1}{a} = \frac{a + d}{a}. \]
Clearing fractions:
\[ 6ad = a + d. \]
Adding these equations, we find $4d = a + 1$, so
\[ 6a(a + 1) = 24ad = 4(a + d) = 5a + 1. \]
The quadratic equation $6a(a + 1) = 5a + 1$ has roots
\[ a = \frac{1}{3}, -\frac{1}{2}, \]
and the corresponding values of the other unknowns are
\[ b = \frac{3}{2}, \frac{2}{3}, \quad c = -2, 3, \quad d = \frac{1}{3}, \frac{1}{8}. \]
The direction vectors of the lines (proportional to $B - A$, $C - A$, and $D - A$) in the two cases are $(3, 6, -2)$ and $(-2, 6, 3)$, respectively. The two lines are given parametrically by
\[ L_1: s \mapsto \left(1, 0, \frac{1}{3} \right) + s(3, 6, -2), \]
and
\[ L_2: t \mapsto \left(1, 0, -\frac{1}{2} \right) + t(-2, 6, 3). \]
These lines cross the given lines (in order) for
\[ s = 0, \frac{1}{6}, -\frac{1}{3}, \frac{1}{3} \quad \text{and} \quad t = 0, \frac{1}{6}, \frac{1}{2}, \frac{1}{8}. \]
In non-parametric form \boxed{$L_1$ is given by
\[ y = 2(x - 1) = 1 - 3z \]
and $L_2$ is given by
\[ y = 3(1 - x) = 2z + 1}. \] | 0 |
1939 | 1939_5 | A heavy particle is attached to the end $A$ of a light rod $AB$ of length $a$. The rod is hinged at $B$ so that it can turn freely in a vertical plane. The rod is balanced in the vertical position above the hinge and then slightly disturbed. Find an expression for the time taken to pass from the horizontal position to the lowest position. | Let $m$ be the mass of the particle, and let $\theta$ be the angular position of the rod, measured from the vertical, at time $t$. The force of gravity $mg$ can be resolved into two components, $mg \cos \theta$ acting along the rod, and $mg \sin \theta$ acting perpendicular to the rod. The former is counterbalanced by the tension (or compression) in the rod and the latter accelerates the particle along the circle of radius $a$. By Newton’s third law we have \[ mg \sin \theta = ma \frac{d^2 \theta}{dt^2}. \] Multiply through by \( \frac{2}{m} \frac{d\theta}{dt} \) to get \[ 2g \sin \theta \frac{d \theta}{dt} = 2a \frac{d \theta}{dt} \frac{d^2 \theta}{dt^2}. \] This can be integrated to give \[ -2g \cos \theta + k = a \left( \frac{d\theta}{dt} \right)^2. \] From the initial conditions, $\theta = \frac{d \theta}{dt} = 0$ when $t = 0$, we find $k = 2g$. Thus we have \[ a \left( \frac{d \theta}{dt} \right)^2 = 2g(1 - \cos \theta) = 4g \sin^2 (\theta/2), \] whence \[ \frac{d \theta}{dt} = 2\sqrt{\frac{g}{a}} \sin (\theta/2). \] We have chosen the positive square root because \( \frac{d\theta}{dt} \) is positive for $0 < \theta \leq \pi$.\nThe time required for the passage from $\theta = \pi/2$ to $\theta = \pi$ is given by \[ \int_{\pi/2}^{\pi} \frac{dt}{d \theta} d \theta = \int_{\pi/2}^\pi \frac{1}{2\sqrt{a/g} \sin (\theta/2)} d\theta = \sqrt{\frac{a}{g}} \int_{\pi/2}^\pi \csc (\theta/2) d \theta. \] Substituting $u = \theta/2$, the limits change to $u = \pi/4$ and $u = \pi/2$, and the integral becomes \[ \sqrt{\frac{a}{g}} \int_{\pi/4}^{\pi/2} \csc u \ d u = \sqrt{\frac{a}{g}} \left[ -\log (\csc u + \cot u) \right]_{\pi/4}^{\pi/2}. \] Evaluating at the limits, \[ \boxed{\sqrt{\frac{a}{g}} \log (1 + \sqrt{2})}. \] | algebraic | putnam (modified boxing) | Calculus Differential Equations | A heavy particle is attached to the end $A$ of a light rod $AB$ of length $a$. The rod is hinged at $B$ so that it can turn freely in a vertical plane. The rod is balanced in the vertical position above the hinge and then slightly disturbed. Prove that the time taken to pass from the horizontal position to the lowest position is \[ \sqrt{\frac{a}{g}} \log (1 + \sqrt{2}). \] | Let $m$ be the mass of the particle, and let $\theta$ be the angular position of the rod, measured from the vertical, at time $t$. The force of gravity $mg$ can be resolved into two components, $mg \cos \theta$ acting along the rod, and $mg \sin \theta$ acting perpendicular to the rod. The former is counterbalanced by the tension (or compression) in the rod and the latter accelerates the particle along the circle of radius $a$. By Newton’s third law we have \[ mg \sin \theta = ma \frac{d^2 \theta}{dt^2}. \] Multiply through by \( \frac{2}{m} \frac{d\theta}{dt} \) to get \[ 2g \sin \theta \frac{d \theta}{dt} = 2a \frac{d \theta}{dt} \frac{d^2 \theta}{dt^2}. \] This can be integrated to give \[ -2g \cos \theta + k = a \left( \frac{d\theta}{dt} \right)^2. \] From the initial conditions, $\theta = \frac{d \theta}{dt} = 0$ when $t = 0$, we find $k = 2g$. Thus we have \[ a \left( \frac{d \theta}{dt} \right)^2 = 2g(1 - \cos \theta) = 4g \sin^2 (\theta/2), \] whence \[ \frac{d \theta}{dt} = 2\sqrt{\frac{g}{a}} \sin (\theta/2). \] We have chosen the positive square root because \( \frac{d\theta}{dt} \) is positive for $0 < \theta \leq \pi$.\nThe time required for the passage from $\theta = \pi/2$ to $\theta = \pi$ is given by \[ \int_{\pi/2}^{\pi} \frac{dt}{d \theta} d \theta = \int_{\pi/2}^\pi \frac{1}{2\sqrt{a/g} \sin (\theta/2)} d\theta = \sqrt{\frac{a}{g}} \int_{\pi/2}^\pi \csc (\theta/2) d \theta. \] Substituting $u = \theta/2$, the limits change to $u = \pi/4$ and $u = \pi/2$, and the integral becomes \[ \sqrt{\frac{a}{g}} \int_{\pi/4}^{\pi/2} \csc u \ d u = \sqrt{\frac{a}{g}} \left[ -\log (\csc u + \cot u) \right]_{\pi/4}^{\pi/2}. \] Evaluating at the limits, \[ \sqrt{\frac{a}{g}} \log (1 + \sqrt{2}). \] | 0 |
1939 | 1939_6_i | A circle of radius $a$ rolls on the inner side of the circumference of a circle of radius $3a$. Find the area contained within the closed curve generated by a point on the circumference of the rolling circle. | Take rectangular coordinates with the origin at the center of the large circle so that the generating point $P$ is in contact with the large circle at $A = (3a, 0)$. It can be seen from the diagram that when the small circle has rolled until the line of centers makes an angle $\theta$ with $OA$, the coordinates of $P$ are \[ x = 2a \cos \theta + a \cos 2\theta \] \[ y = 2a \sin \theta - a \sin 2\theta. \] These are, then, parametric equations for the path of $P$.\nThe area is given by \[ A = \frac{1}{2} \oint (x \ dy - y \ dx). \] \[ = \frac{a^2}{2} \int_0^{2\pi} \{ [2 \cos \theta + \cos 2\theta][2 \cos \theta - 2 \cos 2\theta] - [2 \sin \theta - \sin 2\theta][2 \sin \theta - 2 \sin 2\theta] \} d\theta \] \[ = \frac{a^2}{2} \int_0^{2\pi} (2 - 2 \cos 3\theta) d\theta = \boxed{2\pi a^2}. \] | algebraic | putnam | Geometry Calculus | A circle of radius $a$ rolls on the inner side of the circumference of a circle of radius $3a$. Find the area contained within the closed curve generated by a point on the circumference of the rolling circle. | Take rectangular coordinates with the origin at the center of the large circle so that the generating point $P$ is in contact with the large circle at $A = (3a, 0)$. It can be seen from the diagram that when the small circle has rolled until the line of centers makes an angle $\theta$ with $OA$, the coordinates of $P$ are \[ x = 2a \cos \theta + a \cos 2\theta \] \[ y = 2a \sin \theta - a \sin 2\theta. \] These are, then, parametric equations for the path of $P$.\nThe area is given by \[ A = \frac{1}{2} \oint (x \ dy - y \ dx). \] \[ = \frac{a^2}{2} \int_0^{2\pi} \{ [2 \cos \theta + \cos 2\theta][2 \cos \theta - 2 \cos 2\theta] - [2 \sin \theta - \sin 2\theta][2 \sin \theta - 2 \sin 2\theta] \} d\theta \] \[ = \frac{a^2}{2} \int_0^{2\pi} (2 - 2 \cos 3\theta) d\theta = \boxed{2\pi a^2}. \] | 0 |
1939 | 1939_6_ii | A shell strikes an airplane flying at a height $h$ above the ground. It is known that the shell was fired from a gun on the ground with a muzzle velocity of magnitude $V$, but the position of the gun and its angle of elevation are both unknown. Deduce that the gun is situated within a circle whose center lies directly below the airplane and find an expression for its radius. \] (Neglect the resistance of the atmosphere.) | Choose rectangular coordinates with the $y$-axis vertical, the origin at the position of the gun, and the airplane over a point of the positive $x$-axis. Then the coordinates of the airplane are $(u, h)$ where $u \geq 0$.\nIf the gun is fired at time $t = 0$ with muzzle velocity $V$ and elevation angle $\alpha$, then (neglecting air resistance) the shell's position at time $t$ is given by\n\[ x = Vt \cos \alpha \] \[ y = Vt \sin \alpha - \frac{1}{2}gt^2. \] Since it is given that the shell strikes the airplane, we have \[ u = Vt \cos \alpha \] \[ h = Vt \sin \alpha - \frac{1}{2}gt^2 \] for some $t$ and $\alpha$. Hence \[ u^2 + \left( h + \frac{1}{2}gt^2 \right)^2 = V^2t^2, \] so that \[ \frac{1}{4}g^2t^4 + (gh - V^2)t^2 + h^2 + u^2 = 0. \] In order that this equation have a real root $t$, it is necessary that \[ (gh - V^2)^2 \geq g^2(h^2 + u^2), \] and therefore that \[ g^2u^2 \leq V^2(V^2 - 2gh). \] Thus it is necessary that \[ V^2 \geq 2gh \] and \[ u \leq \frac{V}{g} \sqrt{V^2 - 2gh}. \] This shows that the gun is within distance \[ \boxed{\frac{V}{g} \sqrt{V^2 - 2gh}} \] from the point directly below the airplane when it was hit. | algebraic | putnam (modified boxing) | Calculus Geometry Trigonometry | A shell strikes an airplane flying at a height $h$ above the ground. It is known that the shell was fired from a gun on the ground with a muzzle velocity of magnitude $V$, but the position of the gun and its angle of elevation are both unknown. Deduce that the gun is situated within a circle whose center lies directly below the airplane and whose radius is \[ \frac{V}{g} \sqrt{V^2 - 2gh}. \] (Neglect the resistance of the atmosphere.) | Choose rectangular coordinates with the $y$-axis vertical, the origin at the position of the gun, and the airplane over a point of the positive $x$-axis. Then the coordinates of the airplane are $(u, h)$ where $u \geq 0$.\nIf the gun is fired at time $t = 0$ with muzzle velocity $V$ and elevation angle $\alpha$, then (neglecting air resistance) the shell's position at time $t$ is given by\n\[ x = Vt \cos \alpha \] \[ y = Vt \sin \alpha - \frac{1}{2}gt^2. \] Since it is given that the shell strikes the airplane, we have \[ u = Vt \cos \alpha \] \[ h = Vt \sin \alpha - \frac{1}{2}gt^2 \] for some $t$ and $\alpha$. Hence \[ u^2 + \left( h + \frac{1}{2}gt^2 \right)^2 = V^2t^2, \] so that \[ \frac{1}{4}g^2t^4 + (gh - V^2)t^2 + h^2 + u^2 = 0. \] In order that this equation have a real root $t$, it is necessary that \[ (gh - V^2)^2 \geq g^2(h^2 + u^2), \] and therefore that \[ g^2u^2 \leq V^2(V^2 - 2gh). \] Thus it is necessary that \[ V^2 \geq 2gh \] and \[ u \leq \frac{V}{g} \sqrt{V^2 - 2gh}. \] This shows that the gun is within distance \[ \frac{V}{g} \sqrt{V^2 - 2gh} \] from the point directly below the airplane when it was hit. | 0 |
1939 | 1939_7_i | Find the curve touched by all the curves of the family \[ (y - k^2)^2 = x^2(k^2 - x^2). \] | We may use the graphs of \[ y = x^2(k^2 - x^2) \] and \[ y^2 = x^2(k^2 - x^2) \] as aids in sketching the family of curves:
The function $x^2(k^2 - x^2)$ assumes its maximum when $x^2 = k^2 - x^2$; i.e., when $x = \pm k/\sqrt{2}$. Hence the graph of the curve \[ f(x, y, k) = (y - k^2)^2 - x^2(k^2 - x^2) = 0 \] has lower horizontal tangents at $(\pm k/\sqrt{2}, k^2/2)$ and upper horizontal tangents at $(\pm k/\sqrt{2}, 3k^2/2)$. Because $f$ depends on $k^2$, we need only consider $k \geq 0$. The curve degenerates to a point for $k = 0$, so assume $k$ positive. Clearly, the curve is contained in the strip $-k \leq x \leq k$. There are vertical tangents at $(\pm k, k^2)$. We can check this formally by noting that $\partial f/\partial y$ vanishes at this point, but $\partial f/\partial x$ does not. The curve has a double point at $(0, k^2)$ because both $\partial f/\partial x$ and $\partial f/\partial y$ vanish here. At this point the curve resembles a pair of crossed lines since dropping the terms of degree higher than two in $x$ and $y - k^2$ gives \[ (y - k^2)^2 - k^2x^2 = 0, \] whose graph is the union of the two lines $y - k^2 = \pm kx$.\nTo obtain the equation of the envelope, we eliminate $k$ from the two equations \[ f = (y - k^2)^2 - x^2(k^2 - x^2) = 0 \] and \[ \frac{\partial f}{\partial k} = -4k(y - k^2) - 2kx^2 = 0. \] From the second equation we have written either $k = 0$ or $k^2 = y + \frac{1}{2}x^2$.
The first alternative leads to $y^2 = -x^4$, which is just the origin. The second gives \[ x^2(3x^2 - 4y) = 0 \] which represents the union of the line $x = 0$ and the parabola $\boxed{4y = 3x^2}$.
Although the $y$-axis meets each curve in a double point, it is not tangent to any curve of the family, so it is not part of the envelope. The parabola $4y = 3x^2$, however, is tangent to the curve $f(x, y, k) = 0$ at each of the points $(\pm (2/\sqrt{5})k, (3/5)k^2)$. Hence this parabola is the envelope, provided the one-point “curve” corresponding to $k = 0$ is regarded as tangent to it; otherwise, the envelope is the parabola less the origin. | algebraic | putnam | Geometry | Find the curve touched by all the curves of the family \[ (y - k^2)^2 = x^2(k^2 - x^2). \] Make a rough sketch showing this curve and two curves of the family. | We may use the graphs of \[ y = x^2(k^2 - x^2) \] and \[ y^2 = x^2(k^2 - x^2) \] as aids in sketching the family of curves:
The function $x^2(k^2 - x^2)$ assumes its maximum when $x^2 = k^2 - x^2$; i.e., when $x = \pm k/\sqrt{2}$. Hence the graph of the curve \[ f(x, y, k) = (y - k^2)^2 - x^2(k^2 - x^2) = 0 \] has lower horizontal tangents at $(\pm k/\sqrt{2}, k^2/2)$ and upper horizontal tangents at $(\pm k/\sqrt{2}, 3k^2/2)$. Because $f$ depends on $k^2$, we need only consider $k \geq 0$. The curve degenerates to a point for $k = 0$, so assume $k$ positive. Clearly, the curve is contained in the strip $-k \leq x \leq k$. There are vertical tangents at $(\pm k, k^2)$. We can check this formally by noting that $\partial f/\partial y$ vanishes at this point, but $\partial f/\partial x$ does not. The curve has a double point at $(0, k^2)$ because both $\partial f/\partial x$ and $\partial f/\partial y$ vanish here. At this point the curve resembles a pair of crossed lines since dropping the terms of degree higher than two in $x$ and $y - k^2$ gives \[ (y - k^2)^2 - k^2x^2 = 0, \] whose graph is the union of the two lines $y - k^2 = \pm kx$.\nTo obtain the equation of the envelope, we eliminate $k$ from the two equations \[ f = (y - k^2)^2 - x^2(k^2 - x^2) = 0 \] and \[ \frac{\partial f}{\partial k} = -4k(y - k^2) - 2kx^2 = 0. \] From the second equation we have written either $k = 0$ or $k^2 = y + \frac{1}{2}x^2$.
The first alternative leads to $y^2 = -x^4$, which is just the origin. The second gives \[ x^2(3x^2 - 4y) = 0 \] which represents the union of the line $x = 0$ and the parabola $\boxed{4y = 3x^2}$.
Although the $y$-axis meets each curve in a double point, it is not tangent to any curve of the family, so it is not part of the envelope. The parabola $4y = 3x^2$, however, is tangent to the curve $f(x, y, k) = 0$ at each of the points $(\pm (2/\sqrt{5})k, (3/5)k^2)$. Hence this parabola is the envelope, provided the one-point “curve” corresponding to $k = 0$ is regarded as tangent to it; otherwise, the envelope is the parabola less the origin. | 0 |
1939 | 1939_9 | Evaluate the definite integrals \[ \text{(i)} \int_1^3 \frac{dx}{\sqrt{(x-1)(3-x)}}, \quad \text{(ii)} \int_1^\infty \frac{dx}{e^{x+1} + e^{3-x}} and return their sum. \] | Part (i). Since the integrand is not defined at either bound of integration, one should write \[ \int_1^3 \frac{dx}{\sqrt{(x-1)(3-x)}} = \lim_{\epsilon \to 0^+} \int_{1+\epsilon}^{3-\delta} \frac{dx}{\sqrt{(x-1)(3-x)}} \] \[ = \lim_{\epsilon \to 0^+, \delta \to 0^+} \int_{1+\epsilon}^{3-\delta} \frac{dx}{\sqrt{1 - (x-2)^2}} = \lim_{\epsilon \to 0^+, \delta \to 0^+} \Big[ \arcsin(x-2) \Big]_{1+\epsilon}^{3-\delta}. \] \[ = \lim_{\epsilon \to 0^+, \delta \to 0^+} \Big[ \arcsin(1-\delta) - \arcsin(\epsilon-1) \Big] = \frac{\pi}{2} + \frac{\pi}{2} = \pi. \]
Part (ii). The difficulty here is with the infinite interval of integration. Let $y = x-1$; then \[ \int_1^\infty \frac{dx}{e^{x+1} + e^{3-x}} = \frac{1}{e^2} \int_0^\infty \frac{dy}{e^y + e^{-y}} = \frac{1}{e^2} \int_0^\infty \frac{e^y dy}{e^{2y} + 1}. \]
Using the substitution $v = e^y$, $dv = e^y dy$, the integral becomes \[ \frac{1}{e^2} \int_1^\infty \frac{dv}{v^2 + 1} = \frac{1}{e^2} \arctan(v) \Big|_{e^0}^{e^N}. \]
Hence, \[ \int_1^\infty \frac{dx}{e^{x+1} + e^{3-x}} = \lim_{N \to \infty} \frac{1}{e^2} \Big[ \arctan(e^N) - \arctan(e^0) \Big] = \frac{\pi}{4e^2}. \] Thus the final answer is \boxed{\frac{(4e^2 + 1)\pi}{4e^2}}. | numerical | putnam (modified boxing) | Calculus | Evaluate the definite integrals \[ \text{(i)} \int_1^3 \frac{dx}{\sqrt{(x-1)(3-x)}}, \quad \text{(ii)} \int_1^\infty \frac{dx}{e^{x+1} + e^{3-x}}. \] | Part (i). Since the integrand is not defined at either bound of integration, one should write \[ \int_1^3 \frac{dx}{\sqrt{(x-1)(3-x)}} = \lim_{\epsilon \to 0^+} \int_{1+\epsilon}^{3-\delta} \frac{dx}{\sqrt{(x-1)(3-x)}} \] \[ = \lim_{\epsilon \to 0^+, \delta \to 0^+} \int_{1+\epsilon}^{3-\delta} \frac{dx}{\sqrt{1 - (x-2)^2}} = \lim_{\epsilon \to 0^+, \delta \to 0^+} \Big[ \arcsin(x-2) \Big]_{1+\epsilon}^{3-\delta}. \] \[ = \lim_{\epsilon \to 0^+, \delta \to 0^+} \Big[ \arcsin(1-\delta) - \arcsin(\epsilon-1) \Big] = \frac{\pi}{2} + \frac{\pi}{2} = \pi. \]
Part (ii). The difficulty here is with the infinite interval of integration. Let $y = x-1$; then \[ \int_1^\infty \frac{dx}{e^{x+1} + e^{3-x}} = \frac{1}{e^2} \int_0^\infty \frac{dy}{e^y + e^{-y}} = \frac{1}{e^2} \int_0^\infty \frac{e^y dy}{e^{2y} + 1}. \]
Using the substitution $v = e^y$, $dv = e^y dy$, the integral becomes \[ \frac{1}{e^2} \int_1^\infty \frac{dv}{v^2 + 1} = \frac{1}{e^2} \arctan(v) \Big|_{e^0}^{e^N}. \]
Hence, \[ \int_1^\infty \frac{dx}{e^{x+1} + e^{3-x}} = \lim_{N \to \infty} \frac{1}{e^2} \Big[ \arctan(e^N) - \arctan(e^0) \Big] = \frac{\pi}{4e^2}. \] | 0 |
1939 | 1939_10 | Given the power-series \[ a_0 + a_1 x + a_2 x^2 + \cdots \] in which \[ a_n = (n^2 + 1)3^n, \] show that there is a relation of the form \[ a_n + p a_{n+1} + q a_{n+2} + r a_{n+3} = 0, \] in which $p, q, r$ are constants independent of $n$. Find these constants and the sum of the power-series and given the final answer by adding $p, q, r$ to the sum of the power-series. | The desired relation is \[ (n^2 + 1)3^n + p((n+1)^2 + 1)3^{n+1} + q((n+2)^2 + 1)3^{n+2} + r((n+3)^2 + 1)3^{n+3} = 0, \] which is equivalent to \[ n^2(1 + 3p + 9q + 27r) + n(6p + 36q + 162r) + (1 + 6p + 45q + 270r) = 0. \] Equation (1) holds for all $n$ if and only if \[ 1 + 3p + 9q + 27r = 0, \] \[ p + 6q + 27r = 0, \] \[ 1 + 6p + 45q + 270r = 0. \] These linear equations have the solution $p = -1, q = \frac{1}{3}, r = -\frac{1}{27}$, so \[ a_n - a_{n+1} + \frac{1}{3}a_{n+2} - \frac{1}{27}a_{n+3} = 0. \]
Let $S(x) = a_0 + a_1x + a_2x^2 + \cdots$. Proceeding formally, we have \[ x^3S(x) = a_0x^3 + a_1x^4 + \cdots + a_nx^{n+3} + \cdots, \] \[ px^2S(x) = pa_0x^2 + pa_1x^3 + pa_2x^4 + \cdots, \] \[ qxS(x) = qa_0x + qa_1x^2 + qa_2x^3 + \cdots, \] \[ rS(x) = ra_0 + ra_1x + ra_2x^2 + \cdots. \]
When we sum these we get \[ S(x)[x^3 + px^2 + qx + r] = (pa_0 + qa_1 + ra_2)x^2 + (qa_0 + ra_1)x + ra_0. \] Multiplying through by $-27$, we obtain \[ S(x)[1 - 9x + 27x^2 - 27x^3] = 1 - 3x + 18x^2, \] and therefore \[ S(x) = \frac{1 - 3x + 18x^2}{(1 - 3x)^3}. \] Using the ratio test we conclude that the series converges for $|x| \leq \frac{1}{3}$; hence the formal manipulations above are valid for these values of $x$. The final sum is \boxed{\frac{1 - 3x + 18x^2}{(1 - 3x)^3} - 19/27} | algebraic | putnam (modified boxing) | Algebra Analysis | Given the power-series \[ a_0 + a_1 x + a_2 x^2 + \cdots \] in which \[ a_n = (n^2 + 1)3^n, \] show that there is a relation of the form \[ a_n + p a_{n+1} + q a_{n+2} + r a_{n+3} = 0, \] in which $p, q, r$ are constants independent of $n$. Find these constants and the sum of the power-series. | The desired relation is \[ (n^2 + 1)3^n + p((n+1)^2 + 1)3^{n+1} + q((n+2)^2 + 1)3^{n+2} + r((n+3)^2 + 1)3^{n+3} = 0, \] which is equivalent to \[ n^2(1 + 3p + 9q + 27r) + n(6p + 36q + 162r) + (1 + 6p + 45q + 270r) = 0. \] Equation (1) holds for all $n$ if and only if \[ 1 + 3p + 9q + 27r = 0, \] \[ p + 6q + 27r = 0, \] \[ 1 + 6p + 45q + 270r = 0. \] These linear equations have the solution $p = -1, q = \frac{1}{3}, r = -\frac{1}{27}$, so \[ a_n - a_{n+1} + \frac{1}{3}a_{n+2} - \frac{1}{27}a_{n+3} = 0. \]
Let $S(x) = a_0 + a_1x + a_2x^2 + \cdots$. Proceeding formally, we have \[ x^3S(x) = a_0x^3 + a_1x^4 + \cdots + a_nx^{n+3} + \cdots, \] \[ px^2S(x) = pa_0x^2 + pa_1x^3 + pa_2x^4 + \cdots, \] \[ qxS(x) = qa_0x + qa_1x^2 + qa_2x^3 + \cdots, \] \[ rS(x) = ra_0 + ra_1x + ra_2x^2 + \cdots. \]
When we sum these we get \[ S(x)[x^3 + px^2 + qx + r] = (pa_0 + qa_1 + ra_2)x^2 + (qa_0 + ra_1)x + ra_0. \] Multiplying through by $-27$, we obtain \[ S(x)[1 - 9x + 27x^2 - 27x^3] = 1 - 3x + 18x^2, \] and therefore \[ S(x) = \frac{1 - 3x + 18x^2}{(1 - 3x)^3}. \] Using the ratio test we conclude that the series converges for $|x| \leq \frac{1}{3}$; hence the formal manipulations above are valid for these values of $x$. | 0 |
1939 | 1939_11 | Find the equation of the parabola which touches the $x$-axis at the point $(1, 0)$ and the $y$-axis at the point $(0, 2)$. Find the equation of the axis of the parabola. | Clearly the required parabola does not pass through the origin, and any conic not passing through the origin has an equation of the form \[ Ax^2 + Bxy + Cy^2 + Dx + Ey + 1 = 0. \]
In order that this conic be tangent to the $x$-axis at $(1, 0)$, the equation obtained by setting $y = 0$ must have a double root at $x = 1$. Hence $A = 1$ and $D = -2$. In order that it be tangent to the $y$-axis at $(0, 2)$, we must have $C = \frac{1}{4}$, $E = -1$. In order that the conic be a parabola we must have $B^2 = 4AC$. This leads to two possibilities \[ (1) \quad x^2 + xy + \frac{1}{4}y^2 - 2x - y + 1 = 0, \]
\[ (2) \quad x^2 - xy + \frac{1}{4}y^2 - 2x - y + 1 = 0. \]
Since (1) can be written as $(x + \frac{1}{2}y - 1)^2 = 0$, we see that it represents a degenerate conic, the double line through the two given points. Hence (2) is the equation of the required parabola.
It is convenient to multiply equation (2) by 4 to eliminate the fraction. We obtain \[ 4x^2 - 4xy + y^2 - 8x - 4y + 4 = 0. \]
Since the quadratic terms in (3) can be written in the form $(2x - y)^2$, a transformation is suggested. The orthogonal (but not scale-preserving) transformation \[ u = 2x - y, \quad v = x + 2y \] with inverse \[ x = \frac{1}{5}(2u + v), \quad y = \frac{1}{5}(-u + 2v) \]
transforms (3) into \[ (4) \quad u^2 - \frac{12}{5}u - \frac{16}{5}v + 4 = 0. \]
This has the standard form \[ \left(u - \frac{6}{5}\right)^2 - \frac{16}{5}\left(v - \frac{4}{5}\right) = 0. \]
In this form the axis of the parabola is the line $u = \frac{6}{5}$, and the vertex has $uv$-coordinates $(\frac{6}{5}, \frac{4}{5})$. In terms of the original coordinates, the axis has the equation $\boxed{2x - y = \frac{6}{5}}$ and the vertex is at $(\frac{16}{25}, \frac{22}{25})$. | algebraic | putnam | Algebra Geometry | Find the equation of the parabola which touches the $x$-axis at the point $(1, 0)$ and the $y$-axis at the point $(0, 2)$. Find the equation of the axis of the parabola and the coordinates of its vertex. | Clearly the required parabola does not pass through the origin, and any conic not passing through the origin has an equation of the form \[ Ax^2 + Bxy + Cy^2 + Dx + Ey + 1 = 0. \]
In order that this conic be tangent to the $x$-axis at $(1, 0)$, the equation obtained by setting $y = 0$ must have a double root at $x = 1$. Hence $A = 1$ and $D = -2$. In order that it be tangent to the $y$-axis at $(0, 2)$, we must have $C = \frac{1}{4}$, $E = -1$. In order that the conic be a parabola we must have $B^2 = 4AC$. This leads to two possibilities \[ (1) \quad x^2 + xy + \frac{1}{4}y^2 - 2x - y + 1 = 0, \]
\[ (2) \quad x^2 - xy + \frac{1}{4}y^2 - 2x - y + 1 = 0. \]
Since (1) can be written as $(x + \frac{1}{2}y - 1)^2 = 0$, we see that it represents a degenerate conic, the double line through the two given points. Hence (2) is the equation of the required parabola.
It is convenient to multiply equation (2) by 4 to eliminate the fraction. We obtain \[ 4x^2 - 4xy + y^2 - 8x - 4y + 4 = 0. \]
Since the quadratic terms in (3) can be written in the form $(2x - y)^2$, a transformation is suggested. The orthogonal (but not scale-preserving) transformation \[ u = 2x - y, \quad v = x + 2y \] with inverse \[ x = \frac{1}{5}(2u + v), \quad y = \frac{1}{5}(-u + 2v) \]
transforms (3) into \[ (4) \quad u^2 - \frac{12}{5}u - \frac{16}{5}v + 4 = 0. \]
This has the standard form \[ \left(u - \frac{6}{5}\right)^2 - \frac{16}{5}\left(v - \frac{4}{5}\right) = 0. \]
In this form the axis of the parabola is the line $u = \frac{6}{5}$, and the vertex has $uv$-coordinates $(\frac{6}{5}, \frac{4}{5})$. In terms of the original coordinates, the axis has the equation $2x - y = \frac{6}{5}$ and the vertex is at $(\frac{16}{25}, \frac{22}{25})$. | 0 |
1939 | 1939_12_i | Prove that \[ \int_1^a [x]f'(x) \,dx = [a]f(a) - \{f(1) + \dots + f([a])\}, \] where $a$ is greater than 1 and where $[x]$ denotes the greatest of the integers not exceeding $x$. Obtain a corresponding expression for \[ \int_1^a [x^2]f'(x) \,dx. \] | We have \[ \int_1^a [x]f'(x) \,dx = \int_1^2 1 \cdot f'(x) \,dx + \int_2^3 2 \cdot f'(x) \,dx + \dots + \int_{[a]}^a [a] \cdot f'(x) \,dx \]
\[ = f(2) - f(1) + 2(f(3) - f(2)) + \dots + [a](f(a) - f([a])) \]
\[ = [a]f(a) - \{f(1) + f(2) + \dots + f([a])\}. \]
For the second part, we have \[ \int_1^a [x^2]f'(x) \,dx = \int_1^{\sqrt{2}} 1 \cdot f'(x) \,dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \cdot f'(x) \,dx + \dots + \int_{\sqrt{[a^2]}}^a [a^2] \cdot f'(x) \,dx \]
\[ = (f(\sqrt{2}) - f(1)) + 2(f(\sqrt{3}) - f(\sqrt{2})) + \dots + [a^2](f(a) - f(\sqrt{[a^2]})) \]
\[ = \boxed{[a^2]f(a) - \{f(1) + f(\sqrt{2}) + \dots + f(\sqrt{[a^2]})\}}. \] | algebraic | putnam | Calculus | (i) Prove that \[ \int_1^a [x]f'(x) \,dx = [a]f(a) - \{f(1) + \dots + f([a])\}, \] where $a$ is greater than 1 and where $[x]$ denotes the greatest of the integers not exceeding $x$. Obtain a corresponding expression for \[ \int_1^a [x^2]f'(x) \,dx. \] | Solution. We have \[ \int_1^a [x]f'(x) \,dx = \int_1^2 1 \cdot f'(x) \,dx + \int_2^3 2 \cdot f'(x) \,dx + \dots + \int_{[a]}^a [a] \cdot f'(x) \,dx \]
\[ = f(2) - f(1) + 2(f(3) - f(2)) + \dots + [a](f(a) - f([a])) \]
\[ = [a]f(a) - \{f(1) + f(2) + \dots + f([a])\}. \]
For the second part, we have \[ \int_1^a [x^2]f'(x) \,dx = \int_1^{\sqrt{2}} 1 \cdot f'(x) \,dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \cdot f'(x) \,dx + \dots + \int_{\sqrt{[a^2]}}^a [a^2] \cdot f'(x) \,dx \]
\[ = (f(\sqrt{2}) - f(1)) + 2(f(\sqrt{3}) - f(\sqrt{2})) + \dots + [a^2](f(a) - f(\sqrt{[a^2]})) \]
\[ = \boxed{[a^2]f(a) - \{f(1) + f(\sqrt{2}) + \dots + f(\sqrt{[a^2]})\}}. \] | 0 |
1939 | 1939_12_ii | A particle moves on a straight line, the only force acting on it being a resistance proportional to the velocity. If it started with a velocity of 1,000 ft. per sec. and had a velocity of 900 ft. per sec. when it had travelled 1,200 ft., calculate to the nearest hundredth of a second the time it took to travel this distance. | The differential equation governing the motion is \[ m \frac{d^2x}{dt^2} = -k \frac{dx}{dt}, \] and the boundary conditions are \[ x = 0, \quad \frac{dx}{dt} = 1000, \quad \text{when } t = 0 \] \[ x = 1200, \quad \frac{dx}{dt} = 900, \quad \text{when } t = T, \] where $T$ is the time required.
Let $b = k/m$. Then $d^2x/dt^2 = -b dx/dt$, which implies \[ \frac{dx}{dt} = -bx + c. \] The boundary conditions give \[ 1000 = c, \] \[ 900 = -1200b + c, \] whence $b = \frac{1}{12}$. Using these values and (1), we have \[ T = \int_0^{1200} \frac{dt}{dx} dx = \int_0^{1200} \frac{dx}{1000 - x/12} = 12 \log \frac{10}{9}. \]
To evaluate this it is convenient to write $T = -12 \log \frac{9}{10}$ and use the series expansion \[ -\log (1 - x) = x + \frac{1}{2}x^2 + \frac{1}{3}x^3 + \frac{1}{4}x^4 + \dots. \] Taking $x = \frac{1}{10}$, we have \[ \frac{1}{10} + \frac{1}{200} + \frac{1}{3000} < -\log \frac{9}{10} < \frac{1}{10} + \frac{1}{200} + \frac{1}{2700} + \frac{1}{3} \sum_{n=3}^\infty \left(\frac{1}{10}\right)^n. \] The lower bound exceeds $.1 + .005 + .0003 = .1053$, and the upper bound is $.1 + .005 + 1/2700 < .1054$. Therefore, \[ 1.2636 < -12 \log \frac{9}{10} < 1.2648 \] so, to the nearest hundredth of a second, \[ T = \boxed{1.26} \text{ sec}. \] | numerical | putnam | Differential Equations Calculus | (ii) A particle moves on a straight line, the only force acting on it being a resistance proportional to the velocity. If it started with a velocity of 1,000 ft. per sec. and had a velocity of 900 ft. per sec. when it had travelled 1,200 ft., calculate to the nearest hundredth of a second the time it took to travel this distance. | Solution. The differential equation governing the motion is \[ m \frac{d^2x}{dt^2} = -k \frac{dx}{dt}, \] and the boundary conditions are \[ x = 0, \quad \frac{dx}{dt} = 1000, \quad \text{when } t = 0 \] \[ x = 1200, \quad \frac{dx}{dt} = 900, \quad \text{when } t = T, \] where $T$ is the time required.
Let $b = k/m$. Then $d^2x/dt^2 = -b dx/dt$, which implies \[ \frac{dx}{dt} = -bx + c. \] The boundary conditions give \[ 1000 = c, \] \[ 900 = -1200b + c, \] whence $b = \frac{1}{12}$. Using these values and (1), we have \[ T = \int_0^{1200} \frac{dt}{dx} dx = \int_0^{1200} \frac{dx}{1000 - x/12} = 12 \log \frac{10}{9}. \]
To evaluate this it is convenient to write $T = -12 \log \frac{9}{10}$ and use the series expansion \[ -\log (1 - x) = x + \frac{1}{2}x^2 + \frac{1}{3}x^3 + \frac{1}{4}x^4 + \dots. \] Taking $x = \frac{1}{10}$, we have \[ \frac{1}{10} + \frac{1}{200} + \frac{1}{3000} < -\log \frac{9}{10} < \frac{1}{10} + \frac{1}{200} + \frac{1}{2700} + \frac{1}{3} \sum_{n=3}^\infty \left(\frac{1}{10}\right)^n. \] The lower bound exceeds $.1 + .005 + .0003 = .1053$, and the upper bound is $.1 + .005 + 1/2700 < .1054$. Therefore, \[ 1.2636 < -12 \log \frac{9}{10} < 1.2648 \] so, to the nearest hundredth of a second, \[ T = \boxed{1.26} \text{ sec}. \] | 0 |
1939 | 1939_13_ii | Calculate the mutual gravitational attraction of two uniform rods, each of mass $m$ and length $2a$, placed parallel to one another and perpendicular to the line joining their centers at a distance $b$ apart. | We first find the vertical component of the force of attraction between a particle $P$ of mass $\mu$ situated at the point $(h, b)$ and a uniform rod of mass $m$ lying along the $x$-axis from $(0, 0)$ to $(2a, 0)$.
Consider a short segment $S$ of the rod of length $\Delta x$ and center at $Q = (x, 0)$. Let $\alpha, \beta, \theta$ be the angles shown in the diagram. The mass of $S$ is $m \cdot \Delta x / 2a$. If the mass of $S$ were concentrated at $Q$, the attractive force between $S$ and $P$ would be
\[ G\mu \cdot \frac{m}{2a} \Delta x \cdot \frac{1}{b^2 \csc^2 \theta}, \] and its vertical component would be
\[ G\mu \cdot \frac{m}{2a} \Delta x \cdot \frac{1}{b^2 \csc^2 \theta} \cdot \sin \theta, \] where $G$ is the constant of gravitation.
The vertical component of the total attractive force between $P$ and the rod is therefore
\[ f_y = \frac{G\mu m}{2ab^2} \int_0^{2a} \sin^3 \theta \; dx. \] Now $x$ and $\theta$ are related by $x + b \cot \theta = h$, so if we change the variable of integration to $\theta$, we have
\[ f_y = \frac{G\mu m}{2ab} \int_\alpha^\beta \sin \theta \; d\theta = \frac{G\mu m}{2ab} (\cos \alpha - \cos \beta). \]
\[ = \frac{G\mu m}{2ab} \left( \frac{h}{\sqrt{h^2 + b^2}} - \frac{h - 2a}{\sqrt{(h - 2a)^2 + b^2}} \right). \]
Consider now the two parallel rods, the second running from $(0, b)$ to $(2a, b)$. A short segment of the upper rod of length $\Delta h$ may be considered a particle of mass $m \Delta h / 2a$, and it follows as above that the vertical component of the force of attraction between the rods is
\[ F_y = \frac{Gm^2}{4a^2b} \int_0^{2a} \left( \frac{h}{\sqrt{h^2 + b^2}} - \frac{h - 2a}{\sqrt{(h - 2a)^2 + b^2}} \right) dh. \]
\[ = \frac{Gm^2}{4a^2b} \left[ \sqrt{h^2 + b^2} - \sqrt{(h - 2a)^2 + b^2} \right]_0^{2a}. \]
\[ = \boxed{\frac{Gm^2}{2a^2b} (\sqrt{4a^2 + b^2} - b)}. \]
It is clear from symmetry that the entire force acts along the line of centers, so the force is given by its vertical component. | algebraic | putnam | Calculus | Calculate the mutual gravitational attraction of two uniform rods, each of mass $m$ and length $2a$, placed parallel to one another and perpendicular to the line joining their centers at a distance $b$ apart. In your answer let $a$ approach zero, and comment on the form of the result. | First Solution. We first find the vertical component of the force of attraction between a particle $P$ of mass $\mu$ situated at the point $(h, b)$ and a uniform rod of mass $m$ lying along the $x$-axis from $(0, 0)$ to $(2a, 0)$.
Consider a short segment $S$ of the rod of length $\Delta x$ and center at $Q = (x, 0)$. Let $\alpha, \beta, \theta$ be the angles shown in the diagram. The mass of $S$ is $m \cdot \Delta x / 2a$. If the mass of $S$ were concentrated at $Q$, the attractive force between $S$ and $P$ would be
\[ G\mu \cdot \frac{m}{2a} \Delta x \cdot \frac{1}{b^2 \csc^2 \theta}, \] and its vertical component would be
\[ G\mu \cdot \frac{m}{2a} \Delta x \cdot \frac{1}{b^2 \csc^2 \theta} \cdot \sin \theta, \] where $G$ is the constant of gravitation.
The vertical component of the total attractive force between $P$ and the rod is therefore
\[ f_y = \frac{G\mu m}{2ab^2} \int_0^{2a} \sin^3 \theta \; dx. \] Now $x$ and $\theta$ are related by $x + b \cot \theta = h$, so if we change the variable of integration to $\theta$, we have
\[ f_y = \frac{G\mu m}{2ab} \int_\alpha^\beta \sin \theta \; d\theta = \frac{G\mu m}{2ab} (\cos \alpha - \cos \beta). \]
\[ = \frac{G\mu m}{2ab} \left( \frac{h}{\sqrt{h^2 + b^2}} - \frac{h - 2a}{\sqrt{(h - 2a)^2 + b^2}} \right). \]
Consider now the two parallel rods, the second running from $(0, b)$ to $(2a, b)$. A short segment of the upper rod of length $\Delta h$ may be considered a particle of mass $m \Delta h / 2a$, and it follows as above that the vertical component of the force of attraction between the rods is
\[ F_y = \frac{Gm^2}{4a^2b} \int_0^{2a} \left( \frac{h}{\sqrt{h^2 + b^2}} - \frac{h - 2a}{\sqrt{(h - 2a)^2 + b^2}} \right) dh. \]
\[ = \frac{Gm^2}{4a^2b} \left[ \sqrt{h^2 + b^2} - \sqrt{(h - 2a)^2 + b^2} \right]_0^{2a}. \]
\[ = \boxed{\frac{Gm^2}{2a^2b} (\sqrt{4a^2 + b^2} - b)}. \]
It is clear from symmetry that the entire force acts along the line of centers, so the force is given by its vertical component. | 0 |
1939 | 1939_14_i | If \[ u = 1 + \frac{x^3}{3!} + \frac{x^6}{6!} + \cdots, \]
\[ v = \frac{x}{1!} + \frac{x^4}{4!} + \frac{x^7}{7!} + \cdots, \]
\[ w = \frac{x^2}{2!} + \frac{x^5}{5!} + \frac{x^8}{8!} + \cdots, \]
find the value of
\[ u^3 + v^3 + w^3 - 3uvw. \] | The power series for $u$, $v$, and $w$ converge for all $x$, and
\[ \frac{du}{dx} = w, \quad \frac{dv}{dx} = u, \quad \frac{dw}{dx} = v, \]
as we see by differentiating them. Letting
\[ f = u^3 + v^3 + w^3 - 3uvw, \]
we have
\[ f' = 3u^2u' + 3v^2v' + 3w^2w' - 3uvw' - 3uv'w - 3u'vw \]
\[ = 3u^2w + 3v^2u + 3w^2v - 3uvw - 3uvw - 3uvw = 0. \]
Thus $f$ = constant. But
\[ f(0) = [u(0)]^3 = 1, \]
so $f(x) = \boxed{1}$ for all $x$. | numerical | putnam (modified boxing) | Algebra Analysis | If \[ u = 1 + \frac{x^3}{3!} + \frac{x^6}{6!} + \cdots, \]
\[ v = \frac{x}{1!} + \frac{x^4}{4!} + \frac{x^7}{7!} + \cdots, \]
\[ w = \frac{x^2}{2!} + \frac{x^5}{5!} + \frac{x^8}{8!} + \cdots, \]
prove that
\[ u^3 + v^3 + w^3 - 3uvw = 1. \] | First Solution. The power series for $u$, $v$, and $w$ converge for all $x$, and
\[ \frac{du}{dx} = w, \quad \frac{dv}{dx} = u, \quad \frac{dw}{dx} = v, \]
as we see by differentiating them. Letting
\[ f = u^3 + v^3 + w^3 - 3uvw, \]
we have
\[ f' = 3u^2u' + 3v^2v' + 3w^2w' - 3uvw' - 3uv'w - 3u'vw \]
\[ = 3u^2w + 3v^2u + 3w^2v - 3uvw - 3uvw - 3uvw = 0. \]
Thus $f$ = constant. But
\[ f(0) = [u(0)]^3 = 1, \]
so $f(x) = 1$ for all $x$. | 0 |
1940 | 1940_1 | Given that $f(x)$ is a polynomial with integral coefficients, and there exists an integer $k$ such that none of the integers $f(1), f(2), \ldots, f(k)$ is divisible by $k$, then find the number of integral roots of $f(x)$. | Suppose $f$ has an integral root $r$. Then $f(x) = (x - r)g(x)$ where $g(x)$ is also a polynomial with integral coefficients. Then there are integers $p$ and $q$ such that $r = p + kq$ and $1 \leq p \leq k$. But
\[ f(p) = (p - r)g(p) = -kqg(p), \]
and hence $f(p)$ is divisible by $k$ contrary to the hypothesis. This contradiction shows that $f(x)$ has no integral root making the answer \boxed{0}. | numerical | putnam (modified boxing) | Algebra Number Theory | Prove that if $f(x)$ is a polynomial with integral coefficients, and there exists an integer $k$ such that none of the integers $f(1), f(2), \ldots, f(k)$ is divisible by $k$, then $f(x)$ has no integral root. | Suppose $f$ has an integral root $r$. Then $f(x) = (x - r)g(x)$ where $g(x)$ is also a polynomial with integral coefficients. Then there are integers $p$ and $q$ such that $r = p + kq$ and $1 \leq p \leq k$. But
\[ f(p) = (p - r)g(p) = -kqg(p), \]
and hence $f(p)$ is divisible by $k$ contrary to the hypothesis. This contradiction shows that $f(x)$ has no integral root. | 0 |
1940 | 1940_3 | Find $f(x)$ such that
\[ \int [f(x)]^n dx = \left(\int f(x) dx\right)^n, \]
when constants of integration ($c$) are suitably chosen. | We assume that only real-valued continuous functions $f$ defined on an interval are to be considered. If we put $g(x) = \int f(x)^n dx$ and $h(x) = \int f(x) dx$, we are asked to find all pairs of $C^1$-functions $g$ and $h$ defined on an interval such that
\[ g(x) = h(x)^n \tag{1} \]
and
\[ g'(x) = h'(x)^n. \tag{2} \]
If $n = 1$, then obviously any continuous function $f$ and corresponding functions $g$ and $h$ solve the problem, so we assume from now on that $n \neq 1$.
We proceed formally. Differentiate (1) to get
\[ g'(x) = nh(x)^{n-1}h'(x), \tag{3} \]
whence
\[ h'(x)^n = nh(x)^{n-1}h'(x) \tag{4} \]
\[ h'(x)^{n-1} = nh(x)^{n-1} \tag{5} \]
\[ h'(x) = Ah(x) \tag{6} \]
where $A = n^{1/(n-1)}$. Hence
\[ h(x) = ce^{Ax} \tag{7} \]
for some constant $c$. Finally
\[ f(x) = h'(x) = \boxed{cAe^{Ax}}. \tag{8} \] | algebraic | putnam | Analysis Algebra | Find $f(x)$ such that
\[ \int [f(x)]^n dx = \left(\int f(x) dx\right)^n, \]
when constants of integration are suitably chosen. | We assume that only real-valued continuous functions $f$ defined on an interval are to be considered. If we put $g(x) = \int f(x)^n dx$ and $h(x) = \int f(x) dx$, we are asked to find all pairs of $C^1$-functions $g$ and $h$ defined on an interval such that
\[ g(x) = h(x)^n \tag{1} \]
and
\[ g'(x) = h'(x)^n. \tag{2} \]
If $n = 1$, then obviously any continuous function $f$ and corresponding functions $g$ and $h$ solve the problem, so we assume from now on that $n \neq 1$.
We proceed formally. Differentiate (1) to get
\[ g'(x) = nh(x)^{n-1}h'(x), \tag{3} \]
whence
\[ h'(x)^n = nh(x)^{n-1}h'(x) \tag{4} \]
\[ h'(x)^{n-1} = nh(x)^{n-1} \tag{5} \]
\[ h'(x) = Ah(x) \tag{6} \]
where $A = n^{1/(n-1)}$. Hence
\[ h(x) = ce^{Ax} \tag{7} \]
for some constant $c$. Finally
\[ f(x) = h'(x) = \boxed{cAe^{Ax}}. \tag{8} \] | 0 |
1940 | 1940_4 | The parabola $y^2 = -4px$ rolls without slipping around the parabola $y^2 = 4px$. Find the equation of the locus of the vertex of the rolling parabola. | If the rolling parabola and the fixed parabola are tangent at the point $Q$, it is obvious from symmetry that the vertex $V$ of the rolling parabola is the reflection of the origin (the vertex of the fixed parabola) in the tangent line at $Q$.
In the sketch, we have tacitly assumed $p > 0$. Suppose that $Q$ is the point $(4pt^2, 4pt)$. [Any point on the fixed parabola will have this form for a unique $t$.] The slope of the tangent at $Q$ is $1/(2t)$ and the equation of the tangent is
\[ y = \frac{1}{2t}x + 2pt. \]
The perpendicular on this line through the origin has the equation
\[ y = -2tx. \]
These lines intersect at
\[ \left( \frac{-4pt^2}{1+4t^2}, \frac{8pt^3}{1+4t^2} \right) \]
so the vertex $V$ is at
\[ \left( \frac{-8pt^2}{1+4t^2}, \frac{16pt^3}{1+4t^2} \right). \]
This gives us the parametric equations
\[ x = \frac{-8pt^2}{1+4t^2}, \quad y = \frac{16pt^3}{1+4t^2} \]
for the path of the point $V$. Since $-2t = y/x$ (if $x \neq 0$), the elimination of $t$ gives
\[ x = \frac{-2p\left(\frac{y}{x}\right)^2}{1+\left(\frac{y}{x}\right)^2} \]
so that
\[ \boxed{(x^2 + y^2)x + 2py^2 = 0} \tag{1} \]
is an equation satisfied by all points of the locus (including $(0, 0)$). Conversely, any point $(x, y)$ other than the origin satisfying (1) leads to a value of $t(= -y/2x)$ and is therefore on the locus. Thus (1) describes the locus precisely. | algebraic | putnam | Geometry Analysis | The parabola $y^2 = -4px$ rolls without slipping around the parabola $y^2 = 4px$. Find the equation of the locus of the vertex of the rolling parabola. | If the rolling parabola and the fixed parabola are tangent at the point $Q$, it is obvious from symmetry that the vertex $V$ of the rolling parabola is the reflection of the origin (the vertex of the fixed parabola) in the tangent line at $Q$.
In the sketch, we have tacitly assumed $p > 0$. Suppose that $Q$ is the point $(4pt^2, 4pt)$. [Any point on the fixed parabola will have this form for a unique $t$.] The slope of the tangent at $Q$ is $1/(2t)$ and the equation of the tangent is
\[ y = \frac{1}{2t}x + 2pt. \]
The perpendicular on this line through the origin has the equation
\[ y = -2tx. \]
These lines intersect at
\[ \left( \frac{-4pt^2}{1+4t^2}, \frac{8pt^3}{1+4t^2} \right) \]
so the vertex $V$ is at
\[ \left( \frac{-8pt^2}{1+4t^2}, \frac{16pt^3}{1+4t^2} \right). \]
This gives us the parametric equations
\[ x = \frac{-8pt^2}{1+4t^2}, \quad y = \frac{16pt^3}{1+4t^2} \]
for the path of the point $V$. Since $-2t = y/x$ (if $x \neq 0$), the elimination of $t$ gives
\[ x = \frac{-2p\left(\frac{y}{x}\right)^2}{1+\left(\frac{y}{x}\right)^2} \]
so that
\[ \boxed{(x^2 + y^2)x + 2py^2 = 0} \tag{1} \]
is an equation satisfied by all points of the locus (including $(0, 0)$). Conversely, any point $(x, y)$ other than the origin satisfying (1) leads to a value of $t(= -y/2x)$ and is therefore on the locus. Thus (1) describes the locus precisely. | 0 |
1940 | 1940_5 | The simultaneous equations $x^4 - x^2 = y^4 - y^2 = z^4 - z^2$ are satisfied by the points of $n$ straight lines and $m$ ellipses, and by no other points. Find $m+n$. | Let $L$ denote the locus of the given equations. Then a point is on $L$ if and only if its coordinates $(x, y, z)$ satisfy
\[ (x^2 + y^2 - 1)(x^2 - y^2) = 0 \tag{1} \]
\[ (y^2 + z^2 - 1)(y^2 - z^2) = 0 \tag{2} \]
\[ (z^2 + x^2 - 1)(z^2 - x^2) = 0 \tag{3} \]
Consider the loci $A$, $B$, $C$, $D$ defined as follows:
\[ A: \begin{cases} x^2 + y^2 - 1 = 0, \\ y^2 - z^2 = 0 \end{cases} \]
\[ B: \begin{cases} y^2 + z^2 - 1 = 0, \\ z^2 - x^2 = 0 \end{cases} \]
\[ C: \begin{cases} z^2 + x^2 - 1 = 0, \\ x^2 - y^2 = 0 \end{cases} \]
\[ D: x^2 = y^2 = z^2. \]
$A$ is the intersection of a right circular cylinder with the union of the planes $z = y$ and $z = -y$. Hence $A$ is the union of two ellipses. Similarly $B$ and $C$ are each the union of two ellipses. $D$, on the other hand, is the union of four straight lines, namely:
\[ x = y = z, \quad x = y = -z, \quad x = -y = z, \quad x = -y = -z. \]
Any point common to $A$ and $B$ is in fact also on $D$, so the ellipses of $A$ and $B$ are different. Similarly for $B$ and $C$ and for $C$ and $A$. Hence $A \cup B \cup C \cup D$ consists of the union of 6 (distinct) ellipses and 4 (distinct) lines.
We now show that $L = A \cup B \cup C \cup D$. If $(x, y, z) \in A$, then evidently $(x, y, z)$ satisfies (1), (2), and (3), the latter because
\[ z^2 + x^2 - 1 = (x^2 + y^2 - 1) + (z^2 - y^2) = 0. \]
Thus $A \subseteq L$. Similarly $B \subseteq L$ and $C \subseteq L$. It is immediate that $D \subseteq L$. So $A \cup B \cup C \cup D \subseteq L$.
Now consider a point $p(x, y, z)$ of $L$ that is not on $D$. Assume, therefore,
\[ x^2 \neq y^2 \text{ and } x^2 \neq z^2. \]
Since $p$ satisfies (1) and (3), we have
\[ x^2 + y^2 - 1 = 0 \text{ and } x^2 + z^2 - 1 = 0 \]
and therefore $y^2 = z^2$, so $p \in A$. The other cases of inequalities lead to $p \in B$ or $p \in C$ by the same argument. Hence $L \subseteq A \cup B \cup C \cup D$ and indeed $L = A \cup B \cup C \cup D$ is the union of 4 lines and 6 ellipses. The final answer is \boxed{10}. | numerical | putnam (modified boxing) | Geometry Analysis | Prove that the simultaneous equations $x^4 - x^2 = y^4 - y^2 = z^4 - z^2$ are satisfied by the points of four straight lines and six ellipses, and by no other points. | Let $L$ denote the locus of the given equations. Then a point is on $L$ if and only if its coordinates $(x, y, z)$ satisfy
\[ (x^2 + y^2 - 1)(x^2 - y^2) = 0 \tag{1} \]
\[ (y^2 + z^2 - 1)(y^2 - z^2) = 0 \tag{2} \]
\[ (z^2 + x^2 - 1)(z^2 - x^2) = 0 \tag{3} \]
Consider the loci $A$, $B$, $C$, $D$ defined as follows:
\[ A: \begin{cases} x^2 + y^2 - 1 = 0, \\ y^2 - z^2 = 0 \end{cases} \]
\[ B: \begin{cases} y^2 + z^2 - 1 = 0, \\ z^2 - x^2 = 0 \end{cases} \]
\[ C: \begin{cases} z^2 + x^2 - 1 = 0, \\ x^2 - y^2 = 0 \end{cases} \]
\[ D: x^2 = y^2 = z^2. \]
$A$ is the intersection of a right circular cylinder with the union of the planes $z = y$ and $z = -y$. Hence $A$ is the union of two ellipses. Similarly $B$ and $C$ are each the union of two ellipses. $D$, on the other hand, is the union of four straight lines, namely:
\[ x = y = z, \quad x = y = -z, \quad x = -y = z, \quad x = -y = -z. \]
Any point common to $A$ and $B$ is in fact also on $D$, so the ellipses of $A$ and $B$ are different. Similarly for $B$ and $C$ and for $C$ and $A$. Hence $A \cup B \cup C \cup D$ consists of the union of 6 (distinct) ellipses and 4 (distinct) lines.
We now show that $L = A \cup B \cup C \cup D$. If $(x, y, z) \in A$, then evidently $(x, y, z)$ satisfies (1), (2), and (3), the latter because
\[ z^2 + x^2 - 1 = (x^2 + y^2 - 1) + (z^2 - y^2) = 0. \]
Thus $A \subseteq L$. Similarly $B \subseteq L$ and $C \subseteq L$. It is immediate that $D \subseteq L$. So $A \cup B \cup C \cup D \subseteq L$.
Now consider a point $p(x, y, z)$ of $L$ that is not on $D$. Assume, therefore,
\[ x^2 \neq y^2 \text{ and } x^2 \neq z^2. \]
Since $p$ satisfies (1) and (3), we have
\[ x^2 + y^2 - 1 = 0 \text{ and } x^2 + z^2 - 1 = 0 \]
and therefore $y^2 = z^2$, so $p \in A$. The other cases of inequalities lead to $p \in B$ or $p \in C$ by the same argument. Hence $L \subseteq A \cup B \cup C \cup D$ and indeed $L = A \cup B \cup C \cup D$ is the union of 4 lines and 6 ellipses. | 0 |
1940 | 1940_9 | A projectile, thrown with initial velocity $v_0$ in a direction making angle $\alpha$ with the horizontal, is acted on by no force except gravity. Find an expression for when the flight is the longest. \] | The differential equations of the motion (using $x$ for the horizontal coordinate and $y$ for the vertical coordinate and taking the origin at the initial point) are
\[ \frac{d^2x}{dt^2} = 0, \quad \frac{d^2y}{dt^2} = -g, \]
where $g$ is the acceleration due to gravity. Using the given initial conditions these can be solved to get
\[ x = v_0t \cos \alpha, \quad y = v_0t \sin \alpha - \frac{1}{2}gt^2. \]
The flight lasts from time $t = 0$ to $t = T = (2v_0 \sin \alpha)/g$. The length of the trajectory is given by
\[ S(\alpha) = \int_0^T \sqrt{(v_0 \sin \alpha - gt)^2 + (v_0 \cos \alpha)^2} \, dt. \]
Putting $w = v_0 \sin \alpha - gt$ and $u = v_0 \cos \alpha$, this becomes
\[ S(\alpha) = -\frac{1}{g} \int_{v_0 \sin \alpha}^{-v_0 \sin \alpha} \sqrt{w^2 + u^2} \, dw = \frac{2}{g} \int_0^{v_0 \sin \alpha} \sqrt{w^2 + u^2} \, dw. \]
Differentiating $S(\alpha)$ with respect to $\alpha$ and simplifying yields
\[ S'(\alpha) = \frac{2v_0^2 \cos \alpha}{g} \left(1 - \sin \alpha \log (\sec \alpha + \tan \alpha)\right). \]
Now $\sin \alpha$ increases from $0$ to $1$ as $\alpha$ varies from $0$ to $\pi/2$ and $\log (\sec \alpha + \tan \alpha)$ increases from $0$ to $\infty$, while $\cos \alpha$ is positive except for $\alpha = \pi/2$. It follows that $\boxed{\sin \alpha \log (\sec \alpha + \tan \alpha) = 1}$ for a unique value $\alpha = \alpha_0 \in (0, \pi/2)$ and that $S'(\alpha) > 0$ for $0 < \alpha < \alpha_0$, $S'(\alpha) < 0$ for $\alpha_0 < \alpha < \pi/2$. Since $S$ is a continuous function on $[0, \pi/2]$, it has a unique maximum on this interval at the point $\alpha_0$; i.e., the flight is longest for $\alpha = \alpha_0$. Calculation shows that $\alpha_0 \approx 56^\circ 28'$. | algebraic | putnam (modified boxing) | Algebra Analysis Trigonometry | A projectile, thrown with initial velocity $v_0$ in a direction making angle $\alpha$ with the horizontal, is acted on by no force except gravity. Find the length of its path until it strikes a horizontal plane through the starting point. Show that the flight is longest when
\[ \sin \alpha \log(\sec \alpha + \tan \alpha) = 1. \] | The differential equations of the motion (using $x$ for the horizontal coordinate and $y$ for the vertical coordinate and taking the origin at the initial point) are
\[ \frac{d^2x}{dt^2} = 0, \quad \frac{d^2y}{dt^2} = -g, \]
where $g$ is the acceleration due to gravity. Using the given initial conditions these can be solved to get
\[ x = v_0t \cos \alpha, \quad y = v_0t \sin \alpha - \frac{1}{2}gt^2. \]
The flight lasts from time $t = 0$ to $t = T = (2v_0 \sin \alpha)/g$. The length of the trajectory is given by
\[ S(\alpha) = \int_0^T \sqrt{(v_0 \sin \alpha - gt)^2 + (v_0 \cos \alpha)^2} \, dt. \]
Putting $w = v_0 \sin \alpha - gt$ and $u = v_0 \cos \alpha$, this becomes
\[ S(\alpha) = -\frac{1}{g} \int_{v_0 \sin \alpha}^{-v_0 \sin \alpha} \sqrt{w^2 + u^2} \, dw = \frac{2}{g} \int_0^{v_0 \sin \alpha} \sqrt{w^2 + u^2} \, dw. \]
Differentiating $S(\alpha)$ with respect to $\alpha$ and simplifying yields
\[ S'(\alpha) = \frac{2v_0^2 \cos \alpha}{g} \left(1 - \sin \alpha \log (\sec \alpha + \tan \alpha)\right). \]
Now $\sin \alpha$ increases from $0$ to $1$ as $\alpha$ varies from $0$ to $\pi/2$ and $\log (\sec \alpha + \tan \alpha)$ increases from $0$ to $\infty$, while $\cos \alpha$ is positive except for $\alpha = \pi/2$. It follows that $\sin \alpha \log (\sec \alpha + \tan \alpha) = 1$ for a unique value $\alpha = \alpha_0 \in (0, \pi/2)$ and that $S'(\alpha) > 0$ for $0 < \alpha < \alpha_0$, $S'(\alpha) < 0$ for $\alpha_0 < \alpha < \pi/2$. Since $S$ is a continuous function on $[0, \pi/2]$, it has a unique maximum on this interval at the point $\alpha_0$; i.e., the flight is longest for $\alpha = \alpha_0$. Calculation shows that $\alpha_0 \approx 56^\circ 28'$. | 0 |
1940 | 1940_10_i | A cylindrical hole of radius $r$ is bored through a cylinder of radius $R$ ($r \leq R$) so that the axes intersect at right angles. Find an expression for the area of the larger cylinder which is inside the smaller where $m = \frac{r}{R}$, $x^2 + z^2 = R^2$, $x^2 + y^2 = r^2$ and $v = x/r$. Give the final answer in the form of $a \int_0^1 b$ where bpth $a$ and $b$ are expressions. | Let the two cylindrical surfaces be $x^2+z^2=R^2$ and $x^2+y^2=r^2$, where $r \leq R$. The shaded area shown in the diagram is the part of the required area that lies in one octant. The equation of this surface is \[ z = \sqrt{R^2 - x^2}. \] The required area is \[ S = 8 \int \int \sqrt{1+\left(\frac{\partial z}{\partial y}\right)^2 + \left(\frac{\partial z}{\partial x}\right)^2} \, dy \, dx \] where the double integral is over the region \[ x^2+y^2 \leq r^2, \quad x \geq 0, \quad y \geq 0. \] Converted to an iterated integral, this becomes \[ S = 8 \int_0^r \left(\int_0^{\sqrt{r^2-x^2}} \frac{R}{\sqrt{R^2-x^2}} \, dy\right) dx = 8R \int_0^r \frac{r^2-x^2}{\sqrt{R^2-x^2}} \, dx. \] Now let $x/r = v$ and $r/R = m$ and simplify further to get \[ S = \boxed{8r^2 \int_0^1 \frac{1-v^2}{\sqrt{(1-v^2)(1-m^2v^2)}} \, dv}, \] which completes part (i). | algebraic | putnam (modified boxing) | Geometry Calculus | A cylindrical hole of radius $r$ is bored through a cylinder of radius $R$ ($r \leq R$) so that the axes intersect at right angles. (i) Show that the area of the larger cylinder which is inside the smaller can be expressed in the form \[ S = 8r^2 \int_0^1 \frac{1-v^2}{\sqrt{(1-v^2)(1-m^2v^2)}} dv \] where $m = \frac{r}{R}$. | Let the two cylindrical surfaces be $x^2+z^2=R^2$ and $x^2+y^2=r^2$, where $r \leq R$. The shaded area shown in the diagram is the part of the required area that lies in one octant. The equation of this surface is \[ z = \sqrt{R^2 - x^2}. \] The required area is \[ S = 8 \int \int \sqrt{1+\left(\frac{\partial z}{\partial y}\right)^2 + \left(\frac{\partial z}{\partial x}\right)^2} \, dy \, dx \] where the double integral is over the region \[ x^2+y^2 \leq r^2, \quad x \geq 0, \quad y \geq 0. \] Converted to an iterated integral, this becomes \[ S = 8 \int_0^r \left(\int_0^{\sqrt{r^2-x^2}} \frac{R}{\sqrt{R^2-x^2}} \, dy\right) dx = 8R \int_0^r \frac{r^2-x^2}{\sqrt{R^2-x^2}} \, dx. \] Now let $x/r = v$ and $r/R = m$ and simplify further to get \[ S = 8r^2 \int_0^1 \frac{1-v^2}{\sqrt{(1-v^2)(1-m^2v^2)}} \, dv, \] which completes part (i). | 0 |
1940 | 1940_10_ii | A cylindrical hole of radius $r$ is bored through a cylinder of radius $R$ ($r \leq R$) so that the axes intersect at right angles. We are given that the area of the larger cylinder which is inside the smaller can be expressed in the form
\[ S = 8r^2 \int_0^1 \frac{1 - v^2}{\sqrt{(1 - v^2)(1 - m^2v^2)}} dv \quad \text{where } m = \frac{r}{R}. If
\[ K = \int_0^1 \frac{dv}{\sqrt{(1 - v^2)(1 - m^2v^2)}} \quad \text{and } E = \int_0^1 \sqrt{\frac{1 - m^2v^2}{1 - v^2}} dv, \] find an expression for $S$ in terms of $R, r, E$ and $K$. | Let the two cylindrical surfaces be $x^2 + z^2 = R^2$ and $x^2 + y^2 = r^2$, where $r \leq R$. The shaded area shown in the diagram is the part of the required area that lies in one octant. The equation of this surface is
\[ z = \sqrt{R^2 - x^2}. \]
The required area is
\[ S = 8 \int \int \sqrt{1 + \left( \frac{\partial z}{\partial y} \right)^2 + \left( \frac{\partial z}{\partial x} \right)^2} dy \, dx. \]
Where the double integral is over the region
\[ x^2 + y^2 \leq r^2, \quad x \geq 0, \quad y \geq 0. \]
Converted to an iterated integral, this becomes
\[ S = 8 \int_0^r \left( \int_0^{\sqrt{r^2 - x^2}} \frac{R}{\sqrt{R^2 - x^2}} dy \right) dx \]
\[ = 8R \int_0^r \sqrt{\frac{r^2 - x^2}{R^2 - x^2}} dx. \]
Now let $x/r = v$ and $r/R = m$ and simplify further to get
\[ S = 8r^2 \int_0^1 \frac{1 - v^2}{\sqrt{(1 - v^2)(1 - m^2v^2)}} dv, \]
which completes part (i).
To obtain (ii), write $r^2(1 - v^2) = R^2(1 - m^2v^2) - (R^2 - r^2)$ and substitute to get
\[ S = 8 \int_0^1 \frac{R^2(1 - m^2v^2)}{\sqrt{(1 - v^2)(1 - m^2v^2)}} dv - 8 \int_0^1 \frac{(R^2 - r^2)}{\sqrt{(1 - v^2)(1 - m^2v^2)}} dv \]
\[ = \boxed{8[R^2E - (R^2 - r^2)K]}. \] | algebraic | putnam (modified boxing) | Geometry Calculus | A cylindrical hole of radius $r$ is bored through a cylinder of radius $R$ ($r \leq R$) so that the axes intersect at right angles.
\begin{enumerate}
\item[(i)] Show that the area of the larger cylinder which is inside the smaller can be expressed in the form
\[ S = 8r^2 \int_0^1 \frac{1 - v^2}{\sqrt{(1 - v^2)(1 - m^2v^2)}} dv \quad \text{where } m = \frac{r}{R}. \]
\item[(ii)] If
\[ K = \int_0^1 \frac{dv}{\sqrt{(1 - v^2)(1 - m^2v^2)}} \quad \text{and } E = \int_0^1 \sqrt{\frac{1 - m^2v^2}{1 - v^2}} dv, \]
show that
\[ S = 8\left[R^2E - (R^2 - r^2)K\right]. \]
\end{enumerate} | Let the two cylindrical surfaces be $x^2 + z^2 = R^2$ and $x^2 + y^2 = r^2$, where $r \leq R$. The shaded area shown in the diagram is the part of the required area that lies in one octant. The equation of this surface is
\[ z = \sqrt{R^2 - x^2}. \]
The required area is
\[ S = 8 \int \int \sqrt{1 + \left( \frac{\partial z}{\partial y} \right)^2 + \left( \frac{\partial z}{\partial x} \right)^2} dy \, dx. \]
Where the double integral is over the region
\[ x^2 + y^2 \leq r^2, \quad x \geq 0, \quad y \geq 0. \]
Converted to an iterated integral, this becomes
\[ S = 8 \int_0^r \left( \int_0^{\sqrt{r^2 - x^2}} \frac{R}{\sqrt{R^2 - x^2}} dy \right) dx \]
\[ = 8R \int_0^r \sqrt{\frac{r^2 - x^2}{R^2 - x^2}} dx. \]
Now let $x/r = v$ and $r/R = m$ and simplify further to get
\[ S = 8r^2 \int_0^1 \frac{1 - v^2}{\sqrt{(1 - v^2)(1 - m^2v^2)}} dv, \]
which completes part (i).
To obtain (ii), write $r^2(1 - v^2) = R^2(1 - m^2v^2) - (R^2 - r^2)$ and substitute to get
\[ S = 8 \int_0^1 \frac{R^2(1 - m^2v^2)}{\sqrt{(1 - v^2)(1 - m^2v^2)}} dv - 8 \int_0^1 \frac{(R^2 - r^2)}{\sqrt{(1 - v^2)(1 - m^2v^2)}} dv \]
\[ = 8\left[R^2E - (R^2 - r^2)K\right]. \] | 0 |
1940 | 1940_12 | Find that the locus of the point of intersection of three mutually perpendicular planes tangent to the surface
\begin{equation}
ax^2 + by^2 + cz^2 = 1 \quad (abc \neq 0)
\end{equation} in terms of $x,y,z,a,b,c$. | We first find the conditions on the coefficients in order that the plane
\begin{equation}
\alpha x + \beta y + \gamma z = \delta
\end{equation}
be tangent to the quadric surface $Q$ given by (1).\nThe tangent plane to $Q$ at the point $(x_1, y_1, z_1)$ has the equation
\begin{equation}
ax_1 x + by_1 y + cz_1 z = 1.
\end{equation}
Suppose (3) and (4) are the same plane. Then $\delta \neq 0$ and
\begin{equation}
x_1 = \frac{\alpha}{a\delta}, \quad y_1 = \frac{\beta}{b\delta}, \quad z_1 = \frac{\gamma}{c\delta}.
\end{equation}
Since $(x_1, y_1, z_1)$ is a point of $Q$, we must have
\begin{equation}
\frac{1}{\delta^2} \left( \frac{\alpha^2}{a} + \frac{\beta^2}{b} + \frac{\gamma^2}{c} \right) = 1.
\end{equation}
Hence
\begin{equation}
\frac{\alpha^2}{a} + \frac{\beta^2}{b} + \frac{\gamma^2}{c} = \delta^2
\end{equation}
with $\delta \neq 0$ is a necessary condition in order that the plane (3) be tangent to $Q$. Conversely, if (6) holds and $\delta \neq 0$, then (5) determines a point $P = (x_1, y_1, z_1)$ of $Q$, and the tangent to $Q$ and $P$ has equation (4) which is equivalent to (3). So our condition is both necessary and sufficient.
Later we shall have to know what happens if (6) holds with $\delta = 0$. (We assume, of course, that $\alpha$, $\beta$, and $\gamma$ are not all zero.) Then the plane (3) is asymptotic to $Q$. In the context of projective geometry, (6) is necessary and sufficient that (3) be projectively tangent to $Q$, the point of tangency having homogeneous coordinates $(\alpha/a, \beta/b, \gamma/c, \delta)$. When $\delta \neq 0$, this is an ordinary point with Cartesian coordinates given by (5). When $\delta = 0$, this is a point at infinity on $Q$. (A projective point having homogeneous coordinates $(x, y, z, t)$ is on $Q$ if and only if $ax^2 + by^2 + cz^2 = t^2$. This equation is derived from the equation for $Q$ by multiplying each term by a power of $t$ to bring its degree up to two.) An asymptotic plane is thus "tangent to $Q$ at infinity." We may also view it as the limit of a sequence of tangent planes whose points of tangency approach some point at infinity on $Q$.
Suppose $(\alpha_1, \beta_1, \gamma_1)$, $(\alpha_2, \beta_2, \gamma_2)$, and $(\alpha_3, \beta_3, \gamma_3)$ are mutually orthogonal unit vectors. Then the matrix
\begin{equation}
\begin{pmatrix}
\alpha_1 & \beta_1 & \gamma_1 \\
\alpha_2 & \beta_2 & \gamma_2 \\
\alpha_3 & \beta_3 & \gamma_3
\end{pmatrix}
\end{equation}
is orthogonal, so its columns are also mutually orthogonal unit vectors. Hence
\begin{equation}
\sum \frac{\alpha_i^2}{a} = \sum \frac{\beta_i^2}{b} = \sum \frac{\gamma_i^2}{c} = 1.
\end{equation}
If we have planes tangent to $Q$ with these vectors as normals, these planes are mutually perpendicular; their equations are
\begin{equation}
\alpha_i x + \beta_i y + \gamma_i z = \delta_i, \quad i = 1, 2, 3;
\end{equation}
where
\begin{equation}
\frac{\alpha_i^2}{a} + \frac{\beta_i^2}{b} + \frac{\gamma_i^2}{c} = \delta_i^2.
\end{equation}
Since $|\delta_i|$ is the distance from the origin to the $i$th plane, the Pythagorean Theorem shows that the distance from the origin $O$ to the point $P$ where these planes intersect is given by
\begin{equation}
OP^2 = \delta_1^2 + \delta_2^2 + \delta_3^2 = \frac{1}{a} \sum \alpha_i^2 + \frac{1}{b} \sum \beta_i^2 + \frac{1}{c} \sum \gamma_i^2 = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.
\end{equation}
This can also be derived analytically as follows: If $x$, $y$, and $z$ are the coordinates of $P$, then all of the equations (8) hold. Square these equations and add, using the orthogonality of the columns of (7). This gives
\begin{equation}
x^2 + y^2 + z^2 = \delta_1^2 + \delta_2^2 + \delta_3^2 = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.
\end{equation} Thus the final answer is \boxed{x^2 + y^2 + z^2 = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}}. | algebraic | putnam (modified boxing) | Geometry Linear Algebra | Prove that the locus of the point of intersection of three mutually perpendicular planes tangent to the surface
\begin{equation}
ax^2 + by^2 + cz^2 = 1 \quad (abc \neq 0)
\end{equation}
is the sphere
\begin{equation}
x^2 + y^2 + z^2 = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.
\end{equation} | We first find the conditions on the coefficients in order that the plane
\begin{equation}
\alpha x + \beta y + \gamma z = \delta
\end{equation}
be tangent to the quadric surface $Q$ given by (1).\nThe tangent plane to $Q$ at the point $(x_1, y_1, z_1)$ has the equation
\begin{equation}
ax_1 x + by_1 y + cz_1 z = 1.
\end{equation}
Suppose (3) and (4) are the same plane. Then $\delta \neq 0$ and
\begin{equation}
x_1 = \frac{\alpha}{a\delta}, \quad y_1 = \frac{\beta}{b\delta}, \quad z_1 = \frac{\gamma}{c\delta}.
\end{equation}
Since $(x_1, y_1, z_1)$ is a point of $Q$, we must have
\begin{equation}
\frac{1}{\delta^2} \left( \frac{\alpha^2}{a} + \frac{\beta^2}{b} + \frac{\gamma^2}{c} \right) = 1.
\end{equation}
Hence
\begin{equation}
\frac{\alpha^2}{a} + \frac{\beta^2}{b} + \frac{\gamma^2}{c} = \delta^2
\end{equation}
with $\delta \neq 0$ is a necessary condition in order that the plane (3) be tangent to $Q$. Conversely, if (6) holds and $\delta \neq 0$, then (5) determines a point $P = (x_1, y_1, z_1)$ of $Q$, and the tangent to $Q$ and $P$ has equation (4) which is equivalent to (3). So our condition is both necessary and sufficient.
Later we shall have to know what happens if (6) holds with $\delta = 0$. (We assume, of course, that $\alpha$, $\beta$, and $\gamma$ are not all zero.) Then the plane (3) is asymptotic to $Q$. In the context of projective geometry, (6) is necessary and sufficient that (3) be projectively tangent to $Q$, the point of tangency having homogeneous coordinates $(\alpha/a, \beta/b, \gamma/c, \delta)$. When $\delta \neq 0$, this is an ordinary point with Cartesian coordinates given by (5). When $\delta = 0$, this is a point at infinity on $Q$. (A projective point having homogeneous coordinates $(x, y, z, t)$ is on $Q$ if and only if $ax^2 + by^2 + cz^2 = t^2$. This equation is derived from the equation for $Q$ by multiplying each term by a power of $t$ to bring its degree up to two.) An asymptotic plane is thus "tangent to $Q$ at infinity." We may also view it as the limit of a sequence of tangent planes whose points of tangency approach some point at infinity on $Q$.
Suppose $(\alpha_1, \beta_1, \gamma_1)$, $(\alpha_2, \beta_2, \gamma_2)$, and $(\alpha_3, \beta_3, \gamma_3)$ are mutually orthogonal unit vectors. Then the matrix
\begin{equation}
\begin{pmatrix}
\alpha_1 & \beta_1 & \gamma_1 \\
\alpha_2 & \beta_2 & \gamma_2 \\
\alpha_3 & \beta_3 & \gamma_3
\end{pmatrix}
\end{equation}
is orthogonal, so its columns are also mutually orthogonal unit vectors. Hence
\begin{equation}
\sum \frac{\alpha_i^2}{a} = \sum \frac{\beta_i^2}{b} = \sum \frac{\gamma_i^2}{c} = 1.
\end{equation}
If we have planes tangent to $Q$ with these vectors as normals, these planes are mutually perpendicular; their equations are
\begin{equation}
\alpha_i x + \beta_i y + \gamma_i z = \delta_i, \quad i = 1, 2, 3;
\end{equation}
where
\begin{equation}
\frac{\alpha_i^2}{a} + \frac{\beta_i^2}{b} + \frac{\gamma_i^2}{c} = \delta_i^2.
\end{equation}
Since $|\delta_i|$ is the distance from the origin to the $i$th plane, the Pythagorean Theorem shows that the distance from the origin $O$ to the point $P$ where these planes intersect is given by
\begin{equation}
OP^2 = \delta_1^2 + \delta_2^2 + \delta_3^2 = \frac{1}{a} \sum \alpha_i^2 + \frac{1}{b} \sum \beta_i^2 + \frac{1}{c} \sum \gamma_i^2 = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.
\end{equation}
This can also be derived analytically as follows: If $x$, $y$, and $z$ are the coordinates of $P$, then all of the equations (8) hold. Square these equations and add, using the orthogonality of the columns of (7). This gives
\begin{equation}
x^2 + y^2 + z^2 = \delta_1^2 + \delta_2^2 + \delta_3^2 = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.
\end{equation} | 0 |
1940 | 1940_13 | Determine all rational values for which $a, b, c$ are the roots of \[ x^3 + ax^2 + bx + c = 0. \] Find the sum of all corresponding $a, b, c$ values and return it as a final ordered triple with the sums. | The conditions on the roots are equivalent to\n\n\begin{align} \tag{1} a + b + c &= -a, \\ \tag{2} ab + bc + ca &= b, \\ \tag{3} abc &= -c. \end{align}\n\nIf $c = 0$, then $ab = b$ and $2a + b = 0$, so either $b = 0$, $a = 0$, or $a = 1$, $b = -2$.\n\nIf $c \neq 0$, then $ab = -1$. If $a + b = 0$, then (2) becomes $ab = b$ so that $a = 1$, $b = -1$, $c = -1$. If $a + b \neq 0$, then\n\[ c = \frac{b + 1}{a + b} = \frac{a(b + 1)}{a(a + b)} = \frac{-1 + a}{a^2 - 1} = \frac{1}{a + 1} \]\nand (1) becomes\n\[ 2a - \frac{1}{a} + \frac{1}{a + 1} = 0 \]\nwhence\n\[ 2a^3 + 2a^2 - 1 = 0. \]\nThis equation has no rational roots, since the only possibilities are $\pm 1$, $\pm 1/2$, and these are not roots. There are therefore three solutions\n\n\[ \begin{array}{ccc} a & b & c & \text{corresponding to} \\ 0 & 0 & 0 & x^3 = 0 \\ +1 & -2 & 0 & x^3 + x^2 - 2x = 0 \\ +1 & -1 & -1 & x^3 + x^2 - x - 1 = (x^2 - 1)(x + 1) = 0. \end{array} \] Thus the final triple is \boxed{(2, -3, -1)}. | numerical | putnam (modified boxing) | Algebra | Determine all rational values for which $a, b, c$ are the roots of \[ x^3 + ax^2 + bx + c = 0. \] | The conditions on the roots are equivalent to\n\n\begin{align} \tag{1} a + b + c &= -a, \\ \tag{2} ab + bc + ca &= b, \\ \tag{3} abc &= -c. \end{align}\n\nIf $c = 0$, then $ab = b$ and $2a + b = 0$, so either $b = 0$, $a = 0$, or $a = 1$, $b = -2$.\n\nIf $c \neq 0$, then $ab = -1$. If $a + b = 0$, then (2) becomes $ab = b$ so that $a = 1$, $b = -1$, $c = -1$. If $a + b \neq 0$, then\n\[ c = \frac{b + 1}{a + b} = \frac{a(b + 1)}{a(a + b)} = \frac{-1 + a}{a^2 - 1} = \frac{1}{a + 1} \]\nand (1) becomes\n\[ 2a - \frac{1}{a} + \frac{1}{a + 1} = 0 \]\nwhence\n\[ 2a^3 + 2a^2 - 1 = 0. \]\nThis equation has no rational roots, since the only possibilities are $\pm 1$, $\pm 1/2$, and these are not roots. There are therefore three solutions\n\n\[ \begin{array}{ccc} a & b & c & \text{corresponding to} \\ 0 & 0 & 0 & x^3 = 0 \\ +1 & -2 & 0 & x^3 + x^2 - 2x = 0 \\ +1 & -1 & -1 & x^3 + x^2 - x - 1 = (x^2 - 1)(x + 1) = 0. \end{array} \] | 0 |
1940 | 1940_14 | Prove that \[ \begin{vmatrix} a_1^2 + k & a_1a_2 & a_1a_3 & \cdots & a_1a_n \\ a_2a_1 & a_2^2 + k & a_2a_3 & \cdots & a_2a_n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_na_1 & a_na_2 & a_na_3 & \cdots & a_n^2 + k \end{vmatrix} \] is divisible by $k^{n-1}$ and find its other factor. | Let $B$ be the matrix\n\[ \begin{pmatrix} a_1^2 & a_1a_2 & a_1a_3 & \cdots & a_1a_n \\ a_2a_1 & a_2^2 & a_2a_3 & \cdots & a_2a_n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_na_1 & a_na_2 & a_na_3 & \cdots & a_n^2 \end{pmatrix}. \]\n\n$B$ has rank at most one, since any two rows (or columns) are clearly dependent. So there are $(n-1)$ zeros among the eigenvalues of $B$. Therefore the characteristic polynomial of $B$ is divisible by $x^{n-1}$. Hence\n\[ \det(xI - B) = x^n - (\text{trace } B)x^{n-1} = x^{n-1}(x - a_1^2 - a_2^2 - \cdots - a_n^2) \]\nso\n\[ \det(B + kI) = (-1)^n \det(-kI - B) = k^{n-1}(k + a_1^2 + a_2^2 + \cdots + a_n^2), \]\nand the other factor is $(\boxed{k + a_1^2 + a_2^2 + \cdots + a_n^2})$. | algebraic | putnam | Algebra Linear Algebra | Prove that \[ \begin{vmatrix} a_1^2 + k & a_1a_2 & a_1a_3 & \cdots & a_1a_n \\ a_2a_1 & a_2^2 + k & a_2a_3 & \cdots & a_2a_n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_na_1 & a_na_2 & a_na_3 & \cdots & a_n^2 + k \end{vmatrix} \] is divisible by $k^{n-1}$ and find its other factor. | Let $B$ be the matrix\n\[ \begin{pmatrix} a_1^2 & a_1a_2 & a_1a_3 & \cdots & a_1a_n \\ a_2a_1 & a_2^2 & a_2a_3 & \cdots & a_2a_n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_na_1 & a_na_2 & a_na_3 & \cdots & a_n^2 \end{pmatrix}. \]\n\n$B$ has rank at most one, since any two rows (or columns) are clearly dependent. So there are $(n-1)$ zeros among the eigenvalues of $B$. Therefore the characteristic polynomial of $B$ is divisible by $x^{n-1}$. Hence\n\[ \det(xI - B) = x^n - (\text{trace } B)x^{n-1} = x^{n-1}(x - a_1^2 - a_2^2 - \cdots - a_n^2) \]\nso\n\[ \det(B + kI) = (-1)^n \det(-kI - B) = k^{n-1}(k + a_1^2 + a_2^2 + \cdots + a_n^2), \]\nand the other factor is $(\boxed{k + a_1^2 + a_2^2 + \cdots + a_n^2})$. | 0 |
1941 | 1941_2 | Find the $n$th derivative with respect to $x$ of \[ \int_0^x \left[1 + \frac{(x - t)}{1!} + \frac{(x - t)^2}{2!} + \cdots + \frac{(x - t)^{n-1}}{(n-1)!} \right] e^{nt} dt. \] If there are ranges and in turn multiple expressions return their sum. | Let \[ \phi_k(x) = \int_0^x \frac{(x - t)^k}{k!} e^{nt} dt. \] Then, for $k > 0$, \[ \phi_k'(x) = \phi_{k-1}(x). \] Also \[ \phi_0(x) = \int_0^x e^{nt} dt = \frac{e^{nx} - 1}{n}. \] Therefore, \[ \left(\frac{d}{dx}\right)^n \phi_k(x) = \left(\frac{d}{dx}\right)^{n-k} \phi_0(x) = n^{n-k-1} e^{nx} \text{ for } n > k. \] Accordingly, the $n$th derivative of the given function is \[ \left(\frac{d}{dx}\right)^n [\phi_0(x) + \phi_1(x) + \cdots + \phi_{n-1}(x)] = [n^{n-1} + n^{n-2} + \cdots + 1] e^{nx} \] \[ = \begin{cases} \frac{n^n - 1}{n - 1} e^{nx} & \text{for } n \neq 1, \\ e^x & \text{for } n = 1. \end{cases} \] Thus the final answer is \boxed{\frac{n^n - 1}{n - 1} e^{nx} + e^x}. | algebraic | putnam (modified boxing) | Calculus Analysis | Find the $n$th derivative with respect to $x$ of \[ \int_0^x \left[1 + \frac{(x - t)}{1!} + \frac{(x - t)^2}{2!} + \cdots + \frac{(x - t)^{n-1}}{(n-1)!} \right] e^{nt} dt. \] | Let \[ \phi_k(x) = \int_0^x \frac{(x - t)^k}{k!} e^{nt} dt. \] Then, for $k > 0$, \[ \phi_k'(x) = \phi_{k-1}(x). \] Also \[ \phi_0(x) = \int_0^x e^{nt} dt = \frac{e^{nx} - 1}{n}. \] Therefore, \[ \left(\frac{d}{dx}\right)^n \phi_k(x) = \left(\frac{d}{dx}\right)^{n-k} \phi_0(x) = n^{n-k-1} e^{nx} \text{ for } n > k. \] Accordingly, the $n$th derivative of the given function is \[ \left(\frac{d}{dx}\right)^n [\phi_0(x) + \phi_1(x) + \cdots + \phi_{n-1}(x)] = [n^{n-1} + n^{n-2} + \cdots + 1] e^{nx} \] \[ = \boxed{\begin{cases} \frac{n^n - 1}{n - 1} e^{nx} & \text{for } n \neq 1, \\ e^x & \text{for } n = 1. \end{cases}} \] | 0 |
1941 | 1941_4 | Let the roots $a, b, c$ of \[ f(x) \equiv x^3 + px^2 + qx + r = 0 \] be real, and let $a \leq b \leq c$. Prove that, if the interval $(b, c)$ is divided into six equal parts, a root of $f'(x) = 0$ will lie in the fourth part counting from the end $b$. What will be the form of $f(x)$ if the root in question of $f'(x) = 0$ falls at either end of the fourth part? Add all forms together if there is more than one. | The proposition is valid for $f(x)$ if and only if it is valid for $f(x + b)$ so we can translate all the roots by $-b$ and thus arrange that the middle root is zero. It is no loss of generality, therefore, to assume that $b = 0$ to begin with. Hence we consider \[ f(x) = (x - a)(x - c) = x^3 - (a + c)x^2 + acx. \] The fourth subinterval referred to in the problem is $[c/2, 2c/3]$. \[ f'(x) = 3x^2 - 2(a + c)x + ac. \] From $f'(x)$ we find \[ f'(c/2) = -\frac{1}{4}c^2 \leq 0 \text{ and } f'(2c/3) = -\frac{1}{3}ac \geq 0. \] Hence, since $f'$ is continuous, there is a root of $f'(x) = 0$ on $[c/2, 2c/3]$. \n A root occurs at the left endpoint $c/2$ if and only if $c = 0$; that is, the two largest roots coincide. In this event, \[ f(x) = (x - a)x^2. \] A root occurs at the right endpoint $2c/3$ if and only if $a = 0$ or $c = 0$. If $c = 0$ we have the previous case, and the interval in question has degenerated to a single point. If $c \neq 0$, then $a = 0$ and the two smallest roots coincide. In this case, \[ f(x) = x^2(x - c). \] To answer the second part of the question in terms of the original $a, b, c$: the zero of $f'$ occurs at the left endpoint iff \[ f(x) = (x - a)(x - b)^2 \] and at the right endpoint iff \[ f(x) = (x - b)^2(x - c) \text{ or } f(x) = (x - a)(x - b)^2 which makes the final answer \boxed{(x - b)^2(2x - a - c)}. \] | algebraic | putnam (modified boxing) | Algebra Analysis | Let the roots $a, b, c$ of \[ f(x) \equiv x^3 + px^2 + qx + r = 0 \] be real, and let $a \leq b \leq c$. Prove that, if the interval $(b, c)$ is divided into six equal parts, a root of $f'(x) = 0$ will lie in the fourth part counting from the end $b$. What will be the form of $f(x)$ if the root in question of $f'(x) = 0$ falls at either end of the fourth part? | The proposition is valid for $f(x)$ if and only if it is valid for $f(x + b)$ so we can translate all the roots by $-b$ and thus arrange that the middle root is zero. It is no loss of generality, therefore, to assume that $b = 0$ to begin with. Hence we consider \[ f(x) = (x - a)(x - c) = x^3 - (a + c)x^2 + acx. \] The fourth subinterval referred to in the problem is $[c/2, 2c/3]$. \[ f'(x) = 3x^2 - 2(a + c)x + ac. \] From $f'(x)$ we find \[ f'(c/2) = -\frac{1}{4}c^2 \leq 0 \text{ and } f'(2c/3) = -\frac{1}{3}ac \geq 0. \] Hence, since $f'$ is continuous, there is a root of $f'(x) = 0$ on $[c/2, 2c/3]$. \n A root occurs at the left endpoint $c/2$ if and only if $c = 0$; that is, the two largest roots coincide. In this event, \[ f(x) = (x - a)x^2. \] A root occurs at the right endpoint $2c/3$ if and only if $a = 0$ or $c = 0$. If $c = 0$ we have the previous case, and the interval in question has degenerated to a single point. If $c \neq 0$, then $a = 0$ and the two smallest roots coincide. In this case, \[ f(x) = x^2(x - c). \] To answer the second part of the question in terms of the original $a, b, c$: the zero of $f'$ occurs at the left endpoint iff \[ f(x) = (x - a)(x - b)^2 \] and at the right endpoint iff \[ \boxed{f(x) = (x - b)^2(x - c) \text{ or } f(x) = (x - a)(x - b)^2}. \] | 0 |
1941 | 1941_7_ii | A semi-ellipsoid of revolution is formed by revolving about the $x$-axis the area lying within the first quadrant of the ellipse \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. \] Find the condition for which this semi-ellipsoid will balance in stable equilibrium, with its vertex resting on a horizontal plane in terms of $a$ and $b$. \] | First Solution. Let $C$ be the center of gravity of the solid semi-ellipsoid $S$, and let $V$ be its vertex. Consider the sphere with center $C$ and radius $CV$. Suppose that near $V$ the sphere lies strictly inside $S$ (except for the point $V$, of course). Then if $S$ rests on a horizontal plane with point of contact $V$, any small displacement raises the center of gravity, and therefore $S$ is stably balanced.\n On the other hand, suppose that near $V$ the sphere lies strictly outside $S$ (again except for $V$ itself). Then if $S$ rests on a horizontal plane with point of contact $V$, any small displacement lowers the center of gravity, so $S$ is unstable.\n Consider therefore the function $f(P) = CP$, the distance from $C$ to a variable point $P$ on the surface of the ellipsoid. If this function has a strict local minimum at $V$, the balance will be stable; if it has a strict local maximum at $P$, the balance will be unstable. We may as well consider $f(P)^2$ instead of $f(P)$. \n From the circular symmetry of the problem it is clear that $C$ is at $(c, 0, 0)$ for some $c > 0$; moreover we may restrict ourselves to considering the function $f(P)^2$ where $P$ varies along the generating ellipse \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] instead of the whole surface. If $P = (x, y)$ we have \[ f(P)^2 = (x - c)^2 + y^2 = (x - c)^2 + b^2 \left(1 - \frac{x^2}{a^2}\right) = \left(1 - \frac{b^2}{a^2}\right)x^2 - 2cx + b^2 + c^2. \]\n We want to determine whether $x = a$ is a local minimum for this polynomial relative to the interval $[0, a]$. Considering this polynomial along the whole positive $x$-axis we see that it is strictly decreasing if $b^2 \geq a^2$. If $b^2 < a^2$, then it decreases from $x = 0$ to $x = ca^2/(a^2 - b^2)$ and then increases. Hence $V=(a, 0, 0)$ in the two-dimensional coordinate system is a strict local minimum point if $b^2 \geq a^2$ or if $b^2 < a^2$ and $a \leq ca^2/(a^2 - b^2)$; this is equivalent to $b^2 \geq a^2 - ac$. $V$ is a strict local maximum if $b^2 < a(a - c)$. \n Now we locate the center of gravity. Since nothing is said about the mass distribution, it is presumably uniform, so the center of gravity coincides with the centroid. Hence \[ c = \frac{\int_0^a x \cdot \pi y^2 dx}{\int_0^a \pi y^2 dx} = \frac{\int_0^a x \left(1 - \frac{x^2}{a^2}\right)dx}{\int_0^a \left(1 - \frac{x^2}{a^2}\right)dx} = \frac{3}{8}a, \] and our condition for stability becomes \[ b^2 \geq \frac{8}{5}a^2; \] that is, \[\boxed{b\sqrt{8} \geq a\sqrt{5}} \text{ as required.} \] | algebraic | putnam (modified boxing) | Geometry Calculus | A semi-ellipsoid of revolution is formed by revolving about the $x$-axis the area lying within the first quadrant of the ellipse \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. \] Show that this semi-ellipsoid will balance in stable equilibrium, with its vertex resting on a horizontal plane, when and only when \[ b\sqrt{8} \geq a\sqrt{5}. \] | First Solution. Let $C$ be the center of gravity of the solid semi-ellipsoid $S$, and let $V$ be its vertex. Consider the sphere with center $C$ and radius $CV$. Suppose that near $V$ the sphere lies strictly inside $S$ (except for the point $V$, of course). Then if $S$ rests on a horizontal plane with point of contact $V$, any small displacement raises the center of gravity, and therefore $S$ is stably balanced.\n On the other hand, suppose that near $V$ the sphere lies strictly outside $S$ (again except for $V$ itself). Then if $S$ rests on a horizontal plane with point of contact $V$, any small displacement lowers the center of gravity, so $S$ is unstable.\n Consider therefore the function $f(P) = CP$, the distance from $C$ to a variable point $P$ on the surface of the ellipsoid. If this function has a strict local minimum at $V$, the balance will be stable; if it has a strict local maximum at $P$, the balance will be unstable. We may as well consider $f(P)^2$ instead of $f(P)$. \n From the circular symmetry of the problem it is clear that $C$ is at $(c, 0, 0)$ for some $c > 0$; moreover we may restrict ourselves to considering the function $f(P)^2$ where $P$ varies along the generating ellipse \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] instead of the whole surface. If $P = (x, y)$ we have \[ f(P)^2 = (x - c)^2 + y^2 = (x - c)^2 + b^2 \left(1 - \frac{x^2}{a^2}\right) = \left(1 - \frac{b^2}{a^2}\right)x^2 - 2cx + b^2 + c^2. \]\n We want to determine whether $x = a$ is a local minimum for this polynomial relative to the interval $[0, a]$. Considering this polynomial along the whole positive $x$-axis we see that it is strictly decreasing if $b^2 \geq a^2$. If $b^2 < a^2$, then it decreases from $x = 0$ to $x = ca^2/(a^2 - b^2)$ and then increases. Hence $V=(a, 0, 0)$ in the two-dimensional coordinate system is a strict local minimum point if $b^2 \geq a^2$ or if $b^2 < a^2$ and $a \leq ca^2/(a^2 - b^2)$; this is equivalent to $b^2 \geq a^2 - ac$. $V$ is a strict local maximum if $b^2 < a(a - c)$. \n Now we locate the center of gravity. Since nothing is said about the mass distribution, it is presumably uniform, so the center of gravity coincides with the centroid. Hence \[ c = \frac{\int_0^a x \cdot \pi y^2 dx}{\int_0^a \pi y^2 dx} = \frac{\int_0^a x \left(1 - \frac{x^2}{a^2}\right)dx}{\int_0^a \left(1 - \frac{x^2}{a^2}\right)dx} = \frac{3}{8}a, \] and our condition for stability becomes \[ b^2 \geq \frac{8}{5}a^2; \] that is, \[ b\sqrt{8} \geq a\sqrt{5} \text{ as required.} \] | 0 |
1941 | 1941_9 | Evaluate the following limits:\[\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2^2}} + \cdots + \frac{1}{\sqrt{n^2 + n^2}}\right);\] \[\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} + \cdots + \frac{1}{\sqrt{n^2 + n}}\right);\] \[\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} + \cdots + \frac{1}{\sqrt{n^2 + n^2}}\right).\] Return the answer as a triple with the 3 values. | (i) For the first sum,\[ \frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2^2}} + \cdots + \frac{1}{\sqrt{n^2 + n^2}} = \frac{1}{n}\left[ \frac{1}{\sqrt{1 + (\frac{1}{n})^2}} + \frac{1}{\sqrt{1 + (\frac{2}{n})^2}} + \cdots + \frac{1}{\sqrt{1 + (\frac{n}{n})^2}} \right]. \] This latter form is the lower Riemann sum for \[ \int_0^1 \frac{dx}{\sqrt{1 + x^2}} \] corresponding to the subdivision points \[ \frac{1}{n}, \frac{2}{n}, \ldots, \frac{n-1}{n}. \] Therefore its limit as $n \to \infty$ is \[ \int_0^1 \frac{dx}{\sqrt{1 + x^2}} = \log(x + \sqrt{1 + x^2})\big|_0^1 = \log(1 + \sqrt{2}). \]
(ii) For the second sum, an individual term $\frac{1}{\sqrt{n^2 + i}}$ satisfies\[ \frac{1}{\sqrt{n^2 + n}} \leq \frac{1}{\sqrt{n^2 + i}} \leq \frac{1}{\sqrt{n^2 + 1}}, \] $i = 1, 2, \ldots, n,$ and hence\[ \frac{n}{\sqrt{n^2 + n}} \leq \sum_{i=1}^n \frac{1}{\sqrt{n^2 + i}} \leq \frac{n}{\sqrt{n^2 + 1}}. \] Now as $n \to \infty$ both extremes have the limit $1$; hence \[ \lim_{n \to \infty} \left(\sum_{i=1}^n \frac{1}{\sqrt{n^2 + i}}\right) = 1. \]
(iii) For the third sum,\[ \frac{1}{\sqrt{n^2 + i}} \geq \frac{1}{\sqrt{n^2 + n^2}} = \frac{1}{n\sqrt{2}}, \] $i = 1, 2, \ldots, n^2.$ Hence\[ \sum_{i=1}^{n^2} \frac{1}{\sqrt{n^2 + i}} \geq \frac{n^2}{n\sqrt{2}} = \frac{n}{\sqrt{2}}. \] Therefore \[ \lim_{n \to \infty} \sum_{i=1}^{n^2} \frac{1}{\sqrt{n^2 + i}} = \infty. \] The final answer is \boxed{(\log(1 + \sqrt{2}, 1, \infty)}. | numerical | putnam (modified boxing) | Calculus Analysis | Evaluate the following limits:\[\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2^2}} + \cdots + \frac{1}{\sqrt{n^2 + n^2}}\right);\] \[\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} + \cdots + \frac{1}{\sqrt{n^2 + n}}\right);\] \[\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} + \cdots + \frac{1}{\sqrt{n^2 + n^2}}\right).\] | (i) For the first sum,\[ \frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2^2}} + \cdots + \frac{1}{\sqrt{n^2 + n^2}} = \frac{1}{n}\left[ \frac{1}{\sqrt{1 + (\frac{1}{n})^2}} + \frac{1}{\sqrt{1 + (\frac{2}{n})^2}} + \cdots + \frac{1}{\sqrt{1 + (\frac{n}{n})^2}} \right]. \] This latter form is the lower Riemann sum for \[ \int_0^1 \frac{dx}{\sqrt{1 + x^2}} \] corresponding to the subdivision points \[ \frac{1}{n}, \frac{2}{n}, \ldots, \frac{n-1}{n}. \] Therefore its limit as $n \to \infty$ is \[ \int_0^1 \frac{dx}{\sqrt{1 + x^2}} = \log(x + \sqrt{1 + x^2})\big|_0^1 = \log(1 + \sqrt{2}). \]
(ii) For the second sum, an individual term $\frac{1}{\sqrt{n^2 + i}}$ satisfies\[ \frac{1}{\sqrt{n^2 + n}} \leq \frac{1}{\sqrt{n^2 + i}} \leq \frac{1}{\sqrt{n^2 + 1}}, \] $i = 1, 2, \ldots, n,$ and hence\[ \frac{n}{\sqrt{n^2 + n}} \leq \sum_{i=1}^n \frac{1}{\sqrt{n^2 + i}} \leq \frac{n}{\sqrt{n^2 + 1}}. \] Now as $n \to \infty$ both extremes have the limit $1$; hence \[ \lim_{n \to \infty} \left(\sum_{i=1}^n \frac{1}{\sqrt{n^2 + i}}\right) = 1. \]
(iii) For the third sum,\[ \frac{1}{\sqrt{n^2 + i}} \geq \frac{1}{\sqrt{n^2 + n^2}} = \frac{1}{n\sqrt{2}}, \] $i = 1, 2, \ldots, n^2.$ Hence\[ \sum_{i=1}^{n^2} \frac{1}{\sqrt{n^2 + i}} \geq \frac{n^2}{n\sqrt{2}} = \frac{n}{\sqrt{2}}. \] Therefore \[ \lim_{n \to \infty} \sum_{i=1}^{n^2} \frac{1}{\sqrt{n^2 + i}} = \infty. \] | 0 |
1941 | 1941_10 | Find the differential equation satisfied by the product $z$ of any two linearly independent integrals of the equation\[ y'' + y'P(x) + yQ(x) = 0. \] | Suppose $y_1$ and $y_2$ are two linearly independent solutions of the given differential equation. Then any two solutions have the form $u = ay_1 + by_2$ and $v = cy_1 + dy_2$. Since $uv$ falls in the linear space spanned by $y_1^2, y_1y_2, y_2^2$, we expect to find that it satisfies a linear differential equation of the third order. Letting $z = uv$ we have\[ \text{(1)} \quad u'' + Pu' + Qu = 0 \] \[ \text{(2)} \quad v'' + Pv' + Qv = 0 \] \[ \text{(3)} \quad z' = uv' + u'v \] \[ \text{(4)} \quad z'' = uv'' + 2u'v' + vu''. \]
We find\[ \text{(5)} \quad z'' + Pz' + 2Qz = u(v'' + Pv' + Qv) + v(u'' + Pu' + Qu) + 2u'v' = 2u'v'. \]
Differentiating (5), we get\[ \text{(6)} \quad z''' + Pz'' + (P' + 2Q)z' + 2Q'z = 2u'v' + 2v'u''. \] Next multiply (1) by $2v'$ and (2) by $2u'$ and add to obtain\[ \text{(7)} \quad 2u'v'' + 2v'u'' + 4Pu'v' + 2Qz' = 0. \] Multiply (5) by $2P$ to get\[ \text{(8)} \quad 2Pz'' + 2Pz' + 4PQz = 4Pu'v'. \] Add (6), (7), and (8), cancelling the terms that appear on the right, to get\[ \text{(9)} \quad \boxed{z''' + 3Pz'' + (2P^2 + P' + 4Q)z' + (4PQ + 2Q')z = 0}, \] a third-order differential equation satisfied by the product $z$ of any two solutions of the original equation. | algebraic | putnam | Differential Equations Analysis | Find the differential equation satisfied by the product $z$ of any two linearly independent integrals of the equation\[ y'' + y'P(x) + yQ(x) = 0. \] | Suppose $y_1$ and $y_2$ are two linearly independent solutions of the given differential equation. Then any two solutions have the form $u = ay_1 + by_2$ and $v = cy_1 + dy_2$. Since $uv$ falls in the linear space spanned by $y_1^2, y_1y_2, y_2^2$, we expect to find that it satisfies a linear differential equation of the third order. Letting $z = uv$ we have\[ \text{(1)} \quad u'' + Pu' + Qu = 0 \] \[ \text{(2)} \quad v'' + Pv' + Qv = 0 \] \[ \text{(3)} \quad z' = uv' + u'v \] \[ \text{(4)} \quad z'' = uv'' + 2u'v' + vu''. \]
We find\[ \text{(5)} \quad z'' + Pz' + 2Qz = u(v'' + Pv' + Qv) + v(u'' + Pu' + Qu) + 2u'v' = 2u'v'. \]
Differentiating (5), we get\[ \text{(6)} \quad z''' + Pz'' + (P' + 2Q)z' + 2Q'z = 2u'v' + 2v'u''. \] Next multiply (1) by $2v'$ and (2) by $2u'$ and add to obtain\[ \text{(7)} \quad 2u'v'' + 2v'u'' + 4Pu'v' + 2Qz' = 0. \] Multiply (5) by $2P$ to get\[ \text{(8)} \quad 2Pz'' + 2Pz' + 4PQz = 4Pu'v'. \] Add (6), (7), and (8), cancelling the terms that appear on the right, to get\[ \text{(9)} \quad \boxed{z''' + 3Pz'' + (2P^2 + P' + 4Q)z' + (4PQ + 2Q')z = 0}, \] a third-order differential equation satisfied by the product $z$ of any two solutions of the original equation. | 0 |
1941 | 1941_12 | A car is being driven so that its wheels, all of radius $a$ feet, have an angular velocity of $\omega$ radians per second. A particle is thrown off from the tire of one of these wheels, where it is supposed that $a\omega^2 > g$. Neglecting the resistance of the air, find the expression for the maximum height above the roadway which the particle can reach. \] | If a particle is thrown into motion in a gravitational field starting at height $h$ and with upward component of velocity $v$, it will rise to the height $h + \frac{v^2}{2g}$. [The horizontal components of the motion do not influence the maximum height].
As long as the particle remains attached to the tire, it follows the path of a cycloid, and we may take its equations of motion as\[ x = a\omega t - a \sin \omega t \] \[ y = a(1 - \cos \omega t). \]
If the particle leaves the tire when $\omega t = \theta$, then it starts into gravitational motion with height $a(1 - \cos \theta)$ and upward velocity component $y' = a\omega \sin \theta$. Hence it reaches the height\[ H = a(1 - \cos \theta) + \frac{a^2 \omega^2}{2g} \sin^2 \theta \] provided $0 \leq \theta \leq \pi$ (to ensure that the particle starts upward).
We are asked to maximize $H$ by choice of $\theta$. We set\[ \frac{dH}{d\theta} = a \sin \theta + \frac{a^2 \omega^2}{g} \sin \theta \cos \theta = 0 \] and find that the critical points are $0$, $\pi$, $\theta_0$, where $\theta_0 = \arccos(-g/a\omega^2)$. (Since $a\omega^2 > g$, there is such a $\theta_0$.) The corresponding values of $H$ are $0$, $2a$, and\[ H_0 = a \left(1 + \frac{g}{a\omega^2}\right) + \frac{a^2 \omega^2}{2g} \left(1 - \frac{g^2}{a^2 \omega^4}\right) = \boxed{\frac{1}{2g}(a\omega + g\omega^{-1})^2}. \] | algebraic | putnam (modified boxing) | Calculus | A car is being driven so that its wheels, all of radius $a$ feet, have an angular velocity of $\omega$ radians per second. A particle is thrown off from the tire of one of these wheels, where it is supposed that $a\omega^2 > g$. Neglecting the resistance of the air, show that the maximum height above the roadway which the particle can reach is\[ \frac{(a\omega + g\omega^{-1})^2}{2g}. \] | If a particle is thrown into motion in a gravitational field starting at height $h$ and with upward component of velocity $v$, it will rise to the height $h + \frac{v^2}{2g}$. [The horizontal components of the motion do not influence the maximum height].
As long as the particle remains attached to the tire, it follows the path of a cycloid, and we may take its equations of motion as\[ x = a\omega t - a \sin \omega t \] \[ y = a(1 - \cos \omega t). \]
If the particle leaves the tire when $\omega t = \theta$, then it starts into gravitational motion with height $a(1 - \cos \theta)$ and upward velocity component $y' = a\omega \sin \theta$. Hence it reaches the height\[ H = a(1 - \cos \theta) + \frac{a^2 \omega^2}{2g} \sin^2 \theta \] provided $0 \leq \theta \leq \pi$ (to ensure that the particle starts upward).
We are asked to maximize $H$ by choice of $\theta$. We set\[ \frac{dH}{d\theta} = a \sin \theta + \frac{a^2 \omega^2}{g} \sin \theta \cos \theta = 0 \] and find that the critical points are $0$, $\pi$, $\theta_0$, where $\theta_0 = \arccos(-g/a\omega^2)$. (Since $a\omega^2 > g$, there is such a $\theta_0$.) The corresponding values of $H$ are $0$, $2a$, and\[ H_0 = a \left(1 + \frac{g}{a\omega^2}\right) + \frac{a^2 \omega^2}{2g} \left(1 - \frac{g^2}{a^2 \omega^4}\right) = \frac{1}{2g}(a\omega + g\omega^{-1})^2. \] | 0 |
1942 | 1942_2 | If a polynomial $f(x)$ is divided by $(x - a)^2(x - b)$, where $a \neq b$, derive a formula for the remainder. | Since $f(x)$ is divided by a cubic polynomial, the remainder $R(x)$ will be of degree at most two in $x$, say \[ R(x) = Ax^2 + Bx + C. \]
Then\[ f(x) = (x - a)^2(x - b)Q(x) + Ax^2 + Bx + C \] and \[ f'(x) = 2(x - a)(x - b)Q(x) + (x - a)^2Q(x) + (x - a)^2(x - b)Q'(x) + 2Ax + B. \]
From these relations one gets \[ f(a) = Aa^2 + Ba + C \] \[ f(b) = Ab^2 + Bb + C \] \[ f'(a) = 2Aa + B. \]
Solving for $A$, $B$, $C$ one gets \[ A = \frac{1}{(b - a)^2} \big[f(b) - f(a) - (b - a)f'(a)\big], \] \[ B = \frac{-1}{(b - a)^2} \big[2a\big(f(b) - f(a)\big) - (b^2 - a^2)f'(a)\big], \] \[ C = \frac{1}{(b - a)^2} \big[(b - a)^2f(a) + a^2\big(f(b) - f(a)\big) + ab(a - b)f'(a)\big]. \]
Hence \[ R(x) = \frac{1}{(b - a)^2} \Big\{\big[f(b) - f(a) - (b - a)f'(a)\big]x^2 \] \[ - \big[2a(f(b) - f(a)) + (b^2 - a^2)f'(a)\big]x \] \[ + \big[(b - a)^2f(a) + a^2(f(b) - f(a)) + ab(a - b)f'(a)\big]\Big\}. \]
This is easier to check if written in the form \[ R(x) = \boxed{f(a) + \frac{f(b) - f(a)}{(b - a)^2}(x - a)^2 - \frac{f'(a)}{b - a}(x - a)(x - b)}. \] | algebraic | putnam | Algebra | If a polynomial $f(x)$ is divided by $(x - a)^2(x - b)$, where $a \neq b$, derive a formula for the remainder. | Since $f(x)$ is divided by a cubic polynomial, the remainder $R(x)$ will be of degree at most two in $x$, say \[ R(x) = Ax^2 + Bx + C. \]
Then\[ f(x) = (x - a)^2(x - b)Q(x) + Ax^2 + Bx + C \] and \[ f'(x) = 2(x - a)(x - b)Q(x) + (x - a)^2Q(x) + (x - a)^2(x - b)Q'(x) + 2Ax + B. \]
From these relations one gets \[ f(a) = Aa^2 + Ba + C \] \[ f(b) = Ab^2 + Bb + C \] \[ f'(a) = 2Aa + B. \]
Solving for $A$, $B$, $C$ one gets \[ A = \frac{1}{(b - a)^2} \big[f(b) - f(a) - (b - a)f'(a)\big], \] \[ B = \frac{-1}{(b - a)^2} \big[2a\big(f(b) - f(a)\big) - (b^2 - a^2)f'(a)\big], \] \[ C = \frac{1}{(b - a)^2} \big[(b - a)^2f(a) + a^2\big(f(b) - f(a)\big) + ab(a - b)f'(a)\big]. \]
Hence \[ R(x) = \frac{1}{(b - a)^2} \Big\{\big[f(b) - f(a) - (b - a)f'(a)\big]x^2 \] \[ - \big[2a(f(b) - f(a)) + (b^2 - a^2)f'(a)\big]x \] \[ + \big[(b - a)^2f(a) + a^2(f(b) - f(a)) + ab(a - b)f'(a)\big]\Big\}. \]
This is easier to check if written in the form \[ R(x) = \boxed{f(a) + \frac{f(b) - f(a)}{(b - a)^2}(x - a)^2 - \frac{f'(a)}{b - a}(x - a)(x - b)}. \] | 0 |
1942 | 1942_4 | Find the orthogonal trajectories of the family of conics $(x + 2y)^2 = a(x + y)$. At what angle do the curves of one family cut the curves of the other family at the origin? | The given family is a family of parabolas all tangent to the line $x + y = 0$ at the origin. For $a = 0$ the parabola degenerates to the double line $(x + 2y)^2 = 0$ which should be viewed as two degenerate parabolas, the ray in the fourth quadrant being the limiting case as $a$ goes to zero through positive values and the ray in the second quadrant being the limiting parabola as $a$ goes to zero through negative values.
To find the differential equation of the family we differentiate the given equation and eliminate $a$ between the original equation and its derivative.
\[(x + 2y)^2 = a(x + y), \]
\[2(x + 2y)(1 + 2y') = a(1 + y').\]
We get \[2(x + 2y)(1 + 2y') = (x + 2y)^2(1 + y'),\] which simplifies to \[(3x + 2y)y' + x = 0.\]
The factor $x + 2y$ that was cancelled reflects the degeneracy along the line $x + 2y = 0$. This differential equation is defined along the line $x + y = 0$ (where the original family of parabolas has no members), so in effect the line $x + y = 0$ is another degenerate member of the family corresponding to the case $a = \infty$.
The orthogonal trajectories are obtained by integrating the differential equation \[xy' = 3x + 2y.\]
We write this as \[\frac{d}{dx}(y + 3x) = \frac{2(y + 3x)}{x}.\]
The solution is \[y + 3x = kx^2\] where $k$ is an arbitrary constant. This is a new family of parabolas, with a unique member through every point of the plane except for points on the $y$-axis. The $y$-axis is an integral curve of differential equation $(4)$ rewritten in the form \[x = (3x + 2y)\frac{dx}{dy}.\]
All the curves in the new family $(5)$ are tangent to the line $y + 3x = 0$ at the origin (except the degenerate double parabola made by the $y$-axis).
The angle between the two families at the origin is then the angle $\theta$ between the two lines $x + y = 0$ and $3x + y = 0$. Using the slopes we get \[
\tan \theta = \frac{m_1 - m_2}{1 + m_1m_2} = \frac{-1 - (-3)}{1 + (-1)(-3)} = \frac{2}{4} = \frac{1}{2}.
\]
Hence $\theta = \boxed{\arctan(\frac{1}{2})}.$ | numerical | putnam | Algebra Calculus Geometry | Find the orthogonal trajectories of the family of conics $(x + 2y)^2 = a(x + y)$. At what angle do the curves of one family cut the curves of the other family at the origin? | The given family is a family of parabolas all tangent to the line $x + y = 0$ at the origin. For $a = 0$ the parabola degenerates to the double line $(x + 2y)^2 = 0$ which should be viewed as two degenerate parabolas, the ray in the fourth quadrant being the limiting case as $a$ goes to zero through positive values and the ray in the second quadrant being the limiting parabola as $a$ goes to zero through negative values.
To find the differential equation of the family we differentiate the given equation and eliminate $a$ between the original equation and its derivative.
\[(x + 2y)^2 = a(x + y), \]
\[2(x + 2y)(1 + 2y') = a(1 + y').\]
We get \[2(x + 2y)(1 + 2y') = (x + 2y)^2(1 + y'),\] which simplifies to \[(3x + 2y)y' + x = 0.\]
The factor $x + 2y$ that was cancelled reflects the degeneracy along the line $x + 2y = 0$. This differential equation is defined along the line $x + y = 0$ (where the original family of parabolas has no members), so in effect the line $x + y = 0$ is another degenerate member of the family corresponding to the case $a = \infty$.
The orthogonal trajectories are obtained by integrating the differential equation \[xy' = 3x + 2y.\]
We write this as \[\frac{d}{dx}(y + 3x) = \frac{2(y + 3x)}{x}.\]
The solution is \[y + 3x = kx^2\] where $k$ is an arbitrary constant. This is a new family of parabolas, with a unique member through every point of the plane except for points on the $y$-axis. The $y$-axis is an integral curve of differential equation $(4)$ rewritten in the form \[x = (3x + 2y)\frac{dx}{dy}.\]
All the curves in the new family $(5)$ are tangent to the line $y + 3x = 0$ at the origin (except the degenerate double parabola made by the $y$-axis).
The angle between the two families at the origin is then the angle $\theta$ between the two lines $x + y = 0$ and $3x + y = 0$. Using the slopes we get \[
\tan \theta = \frac{m_1 - m_2}{1 + m_1m_2} = \frac{-1 - (-3)}{1 + (-1)(-3)} = \frac{2}{4} = \frac{1}{2}.
\]
Hence $\theta = \boxed{\arctan(\frac{1}{2})}.$ | 0 |
1942 | 1942_5 | A circle of radius $a$ is revolved through $180^\circ$ about a line in its plane, distant $b$ from the center of the circle, where $b > a$. For what value of the ratio $b/a$ does the center of gravity of the solid thus generated lie on the surface of the solid? | We choose axes so that the generating circle starts in the $x$-$z$ plane and is revolved about the $z$-axis. The generated solid is half of a toroid (i.e., a solid bounded by a torus).
It is clear from symmetry that the centroid lies at a point $(0, \bar{y}, 0)$ on the $y$-axis, and the requirement of the problem is that $\bar{y} = b - a$.
To find the centroid we introduce polar coordinates in the $x$-$y$ plane. Corresponding to the element of area $r \, d\theta$ in the plane there is the element of volume
\[
2\sqrt{a^2 - (r - b)^2} \, r \, dr \, d\theta
\]
which contributes
\[
2r \sin \theta \sqrt{a^2 - (r - b)^2} \, r \, dr \, d\theta
\]
to the moment $M_y$ of the solid in the $y$ direction. We have $\bar{y} = M_y/V$ where $V$ is the volume of the semi-toroid.
\[
V = \int_{b-a}^{b+a} 2 \int_0^{\pi} \sqrt{a^2 - (r - b)^2} \, r \, dr \, d\theta \\
= 2\pi \int_{b-a}^{b+a} \sqrt{a^2 - (r - b)^2} \, r \, dr \\
= 2\pi a^2 b.
\]
\[
M_y = \int_{b-a}^{b+a} 2 \int_0^{\pi} \sqrt{a^2 - (r - b)^2} \, r^2 \sin \theta \, dr \, d\theta \\
= \pi a^2 b^2 + \frac{\pi}{2} a^4.
\]
In both integrals we used the substitution $r = b + a \sin \phi$. Hence
\[
\bar{y} = \frac{M_y}{V} = \frac{a^2 + 4b^2}{2\pi b}.
\]
But we require $\bar{y} = b - a$, so
\[
2\pi b^2 - 2\pi ab = a^2 + 4b^2.
\]
If $c = b/a$, then
\[
(2\pi - 4)c^2 - 2\pi c - 1 = 0
\]
and
\[
c = \boxed{\frac{\pi + \sqrt{\pi^2 + 2\pi - 4}}{2\pi - 4}}.
\]
We chose the positive sign since $c$ must be positive. | algebraic | putnam | Geometry Calculus | A circle of radius $a$ is revolved through $180^\circ$ about a line in its plane, distant $b$ from the center of the circle, where $b > a$. For what value of the ratio $b/a$ does the center of gravity of the solid thus generated lie on the surface of the solid? | We choose axes so that the generating circle starts in the $x$-$z$ plane and is revolved about the $z$-axis. The generated solid is half of a toroid (i.e., a solid bounded by a torus).
It is clear from symmetry that the centroid lies at a point $(0, \bar{y}, 0)$ on the $y$-axis, and the requirement of the problem is that $\bar{y} = b - a$.
To find the centroid we introduce polar coordinates in the $x$-$y$ plane. Corresponding to the element of area $r \, d\theta$ in the plane there is the element of volume
\[
2\sqrt{a^2 - (r - b)^2} \, r \, dr \, d\theta
\]
which contributes
\[
2r \sin \theta \sqrt{a^2 - (r - b)^2} \, r \, dr \, d\theta
\]
to the moment $M_y$ of the solid in the $y$ direction. We have $\bar{y} = M_y/V$ where $V$ is the volume of the semi-toroid.
\[
V = \int_{b-a}^{b+a} 2 \int_0^{\pi} \sqrt{a^2 - (r - b)^2} \, r \, dr \, d\theta \\
= 2\pi \int_{b-a}^{b+a} \sqrt{a^2 - (r - b)^2} \, r \, dr \\
= 2\pi a^2 b.
\]
\[
M_y = \int_{b-a}^{b+a} 2 \int_0^{\pi} \sqrt{a^2 - (r - b)^2} \, r^2 \sin \theta \, dr \, d\theta \\
= \pi a^2 b^2 + \frac{\pi}{2} a^4.
\]
In both integrals we used the substitution $r = b + a \sin \phi$. Hence
\[
\bar{y} = \frac{M_y}{V} = \frac{a^2 + 4b^2}{2\pi b}.
\]
But we require $\bar{y} = b - a$, so
\[
2\pi b^2 - 2\pi ab = a^2 + 4b^2.
\]
If $c = b/a$, then
\[
(2\pi - 4)c^2 - 2\pi c - 1 = 0
\]
and
\[
c = \boxed{\frac{\pi + \sqrt{\pi^2 + 2\pi - 4}}{2\pi - 4}}.
\]
We chose the positive sign since $c$ must be positive. | 0 |
1942 | 1942_10 | A particle moves under a central force inversely proportional to the $k$th power of the distance. If the particle describes a circle (the central force proceeding from a point on the circumference of the circle), find $k$. | Choose polar coordinates with pole at the center of force and initial ray a diameter of the circular orbit of the particle. The equation of the orbit is then
\[
r = A \cos \theta,
\]
where $A$ is the diameter of the circle.
The equations of motion are
\[
\frac{d^2r}{dt^2} - r\left(\frac{d\theta}{dt}\right)^2 = -\frac{1}{m}f
\]
\[
r \frac{d^2\theta}{dt^2} + 2 \frac{dr}{dt} \frac{d\theta}{dt} = 0,
\]
where $m$ is the mass of the particle and $f$ is the magnitude of the central force. Since the sign is taken as negative, a positive $f$ means an attractive force.
After multiplication by $r$, equation (3) can be integrated to give
\[
r^2 \frac{d\theta}{dt} = h,
\]
which asserts that the angular momentum of the particle is constant.
Differentiating (1) twice, using (4), we get
\[
\frac{dr}{dt} = -A \sin \theta \frac{d\theta}{dt} = -\frac{Ah \sin \theta}{r^2},
\]
\[
\frac{d^2r}{dt^2} = -\frac{Ah \cos \theta}{r^2} \frac{d\theta}{dt} + \frac{2Ah \sin \theta}{r^3} \frac{dr}{dt}.
\]
Substituting in (2), we find
\[
-\frac{f}{m} = -\frac{h^2}{r^3} - \frac{2A^2h^2 \sin^2 \theta}{r^5},
\]
which simplifies to
\[
f = \frac{2mA^2h^2}{r^5}.
\]
Thus, $f \propto r^{-5}$ and $k = \boxed{5}$. | numerical | putnam | Calculus | A particle moves under a central force inversely proportional to the $k$th power of the distance. If the particle describes a circle (the central force proceeding from a point on the circumference of the circle), find $k$. | Choose polar coordinates with pole at the center of force and initial ray a diameter of the circular orbit of the particle. The equation of the orbit is then
\[
r = A \cos \theta,
\]
where $A$ is the diameter of the circle.
The equations of motion are
\[
\frac{d^2r}{dt^2} - r\left(\frac{d\theta}{dt}\right)^2 = -\frac{1}{m}f
\]
\[
r \frac{d^2\theta}{dt^2} + 2 \frac{dr}{dt} \frac{d\theta}{dt} = 0,
\]
where $m$ is the mass of the particle and $f$ is the magnitude of the central force. Since the sign is taken as negative, a positive $f$ means an attractive force.
After multiplication by $r$, equation (3) can be integrated to give
\[
r^2 \frac{d\theta}{dt} = h,
\]
which asserts that the angular momentum of the particle is constant.
Differentiating (1) twice, using (4), we get
\[
\frac{dr}{dt} = -A \sin \theta \frac{d\theta}{dt} = -\frac{Ah \sin \theta}{r^2},
\]
\[
\frac{d^2r}{dt^2} = -\frac{Ah \cos \theta}{r^2} \frac{d\theta}{dt} + \frac{2Ah \sin \theta}{r^3} \frac{dr}{dt}.
\]
Substituting in (2), we find
\[
-\frac{f}{m} = -\frac{h^2}{r^3} - \frac{2A^2h^2 \sin^2 \theta}{r^5},
\]
which simplifies to
\[
f = \frac{2mA^2h^2}{r^5}.
\]
Thus, $f \propto r^{-5}$ and $k = \boxed{5}$. | 0 |
1946 | 1946_1 | Suppose that the function $f(x) = ax^2 + bx + c$, where $a$, $b$, $c$ are real constants, satisfies the condition $|f(x)| \leq 1$ for $|x| \leq 1$. Find the upper bound of $|f'(x)|$ for $|x| \leq 1$. | If $a \neq 0$, the graph of $y = ax^2 + bx + c$ is a parabola that opens upward, i.e., $a > 0$. Without loss of generality, assume $b \geq 0$. The vertex falls in the left half-plane, and the maximum value of $|f'(x)|$ for $|x| \leq 1$ occurs at $x = 1$. Therefore, $|f'(1)| = 2a + b$.
Now evaluate $f(1)$ and $f(0)$:
\[
f(1) = a + b + c \leq 1, \quad f(0) = c \geq -1.
\]
From these, $a + b \leq 2$, and since $a, b \geq 0$, we also have $a \leq 2$ and $2a + b \leq 4$.
For the case $a = 0$, the function is linear: $f(x) = bx + c$, and $f'(x) = b$. Using the conditions $f(1), f(-1)$:
\[
|f'(x)| = |b| = \frac{|f(1) - f(-1)|}{2} \leq 1.
\]
Thus, $|f'(x)| \leq \boxed{4}$ for all $|x| \leq 1$ as required. | numerical | putnam (modified boxing) | Algebra Analysis | Suppose that the function $f(x) = ax^2 + bx + c$, where $a$, $b$, $c$ are real constants, satisfies the condition $|f(x)| \leq 1$ for $|x| \leq 1$. Prove that $|f'(x)| \leq 4$ for $|x| \leq 1$. | If $a \neq 0$, the graph of $y = ax^2 + bx + c$ is a parabola that can be assumed to open upward, i.e., $a > 0$. Without loss of generality, assume $b \geq 0$. By symmetry, the vertex falls in the left half-plane, and the maximum value of $|f'(x)|$ for $|x| \leq 1$ occurs at $x = 1$. Therefore, $|f'(1)| = 2a + b$.
Now evaluate $f(1)$ and $f(0)$:
\[
f(1) = a + b + c \leq 1, \quad f(0) = c \geq -1.
\]
From these, $a + b \leq 2$, and since $a, b \geq 0$, we also have $a \leq 2$ and $2a + b \leq 4$.
For the case $a = 0$, the function is linear: $f(x) = bx + c$, and $f'(x) = b$. Using the conditions $f(1), f(-1)$:
\[
|f'(x)| = |b| = \frac{|f(1) - f(-1)|}{2} \leq 1.
\]
Thus, $|f'(x)| \leq 4$ for all $|x| \leq 1$ as required. | 0 |
1946 | 1946_5 | Find the smallest volume bounded by the coordinate planes and by a tangent plane to the ellipsoid
\[
\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1.
\] | The tangent plane to the ellipsoid at the point $(x_1, y_1, z_1)$ has the equation
\[
\frac{xx_1}{a^2} + \frac{yy_1}{b^2} + \frac{zz_1}{c^2} = 1.
\]
Its intercepts on the $x$, $y$, and $z$-axes are respectively $\frac{a^2}{x_1}$, $\frac{b^2}{y_1}$, and $\frac{c^2}{z_1}$. The volume of the solid cut off by the tangent plane and the three coordinate planes is
\[
V = \frac{1}{6} \left| \frac{a^2 b^2 c^2}{x_1 y_1 z_1} \right|.
\]
Using the constraint that $(x_1, y_1, z_1)$ is a point on the ellipsoid, we express the volume as:
\[
V^2 = \frac{1}{36} a^2 b^2 c^2 \left( \frac{x_1^2}{a^2} \cdot \frac{y_1^2}{b^2} \cdot \frac{z_1^2}{c^2} \right)^{-1}.
\]
Using the arithmetic-geometric mean inequality:
\[
\left( \frac{x_1^2}{a^2} \cdot \frac{y_1^2}{b^2} \cdot \frac{z_1^2}{c^2} \right)^{1/3} \leq \frac{1}{3} \left( \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} + \frac{z_1^2}{c^2} \right) = \frac{1}{3}.
\]
Equality holds if and only if $\frac{x_1^2}{a^2} = \frac{y_1^2}{b^2} = \frac{z_1^2}{c^2} = \frac{1}{3}$. Thus,
\[
V^2 \geq \frac{27}{36} a^2 b^2 c^2 \quad \text{and} \quad V \geq \boxed{\frac{1}{2} \sqrt{3} abc}.
\]
Equality is achieved when $(x_1, y_1, z_1)$ is one of the eight points for which $\frac{x_1^2}{a^2} = \frac{y_1^2}{b^2} = \frac{z_1^2}{c^2} = \frac{1}{3}$, namely
\[
(\pm \frac{a}{\sqrt{3}}, \pm \frac{b}{\sqrt{3}}, \pm \frac{c}{\sqrt{3}}).
\] | algebraic | putnam | Geometry Analysis | Find the smallest volume bounded by the coordinate planes and by a tangent plane to the ellipsoid
\[
\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1.
\] | The tangent plane to the ellipsoid at the point $(x_1, y_1, z_1)$ has the equation
\[
\frac{xx_1}{a^2} + \frac{yy_1}{b^2} + \frac{zz_1}{c^2} = 1.
\]
Its intercepts on the $x$, $y$, and $z$-axes are respectively $\frac{a^2}{x_1}$, $\frac{b^2}{y_1}$, and $\frac{c^2}{z_1}$. The volume of the solid cut off by the tangent plane and the three coordinate planes is
\[
V = \frac{1}{6} \left| \frac{a^2 b^2 c^2}{x_1 y_1 z_1} \right|.
\]
Using the constraint that $(x_1, y_1, z_1)$ is a point on the ellipsoid, we express the volume as:
\[
V^2 = \frac{1}{36} a^2 b^2 c^2 \left( \frac{x_1^2}{a^2} \cdot \frac{y_1^2}{b^2} \cdot \frac{z_1^2}{c^2} \right)^{-1}.
\]
Using the arithmetic-geometric mean inequality:
\[
\left( \frac{x_1^2}{a^2} \cdot \frac{y_1^2}{b^2} \cdot \frac{z_1^2}{c^2} \right)^{1/3} \leq \frac{1}{3} \left( \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} + \frac{z_1^2}{c^2} \right) = \frac{1}{3}.
\]
Equality holds if and only if $\frac{x_1^2}{a^2} = \frac{y_1^2}{b^2} = \frac{z_1^2}{c^2} = \frac{1}{3}$. Thus,
\[
V^2 \geq \frac{27}{36} a^2 b^2 c^2 \quad \text{and} \quad V \geq \boxed{\frac{1}{2} \sqrt{3} abc}.
\]
Equality is achieved when $(x_1, y_1, z_1)$ is one of the eight points for which $\frac{x_1^2}{a^2} = \frac{y_1^2}{b^2} = \frac{z_1^2}{c^2} = \frac{1}{3}$, namely
\[
(\pm \frac{a}{\sqrt{3}}, \pm \frac{b}{\sqrt{3}}, \pm \frac{c}{\sqrt{3}}).
\] | 0 |
1946 | 1946_7 | Let $K$ denote the circumference of a circular disc of radius one, and let $k$ denote a circular arc that joins two points $a$, $b$ on $K$ and lies otherwise in the given circular disc. Suppose that $k$ divides the circular disc into two parts of equal area. Find the lower bound of the length of $k$. | If $a$ and $b$ were diametrically opposite on $K$, there would exist no circular arc from $a$ to $b$ that bisects $K$. Hence we may choose coordinates such that $K$ is the unit circle $x^2 + y^2 = 1$ and $a$ and $b$ have coordinates $(c, d)$ and $(c, -d)$, respectively, where $c < 0$.
Now the arc $k$ divides the circular disc into two parts of equal area, and hence it must intersect the positive $x$-axis at a point $e$. If $O$ is the origin, we get length $k > 2ae > 2aO = \boxed{2}$. | numerical | putnam (modified boxing) | Geometry | Let $K$ denote the circumference of a circular disc of radius one, and let $k$ denote a circular arc that joins two points $a$, $b$ on $K$ and lies otherwise in the given circular disc. Suppose that $k$ divides the circular disc into two parts of equal area. Prove that the length of $k$ exceeds 2. | If $a$ and $b$ were diametrically opposite on $K$, there would exist no circular arc from $a$ to $b$ that bisects $K$. Hence we may choose coordinates such that $K$ is the unit circle $x^2 + y^2 = 1$ and $a$ and $b$ have coordinates $(c, d)$ and $(c, -d)$, respectively, where $c < 0$.
Now the arc $k$ divides the circular disc into two parts of equal area, and hence it must intersect the positive $x$-axis at a point $e$. If $O$ is the origin, we get length $k > 2ae > 2aO = 2$. | 0 |
1946 | 1946_8 | Let $A$, $B$ be variable points on a parabola $P$, such that the tangents at $A$ and $B$ are perpendicular to each other. Show that the locus of the centroid of the triangle formed by $A$, $B$ and the vertex of $P$ is a parabola $P_1$. Apply the same process to $P_1$, obtaining a parabola $P_2$, and repeat the process, obtaining altogether the sequence of parabolas $P$, $P_1$, $P_2$, \dots, $P_n$. If the equation of $P$ is $y^2 = mx$, find the equation of $P_n$. | Since $P$ is a parabola with equation $y^2 = mx$, any point of $P$ has coordinates of the form $(mt^2, mt)$ for some real $t$, and conversely, every such point is on $P$. The slope of the line tangent to $P$ at $(mt^2, mt)$ is $1/2t$.
Let $A$ and $B$ be the points $(ms^2, ms)$ and $(mt^2, mt)$, respectively. The tangents to $P$ at $A$ and $B$ are perpendicular if and only if $(1/2s)(1/2t) = -1$, i.e., if and only if
\[(1)\quad st = -\frac{1}{4}.\]
The centroid of the points $A$, $B$, and the vertex $(0, 0)$ of $P$ is
\[(2)\quad \left( \frac{1}{3}m(s^2 + t^2), \frac{1}{3}m(s + t) \right)\]
and this centroid lies on a new parabola
\[(3)\quad y^2 = \frac{1}{3}m\left(x - \frac{m}{6}\right)\]
if the perpendicularity condition (1) is satisfied.
Conversely, any point $(x, y)$ on the parabola (3) has the form (2), since the equations
\[\frac{1}{3}m(s + t) = y\]
\[st = -\frac{1}{4}\]
can always be solved to give real $s$ and $t$. Indeed, $s$ and $t$ are zeros of $S^2 - (3y/m)S - \frac{1}{4}$, which has positive discriminant. Hence the locus in question is the entire parabola $P_1$ given by (3).
Now $P_1$ is obtained from $P$ by changing the constant from $m$ to $m/3$, and displacing the vertex to the right by $m/6$. Consequently, $P_2$ can be obtained from $P_1$ by changing the constant from $m/3$ to $(m/3)/3$ and displacing the vertex $(m/3)/6$ further to the right. The equation of $P_2$ is therefore
\[y^2 = \frac{1}{9}m\left(x - \frac{1}{6}m - \frac{1}{18}m\right).\]
Continuing this reasoning, we see that $P_n$ has the equation
\[y^2 = \frac{1}{3^n}m\left(x - \frac{m}{6} - \frac{1}{6\cdot3}m - \frac{1}{6\cdot3^2}m - \cdots - \frac{1}{6\cdot3^{n-1}}m\right)\]
\[= \boxed{\frac{1}{3^n}m\left(x - \frac{m}{4}\left(1 - \frac{1}{3^n}\right)\right)}.\] | algebraic | putnam | Geometry Algebra | Let $A$, $B$ be variable points on a parabola $P$, such that the tangents at $A$ and $B$ are perpendicular to each other. Show that the locus of the centroid of the triangle formed by $A$, $B$ and the vertex of $P$ is a parabola $P_1$. Apply the same process to $P_1$, obtaining a parabola $P_2$, and repeat the process, obtaining altogether the sequence of parabolas $P$, $P_1$, $P_2$, \dots, $P_n$. If the equation of $P$ is $y^2 = mx$, find the equation of $P_n$. | Since $P$ is a parabola with equation $y^2 = mx$, any point of $P$ has coordinates of the form $(mt^2, mt)$ for some real $t$, and conversely, every such point is on $P$. The slope of the line tangent to $P$ at $(mt^2, mt)$ is $1/2t$.
Let $A$ and $B$ be the points $(ms^2, ms)$ and $(mt^2, mt)$, respectively. The tangents to $P$ at $A$ and $B$ are perpendicular if and only if $(1/2s)(1/2t) = -1$, i.e., if and only if
\[(1)\quad st = -\frac{1}{4}.\]
The centroid of the points $A$, $B$, and the vertex $(0, 0)$ of $P$ is
\[(2)\quad \left( \frac{1}{3}m(s^2 + t^2), \frac{1}{3}m(s + t) \right)\]
and this centroid lies on a new parabola
\[(3)\quad y^2 = \frac{1}{3}m\left(x - \frac{m}{6}\right)\]
if the perpendicularity condition (1) is satisfied.
Conversely, any point $(x, y)$ on the parabola (3) has the form (2), since the equations
\[\frac{1}{3}m(s + t) = y\]
\[st = -\frac{1}{4}\]
can always be solved to give real $s$ and $t$. Indeed, $s$ and $t$ are zeros of $S^2 - (3y/m)S - \frac{1}{4}$, which has positive discriminant. Hence the locus in question is the entire parabola $P_1$ given by (3).
Now $P_1$ is obtained from $P$ by changing the constant from $m$ to $m/3$, and displacing the vertex to the right by $m/6$. Consequently, $P_2$ can be obtained from $P_1$ by changing the constant from $m/3$ to $(m/3)/3$ and displacing the vertex $(m/3)/6$ further to the right. The equation of $P_2$ is therefore
\[y^2 = \frac{1}{9}m\left(x - \frac{1}{6}m - \frac{1}{18}m\right).\]
Continuing this reasoning, we see that $P_n$ has the equation
\[y^2 = \frac{1}{3^n}m\left(x - \frac{m}{6} - \frac{1}{6\cdot3}m - \frac{1}{6\cdot3^2}m - \cdots - \frac{1}{6\cdot3^{n-1}}m\right)\]
\[= \boxed{\frac{1}{3^n}m\left(x - \frac{m}{4}\left(1 - \frac{1}{3^n}\right)\right)}.\] | 0 |
1946 | 1946_9 | In a solid sphere of radius $R$ the density $\rho$ is a function of $r$, the distance from the center of the sphere. If the magnitude of the gravitational force of attraction due to the sphere at any point inside the sphere is $kr^2$, where $k$ is a constant, find $\rho$ as a function of $r$. Find also the magnitude of the force of attraction at a point outside the sphere at a distance $r$ from the center and return this as the final answer. (Assume that the magnitude of the force of attraction at a point $P$ due to a thin spherical shell is zero if $P$ is inside the shell, and is $m/r^2$ if $P$ is outside the shell, $m$ being the mass of the shell, and $r$ the distance of $P$ from the center.) | Let $P$ be a point at distance $r$ from the center. We regard the solid sphere as the union of many thin concentric spherical shells. The force of attraction at $P$ is the sum of the forces of attraction due to the shells. All of these forces act in the same direction so we can simply add their magnitudes.
The shell at distance $s$ from the center and thickness $\Delta s$ has (approximately) volume $4\pi s^2 \Delta s$ and mass $4\pi \rho(s)s^2 \Delta s$, so the force of attraction at $P$ due to this shell is approximately
\[\frac{4\pi \rho(s)s^2 \Delta s}{r^2},\]
if $r > s$. It is zero if $r < s$. The total force of attraction at $P$ is therefore (exactly)
\[(1)\quad F = \begin{cases} \frac{\int_0^r 4\pi \rho(s)s^2 ds}{r^2} & \text{if } r \leq R, \\
\frac{\int_0^R 4\pi \rho(s)s^2 ds}{r^2} & \text{if } r > R. \end{cases}\]
Since we are given that $F = kr^2$ for $r \leq R$, we have
\[kr^2 = \frac{\int_0^r 4\pi \rho(s)s^2 ds}{r^2} \quad \text{for } r \leq R,\]
and we must solve for $\rho$. Multiplying by $r^2$, and then differentiating with respect to $r$, we obtain
\[4kr^3 = 4\pi \rho(r)r^2.\]
Therefore
\[\rho(r) = \frac{k}{\pi}r\]
is the required formula for $\rho$.
Substituting this in (1), we have
\[F = \frac{4\pi}{r^2}\int_0^R \frac{ks}{\pi}s^2 ds = \boxed{\frac{kR^4}{r^2}} \quad \text{for } r > R.\] | algebraic | putnam | Calculus Differential Equations | In a solid sphere of radius $R$ the density $\rho$ is a function of $r$, the distance from the center of the sphere. If the magnitude of the gravitational force of attraction due to the sphere at any point inside the sphere is $kr^2$, where $k$ is a constant, find $\rho$ as a function of $r$. Find also the magnitude of the force of attraction at a point outside the sphere at a distance $r$ from the center. (Assume that the magnitude of the force of attraction at a point $P$ due to a thin spherical shell is zero if $P$ is inside the shell, and is $m/r^2$ if $P$ is outside the shell, $m$ being the mass of the shell, and $r$ the distance of $P$ from the center.) | Let $P$ be a point at distance $r$ from the center. We regard the solid sphere as the union of many thin concentric spherical shells. The force of attraction at $P$ is the sum of the forces of attraction due to the shells. All of these forces act in the same direction so we can simply add their magnitudes.
The shell at distance $s$ from the center and thickness $\Delta s$ has (approximately) volume $4\pi s^2 \Delta s$ and mass $4\pi \rho(s)s^2 \Delta s$, so the force of attraction at $P$ due to this shell is approximately
\[\frac{4\pi \rho(s)s^2 \Delta s}{r^2},\]
if $r > s$. It is zero if $r < s$. The total force of attraction at $P$ is therefore (exactly)
\[(1)\quad F = \begin{cases} \frac{\int_0^r 4\pi \rho(s)s^2 ds}{r^2} & \text{if } r \leq R, \\
\frac{\int_0^R 4\pi \rho(s)s^2 ds}{r^2} & \text{if } r > R. \end{cases}\]
Since we are given that $F = kr^2$ for $r \leq R$, we have
\[kr^2 = \frac{\int_0^r 4\pi \rho(s)s^2 ds}{r^2} \quad \text{for } r \leq R,\]
and we must solve for $\rho$. Multiplying by $r^2$, and then differentiating with respect to $r$, we obtain
\[4kr^3 = 4\pi \rho(r)r^2.\]
Therefore
\[\rho(r) = \frac{k}{\pi}r\]
is the required formula for $\rho$.
Substituting this in (1), we have
\[F = \frac{4\pi}{r^2}\int_0^R \frac{ks}{\pi}s^2 ds = \boxed{\frac{kR^4}{r^2}} \quad \text{for } r > R.\] | 0 |
1947 | 1947_1 | If $\{a_n\}$ is a sequence of numbers such that for $n \geq 1$
\[ (2 - a_n)a_{n+1} = 1, \]
prove that $\lim a_n$, as $n \to \infty$, exists and find its value. | We begin by describing a graphical method of great utility in analyzing recursions of the form $a_{n+1} = f(a_n)$.
Draw the graph of $f$ and the line $y = x$ on the same axes. (In this case $f(x) = 1/(2 - x)$.) Then start from the point $(a_1, a_1)$ on the line and move up or down to the point $(a_1, a_2)$ on the graph. Then move horizontally to the point $(a_2, a_2)$ on the line, then vertically to $(a_2, a_3)$, etc.
Connect the successive points to form a polygonal line. If the sequence $\{a_n\}$ is convergent with limit $L$, then the polygonal line must converge to $(L, L)$, and if $L$ is a point of continuity for $f$, then $f(L) = L$. Often it is possible to see at a glance how the sequence behaves. In this case, for example, it is clear that with any start (as long as we do not encounter the point 2 where $f$ is undefined) the polygon must eventually reach the region below and to the left of $(1, 1)$, after which it must work its way up toward the point $(1, 1)$.
To make this precise, we prove first
\[ x \leq \frac{1}{2 - x} \]
for $x < 2$, which follows immediately from
\[ \frac{1}{2 - x} - x = \frac{(1 - x)^2}{2 - x}. \]
Also, if $x \leq 1$, then $1/(2 - x) \leq 1$.
Suppose $1 < a_1 \leq \frac{3}{2}$. We claim that for some $n$, $a_n \geq \frac{3}{2}$. For if not, then
\[ 1 \leq a_1 \leq a_2 \leq \cdots \leq \frac{3}{2} \]
by (1), so $\{a_n\}$ is convergent, say to $L$; then $1 \leq L \leq \frac{3}{2}$ and $f(L) = L$; but there is no such $L$. If $a_n \geq \frac{3}{2}$, then $a_{n+2} \leq 1$. (Note. We need not worry about the possibility that $a_k = 2$ for some $k$ since we are given that the sequence does satisfy the recursion.) So in any event there is an index $p$ such that $a_p \leq 1$. Then we have
\[ a_p \leq a_{p+1} \leq a_{p+2} \leq \cdots \leq 1. \]
Hence the sequence converges to a number $M$ and $f(M) = M$. This implies $M = \boxed{1}$. | numerical | putnam (modified boxing) | Analysis Calculus | If $\{a_n\}$ is a sequence of numbers such that for $n \geq 1$
\[ (2 - a_n)a_{n+1} = 1, \]
prove that $\lim a_n$, as $n \to \infty$, exists and is equal to one. | We begin by describing a graphical method of great utility in analyzing recursions of the form $a_{n+1} = f(a_n)$.
Draw the graph of $f$ and the line $y = x$ on the same axes. (In this case $f(x) = 1/(2 - x)$.) Then start from the point $(a_1, a_1)$ on the line and move up or down to the point $(a_1, a_2)$ on the graph. Then move horizontally to the point $(a_2, a_2)$ on the line, then vertically to $(a_2, a_3)$, etc.
Connect the successive points to form a polygonal line. If the sequence $\{a_n\}$ is convergent with limit $L$, then the polygonal line must converge to $(L, L)$, and if $L$ is a point of continuity for $f$, then $f(L) = L$. Often it is possible to see at a glance how the sequence behaves. In this case, for example, it is clear that with any start (as long as we do not encounter the point 2 where $f$ is undefined) the polygon must eventually reach the region below and to the left of $(1, 1)$, after which it must work its way up toward the point $(1, 1)$.
To make this precise, we prove first
\[ x \leq \frac{1}{2 - x} \]
for $x < 2$, which follows immediately from
\[ \frac{1}{2 - x} - x = \frac{(1 - x)^2}{2 - x}. \]
Also, if $x \leq 1$, then $1/(2 - x) \leq 1$.
Suppose $1 < a_1 \leq \frac{3}{2}$. We claim that for some $n$, $a_n \geq \frac{3}{2}$. For if not, then
\[ 1 \leq a_1 \leq a_2 \leq \cdots \leq \frac{3}{2} \]
by (1), so $\{a_n\}$ is convergent, say to $L$; then $1 \leq L \leq \frac{3}{2}$ and $f(L) = L$; but there is no such $L$. If $a_n \geq \frac{3}{2}$, then $a_{n+2} \leq 1$. (Note. We need not worry about the possibility that $a_k = 2$ for some $k$ since we are given that the sequence does satisfy the recursion.) So in any event there is an index $p$ such that $a_p \leq 1$. Then we have
\[ a_p \leq a_{p+1} \leq a_{p+2} \leq \cdots \leq 1. \]
Hence the sequence converges to a number $M$ and $f(M) = M$. This implies $M = 1$. | 0 |
1947 | 1947_2 | A real valued continuous function satisfies for all real $x$ and $y$ the functional equation
\[ f(\sqrt{x^2 + y^2}) = f(x)f(y). \]
Find the value of f(x) in terms of f(1). \] | A slight qualification in the statement of the problem is needed since the real valued continuous function $f(x) \equiv 0$ satisfies the functional equation for all real $x$ and $y$, but does not satisfy the relation $f(0) = [f(1)]^0$, since $0^0$ is undefined.
Assume then that for some $y_0$, $f(y_0) \neq 0$. Since
\[ f(x)f(y_0) = f(\sqrt{x^2 + y_0^2}) = f(-x)f(y_0), \]
we have $f(x) = f(-x) = f(|x|)$ for all $x$. We now show by induction that for any positive integer $n$ and any real number $x$, we have
\[ f(\sqrt{n}x) = [f(x)]^n. \]
This is certainly true for $n = 1$, and assuming it true for $n = k$ we have
\[ f(\sqrt{k+1}x) = f(\sqrt{k+1}|x|) = f(\sqrt{(\sqrt{k}x)^2 + x^2}) = f(\sqrt{k}x)f(x) \]
\[ = [f(x)]^k f(x) = [f(x)]^{k+1}. \]
Therefore (1) is true for all positive integers $n$.
If $p$ and $q$ are non-zero integers, then
\[ f(p) = f(|p|) = f(\sqrt{p^2 \cdot 1}) = [f(1)]^{p^2} \]
and
\[ f(|p|) = f\left(\sqrt{\frac{q^2}{q^2}|p|^2}\right) = \left[f\left(\frac{|p|}{q}\right)\right]^{q^2}. \]
From these two relations it follows that
\[ \left[f\left(\frac{p}{q}\right)\right]^{q^2} = [f(1)]^{p^2}. \]
If $f(1) > 0$ it then follows that
\[ \left[f\left(\frac{p}{q}\right)\right] = [f(1)]^{p^2/q^2}; \]
that is, the required equation is valid for all rational values of $x$ except, perhaps, $x = 0$. By continuity it follows for all values of $x$.
If $f(1) = 0$, then (2) implies that $f(\frac{p}{q}) = 0$ for all non-zero integers $p$ and $q$, and thus $f(x) = 0$ for all rational $x$, hence for all real $x$.
Finally we show that $f(1) < 0$ is impossible. If $p$ is even and $q$ is odd, equation (2) implies that $f(\frac{p}{q}) > 0$. Hence $f(x) > 0$ for a dense set of $x$, and therefore $f(x) \geq 0$ for all $x$; in particular, $f(1) \geq 0$. This makes the final answer \boxed{f(1)^{x^2}}. | algebraic | putnam (modified boxing) | Analysis Algebra | A real valued continuous function satisfies for all real $x$ and $y$ the functional equation
\[ f(\sqrt{x^2 + y^2}) = f(x)f(y). \]
Prove that
\[ f(x) = [f(1)]^{x^2}. \] | A slight qualification in the statement of the problem is needed since the real valued continuous function $f(x) \equiv 0$ satisfies the functional equation for all real $x$ and $y$, but does not satisfy the relation $f(0) = [f(1)]^0$, since $0^0$ is undefined.
Assume then that for some $y_0$, $f(y_0) \neq 0$. Since
\[ f(x)f(y_0) = f(\sqrt{x^2 + y_0^2}) = f(-x)f(y_0), \]
we have $f(x) = f(-x) = f(|x|)$ for all $x$. We now show by induction that for any positive integer $n$ and any real number $x$, we have
\[ f(\sqrt{n}x) = [f(x)]^n. \]
This is certainly true for $n = 1$, and assuming it true for $n = k$ we have
\[ f(\sqrt{k+1}x) = f(\sqrt{k+1}|x|) = f(\sqrt{(\sqrt{k}x)^2 + x^2}) = f(\sqrt{k}x)f(x) \]
\[ = [f(x)]^k f(x) = [f(x)]^{k+1}. \]
Therefore (1) is true for all positive integers $n$.
If $p$ and $q$ are non-zero integers, then
\[ f(p) = f(|p|) = f(\sqrt{p^2 \cdot 1}) = [f(1)]^{p^2} \]
and
\[ f(|p|) = f\left(\sqrt{\frac{q^2}{q^2}|p|^2}\right) = \left[f\left(\frac{|p|}{q}\right)\right]^{q^2}. \]
From these two relations it follows that
\[ \left[f\left(\frac{p}{q}\right)\right]^{q^2} = [f(1)]^{p^2}. \]
If $f(1) > 0$ it then follows that
\[ \left[f\left(\frac{p}{q}\right)\right] = [f(1)]^{p^2/q^2}; \]
that is, the required equation is valid for all rational values of $x$ except, perhaps, $x = 0$. By continuity it follows for all values of $x$.
If $f(1) = 0$, then (2) implies that $f(\frac{p}{q}) = 0$ for all non-zero integers $p$ and $q$, and thus $f(x) = 0$ for all rational $x$, hence for all real $x$.
Finally we show that $f(1) < 0$ is impossible. If $p$ is even and $q$ is odd, equation (2) implies that $f(\frac{p}{q}) > 0$. Hence $f(x) > 0$ for a dense set of $x$, and therefore $f(x) \geq 0$ for all $x$; in particular, $f(1) \geq 0$. | 0 |
1947 | 1947_5 | $a_1, b_1, c_1$ are positive numbers whose sum is 1, and for $n = 1, 2, \dots$ we define
\[ a_{n+1} = a_n^2 + 2b_nc_n, \quad b_{n+1} = b_n^2 + 2c_na_n, \quad c_{n+1} = c_n^2 + 2a_nb_n. \]
Show that $a_n, b_n, c_n$ approach limits as $n \to \infty$ and find these limits. | First note that
\[ a_{n+1} + b_{n+1} + c_{n+1} = (a_n + b_n + c_n)^2 \]
so $a_k + b_k + c_k = 1$ for all $k$ by induction. Also it is clear that the $a_n$'s, $b_n$'s, and $c_n$'s are all positive.
Define $E_n = \max (a_n, b_n, c_n)$ and $F_n = \min (a_n, b_n, c_n)$. We will show
\[ F_1 \leq F_2 \leq F_3 \leq \cdots \leq F_n \leq F_{n+1} \leq \cdots \]
\[ \leq E_{n+1} \leq E_n \leq \cdots \leq E_3 \leq E_2 \leq E_1 \]
and also that
\[ \lim_{n \to \infty} (E_n - F_n) = 0. \]
It follows from (1) and (2) that $E_n$ decreases weakly to some limit $L$ and that $F_n$ increases weakly to the same limit $L$. Since $F_n \leq a_n \leq E_n$, this implies $a_n \to L$. Similarly $b_n \to L$ and $c_n \to L$. Then since $a_n + b_n + c_n = 1$, we have $L = \frac{1}{3}$.
To prove (1), assume $a_n \geq b_n \geq c_n$ for some value of $n$, then
\[ a_{n+1} = a_n^2 + b_nc_n + b_nc_n, \]
\[ b_{n+1} = a_nc_n + b_n^2 + a_nc_n, \]
\[ c_{n+1} = a_nb_n + a_nb_n + c_n^2. \]
In all equations in (3), the right member is less than or equal to $a_n^2 + a_nb_n + a_nc_n = a_n$, and greater than or equal to $a_nc_n + b_nc_n + c_n^2 = c_n$. Hence $E_{n+1} \leq E_n$ and $F_{n+1} \geq F_n$, which proves (1).
To prove (2), we again assume $a_n \geq b_n \geq c_n$ for some $n$. Set
\[ a_n - b_n = \alpha \geq 0 \]
\[ b_n - c_n = \beta \geq 0 \]
\[ a_n - c_n = \delta = \alpha + \beta \geq 0. \]
Then
\[ |a_{n+1} - b_{n+1}| = |a_n - b_n| |a_n + b_n - 2c_n| = \alpha(\delta + \beta) = (\delta - \beta)(\delta + \beta) \leq \delta^2 \]
\[ |a_{n+1} - c_{n+1}| = |a_n - c_n| |a_n + c_n - 2b_n| = \delta |\alpha - \beta| \leq \delta(\alpha + \beta) = \delta^2 \]
\[ |c_{n+1} - b_{n+1}| = |b_n - c_n| |2a_n - b_n - c_n| = \beta(\alpha + \delta) = (\delta - \alpha)(\delta + \alpha) \leq \delta^2. \]
This set of inequalities shows that
\[ E_{n+1} - F_{n+1} \leq (E_n - F_n)^2 \]
for all $n$. Therefore
\[ E_{n+1} - F_{n+1} \leq (E_1 - F_1)2^n \]
for all $n$.
Since we are given that $E_1 < 1$ and $F_1 > 0$, we have $E_1 - F_1 < 1$, and (2) follows. This concludes the proof. The final answer is \boxed{\frac{1}{3}}. | numerical | putnam | Algebra Analysis | $a_1, b_1, c_1$ are positive numbers whose sum is 1, and for $n = 1, 2, \dots$ we define
\[ a_{n+1} = a_n^2 + 2b_nc_n, \quad b_{n+1} = b_n^2 + 2c_na_n, \quad c_{n+1} = c_n^2 + 2a_nb_n. \]
Show that $a_n, b_n, c_n$ approach limits as $n \to \infty$ and find these limits. | First note that
\[ a_{n+1} + b_{n+1} + c_{n+1} = (a_n + b_n + c_n)^2 \]
so $a_k + b_k + c_k = 1$ for all $k$ by induction. Also it is clear that the $a_n$'s, $b_n$'s, and $c_n$'s are all positive.
Define $E_n = \max (a_n, b_n, c_n)$ and $F_n = \min (a_n, b_n, c_n)$. We will show
\[ F_1 \leq F_2 \leq F_3 \leq \cdots \leq F_n \leq F_{n+1} \leq \cdots \]
\[ \leq E_{n+1} \leq E_n \leq \cdots \leq E_3 \leq E_2 \leq E_1 \]
and also that
\[ \lim_{n \to \infty} (E_n - F_n) = 0. \]
It follows from (1) and (2) that $E_n$ decreases weakly to some limit $L$ and that $F_n$ increases weakly to the same limit $L$. Since $F_n \leq a_n \leq E_n$, this implies $a_n \to L$. Similarly $b_n \to L$ and $c_n \to L$. Then since $a_n + b_n + c_n = 1$, we have $L = \boxed{\frac{1}{3}}$.
To prove (1), assume $a_n \geq b_n \geq c_n$ for some value of $n$, then
\[ a_{n+1} = a_n^2 + b_nc_n + b_nc_n, \]
\[ b_{n+1} = a_nc_n + b_n^2 + a_nc_n, \]
\[ c_{n+1} = a_nb_n + a_nb_n + c_n^2. \]
In all equations in (3), the right member is less than or equal to $a_n^2 + a_nb_n + a_nc_n = a_n$, and greater than or equal to $a_nc_n + b_nc_n + c_n^2 = c_n$. Hence $E_{n+1} \leq E_n$ and $F_{n+1} \geq F_n$, which proves (1).
To prove (2), we again assume $a_n \geq b_n \geq c_n$ for some $n$. Set
\[ a_n - b_n = \alpha \geq 0 \]
\[ b_n - c_n = \beta \geq 0 \]
\[ a_n - c_n = \delta = \alpha + \beta \geq 0. \]
Then
\[ |a_{n+1} - b_{n+1}| = |a_n - b_n| |a_n + b_n - 2c_n| = \alpha(\delta + \beta) = (\delta - \beta)(\delta + \beta) \leq \delta^2 \]
\[ |a_{n+1} - c_{n+1}| = |a_n - c_n| |a_n + c_n - 2b_n| = \delta |\alpha - \beta| \leq \delta(\alpha + \beta) = \delta^2 \]
\[ |c_{n+1} - b_{n+1}| = |b_n - c_n| |2a_n - b_n - c_n| = \beta(\alpha + \delta) = (\delta - \alpha)(\delta + \alpha) \leq \delta^2. \]
This set of inequalities shows that
\[ E_{n+1} - F_{n+1} \leq (E_n - F_n)^2 \]
for all $n$. Therefore
\[ E_{n+1} - F_{n+1} \leq (E_1 - F_1)2^n \]
for all $n$.
Since we are given that $E_1 < 1$ and $F_1 > 0$, we have $E_1 - F_1 < 1$, and (2) follows. This concludes the proof. | 0 |
1947 | 1947_6 | A three-by-three matrix has determinant zero, and has the further property that the cofactor of any element is equal to the square of that element. (The cofactor of $a_{ij}$ is $(-1)^{i+j}$ multiplied by the determinant obtained by striking out the $i$th row and $j$th column.) Find the maximum value across each element in the matrix. | Let $A$ be an $n \times n$ matrix ($n > 1$), and let $B$ be the transpose of the matrix of its cofactors. Classically $B$ is called the adjoint of $A$. Then
\[ AB = BA = (\det A)I, \]
where $I$ is the identity matrix. Furthermore, the adjoint of $B$ is $(\det A)^{n-2}A$. (For the case $n = 3$, this can be verified quite directly.) Assuming that $n > 2$, this implies that if $A$ is singular, then $B$ has rank at most $n - 2$; that is, all $(n-1) \times (n-1)$ minors of $B$ are zero.
In the present problem, let
\[ A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}. \]
Then the conditions imply that
\[ B = \begin{pmatrix} a^2 & d^2 & g^2 \\ b^2 & e^2 & h^2 \\ c^2 & f^2 & i^2 \end{pmatrix}. \]
We shall show that at least one entry of $B$ is zero. We know that all $2 \times 2$ minors of $B$ are zero; in particular
\[ a^2e^2 - b^2d^2 = 0 \\
b^2f^2 - c^2e^2 = 0 \\
c^2d^2 - a^2f^2 = 0. \]
These equations imply
\[ ae = \pm bd \\
bf = \pm ce \\
cd = \pm af. \]
If the minus sign is correct in all of these equations, then multiplying them all together, we find $abcdef = -abcdef$ and conclude that one of $a, b, c, d, e, f$ is zero, so $B$ has a zero entry. On the other hand, if the plus sign is correct in at least one of the equations (1), then a $2 \times 2$ minor of $A$ is zero, and again $B$ has a zero entry.
Any matrix of rank zero or one which has at least one entry zero has either a whole row or a whole column of zeros. If the column of the zero entry is not all zero, then this column spans the column space, and every other column is a multiple of it; so a whole row of zeros must appear. Thus we conclude that $B$ has either a whole row or a whole column of zeros. Correspondingly, $A$ has either a whole column or a whole row of zeros. Then the cofactors of all other entries of $A$ are zeros, and this shows that all the rest of $B$ is zeros. Hence $A = \boxed{0}$, as required. | numerical | putnam (modified boxing) | Linear Algebra | A three-by-three matrix has determinant zero, and has the further property that the cofactor of any element is equal to the square of that element. (The cofactor of $a_{ij}$ is $(-1)^{i+j}$ multiplied by the determinant obtained by striking out the $i$th row and $j$th column.) Show that every element in the matrix is zero. | Let $A$ be an $n \times n$ matrix ($n > 1$), and let $B$ be the transpose of the matrix of its cofactors. Classically $B$ is called the adjoint of $A$. Then
\[ AB = BA = (\det A)I, \]
where $I$ is the identity matrix. Furthermore, the adjoint of $B$ is $(\det A)^{n-2}A$. (For the case $n = 3$, this can be verified quite directly.) Assuming that $n > 2$, this implies that if $A$ is singular, then $B$ has rank at most $n - 2$; that is, all $(n-1) \times (n-1)$ minors of $B$ are zero.
In the present problem, let
\[ A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}. \]
Then the conditions imply that
\[ B = \begin{pmatrix} a^2 & d^2 & g^2 \\ b^2 & e^2 & h^2 \\ c^2 & f^2 & i^2 \end{pmatrix}. \]
We shall show that at least one entry of $B$ is zero. We know that all $2 \times 2$ minors of $B$ are zero; in particular
\[ a^2e^2 - b^2d^2 = 0 \\
b^2f^2 - c^2e^2 = 0 \\
c^2d^2 - a^2f^2 = 0. \]
These equations imply
\[ ae = \pm bd \\
bf = \pm ce \\
cd = \pm af. \]
If the minus sign is correct in all of these equations, then multiplying them all together, we find $abcdef = -abcdef$ and conclude that one of $a, b, c, d, e, f$ is zero, so $B$ has a zero entry. On the other hand, if the plus sign is correct in at least one of the equations (1), then a $2 \times 2$ minor of $A$ is zero, and again $B$ has a zero entry.
Any matrix of rank zero or one which has at least one entry zero has either a whole row or a whole column of zeros. If the column of the zero entry is not all zero, then this column spans the column space, and every other column is a multiple of it; so a whole row of zeros must appear. Thus we conclude that $B$ has either a whole row or a whole column of zeros. Correspondingly, $A$ has either a whole column or a whole row of zeros. Then the cofactors of all other entries of $A$ are zeros, and this shows that all the rest of $B$ is zeros. Hence $A = 0$, as required. | 0 |
1947 | 1947_7 | Let $f(x)$ be a function such that $f(1) = 1$ and for $x \geq 1$
\[ f'(x) = \frac{1}{x^2 + f^2(x)}. \]
Prove that
\[ \lim_{x \to \infty} f(x) \]
exists and find its upper bound. | Since $f'$ is everywhere positive, $f$ is strictly increasing and therefore
\[ f(t) > f(1) = 1 \quad \text{for } t > 1. \]
Therefore
\[ f'(t) = \frac{1}{t^2 + f^2(t)} < \frac{1}{t^2 + 1} \quad \text{for } t > 1. \]
So
\[ f(x) = 1 + \int_1^x f'(t) dt \]
\[ < 1 + \int_1^x \frac{1}{1 + t^2} dt < 1 + \int_1^\infty \frac{dt}{1 + t^2} = 1 + \pi/4. \]
Since $f$ is increasing and bounded,
\[ \lim_{x \to \infty} f(x) \]
exists and is at most $1 + \pi/4$. Strict inequality also follows from (1) because
\[ \lim_{x \to \infty} f(x) = 1 + \int_1^\infty f'(t) dt < 1 + \int_1^\infty \frac{1}{1 + t^2} dt = \boxed{1 + \pi/4}. \] | numerical | putnam (modified boxing) | Calculus Analysis | Let $f(x)$ be a function such that $f(1) = 1$ and for $x \geq 1$
\[ f'(x) = \frac{1}{x^2 + f^2(x)}. \]
Prove that
\[ \lim_{x \to \infty} f(x) \]
exists and is less than $1 + \pi/4$. | Since $f'$ is everywhere positive, $f$ is strictly increasing and therefore
\[ f(t) > f(1) = 1 \quad \text{for } t > 1. \]
Therefore
\[ f'(t) = \frac{1}{t^2 + f^2(t)} < \frac{1}{t^2 + 1} \quad \text{for } t > 1. \]
So
\[ f(x) = 1 + \int_1^x f'(t) dt \]
\[ < 1 + \int_1^x \frac{1}{1 + t^2} dt < 1 + \int_1^\infty \frac{dt}{1 + t^2} = 1 + \pi/4. \]
Since $f$ is increasing and bounded,
\[ \lim_{x \to \infty} f(x) \]
exists and is at most $1 + \pi/4$. Strict inequality also follows from (1) because
\[ \lim_{x \to \infty} f(x) = 1 + \int_1^\infty f'(t) dt < 1 + \int_1^\infty \frac{1}{1 + t^2} dt = 1 + \pi/4. \] | 0 |
1947 | 1947_8 | Let $f(x)$ be a differentiable function defined in the closed interval $(0, 1)$ and such that
\[ |f'(x)| \leq M, \quad 0 < x < 1. \]
Find the upper bound of
\[ \left| \int_0^1 f(x) \, dx - \frac{1}{n} \sum_{k=1}^n f\left(\frac{k}{n}\right) \right| in terms of M. \] | Let
\[ E_k = \int_{(k-1)/n}^{k/n} f(x)dx - \frac{1}{n} f\left(\frac{k}{n}\right) \]
for $k = 1, 2, \ldots, n$. Since $f$ is differentiable, it is continuous, and therefore by the mean value theorem for integrals there exists a number $\eta_k$ such that
\[ \frac{k-1}{n} < \eta_k < \frac{k}{n} \]
and
\[ \int_{(k-1)/n}^{k/n} f(x)dx = \frac{1}{n} f(\eta_k). \]
By the mean value theorem for derivatives there exists a number $\xi_k$ such that $\eta_k < \xi_k < (k/n)$ and
\[ f(\eta_k) - f\left(\frac{k}{n}\right) = \left(\eta_k - \frac{k}{n}\right) f'(\xi_k). \]
Then
\[ |E_k| = \frac{1}{n} \left| f(\eta_k) - f\left(\frac{k}{n}\right) \right| \]
\[ = \frac{1}{n} \left| \eta_k - \frac{k}{n} \right| \cdot |f'(\xi_k)| \leq \frac{1}{n^2} M. \]
Hence
\[ \left| \int_0^1 f(x)dx - \frac{1}{n} \sum_{k=1}^n f\left(\frac{k}{n}\right) \right| = \left| \sum_{k=1}^n E_k \right| \leq \sum_{k=1}^n |E_k| \leq \boxed{\frac{M}{n}}. \] | algebraic | putnam (modified boxing) | Calculus Analysis | Let $f(x)$ be a differentiable function defined in the closed interval $(0, 1)$ and such that
\[ |f'(x)| \leq M, \quad 0 < x < 1. \]
Prove that
\[ \left| \int_0^1 f(x) \, dx - \frac{1}{n} \sum_{k=1}^n f\left(\frac{k}{n}\right) \right| \leq \frac{M}{n}. \] | Let
\[ E_k = \int_{(k-1)/n}^{k/n} f(x)dx - \frac{1}{n} f\left(\frac{k}{n}\right) \]
for $k = 1, 2, \ldots, n$. Since $f$ is differentiable, it is continuous, and therefore by the mean value theorem for integrals there exists a number $\eta_k$ such that
\[ \frac{k-1}{n} < \eta_k < \frac{k}{n} \]
and
\[ \int_{(k-1)/n}^{k/n} f(x)dx = \frac{1}{n} f(\eta_k). \]
By the mean value theorem for derivatives there exists a number $\xi_k$ such that $\eta_k < \xi_k < (k/n)$ and
\[ f(\eta_k) - f\left(\frac{k}{n}\right) = \left(\eta_k - \frac{k}{n}\right) f'(\xi_k). \]
Then
\[ |E_k| = \frac{1}{n} \left| f(\eta_k) - f\left(\frac{k}{n}\right) \right| \]
\[ = \frac{1}{n} \left| \eta_k - \frac{k}{n} \right| \cdot |f'(\xi_k)| \leq \frac{1}{n^2} M. \]
Hence
\[ \left| \int_0^1 f(x)dx - \frac{1}{n} \sum_{k=1}^n f\left(\frac{k}{n}\right) \right| = \left| \sum_{k=1}^n E_k \right| \leq \sum_{k=1}^n |E_k| \leq \frac{M}{n}. \] | 0 |
1947 | 1947_10 | Given $P(z) = z^2 + az + b$, a quadratic polynomial of the complex variable $z$ with complex coefficients $a, b$. Suppose that $|P(z)| = 1$ for every $z$ such that $|z| = 1$. Find the values of $a$ and $b$ and give their sum. | In particular, \( |p(1)| = |p(-1)| = 1 \), so \( 1 + a + b \) and \( 1 - a + b \) lie on the unit circle. Hence their midpoint \( 1 + b \) lies in the unit disk. Similarly, \( |p(i)| = |p(-i)| = 1 \), so \( -1 + ia + b \) and \( -1 - ia + b \) lie on the unit circle and hence their midpoint \( -1 + b \) lies in the unit disk. But \( 1 + b \) and \( -1 + b \) are a distance 2 apart, so they must lie at either end of a diameter of the unit circle and hence \( b = 0 \). Now \( 1 + a \) and \( 1 - a \) lie on the unit circle, as does their midpoint \( 1 \). Hence they must coincide and so \( a = 0 \). Thus, \( a + b = \boxed{0} \). | numerical | putnam (modified boxing) | Complex Numbers Algebra | Given $P(z) = z^2 + az + b$, a quadratic polynomial of the complex variable $z$ with complex coefficients $a, b$. Suppose that $|P(z)| = 1$ for every $z$ such that $|z| = 1$. Prove that $a = b = 0$. | Let $P(z) = z^2 + az + b$. Let $a = p + iq, b = r + is$, where $p, q, r, s$ are real. We are given that $|P(1)| = |P(-1)| = |P(i)| = |P(-i)| = 1$, and we find
\[ |P(1)|^2 = |1 + p + iq + r + is|^2 \]
\[ = 1 + p^2 + q^2 + r^2 + s^2 + 2p + 2r + 2pq + 2qs, \]
\[ |P(-1)|^2 = |1 - p - iq + r + is|^2 \]
\[ = 1 + p^2 + q^2 + r^2 + s^2 - 2p + 2r - 2pq - 2qs, \]
\[ |P(i)|^2 = |-1 + ip - q + r + is|^2 \]
\[ = 1 + p^2 + q^2 + r^2 + s^2 + 2q - 2r - 2qr + 2ps, \]
\[ |P(-i)|^2 = |-1 - ip + q + r + is|^2 \]
\[ = 1 + p^2 + q^2 + r^2 + s^2 - 2q - 2r + 2qr - 2ps. \]
Adding these equations, we get
\[ 4 = 4 + 4(p^2 + q^2 + r^2 + s^2). \]
It follows that $p = q = r = s = 0$, so $a = b = 0$.
We can equally well use a different set of roots of one. If $\xi$ is a primitive $n$th root of one, $n > 2$, then
\[ n = \sum_{k=1}^n |P(\xi^k)|^2 = n(1 + |a|^2 + |b|^2), \]
and $a = b = 0$ follows. We can also replace the sum by an integral. For all real $\theta$, we have
\[ 1 = |P(e^{i\theta})|^2 = (e^{2i\theta} + ae^{i\theta} + b)(e^{-2i\theta} + \overline{a}e^{-i\theta} + \overline{b}) \]
\[ = b\overline{b} + (\overline{a} + a\overline{b})e^{i\theta} + 1 + |a|^2 + |b|^2 + (a + a\overline{b})e^{-i\theta} + b\overline{b}e^{-2i\theta}. \]
If we integrate this over $[0, 2\pi]$, we get
\[ 2\pi = 2\pi(1 + |a|^2 + |b|^2), \]
and $a = b = 0$, as before. | 0 |
1947 | 1947_11 | a, b, c, d are distinct integers such that
\[ (x - a)(x - b)(x - c)(x - d) - 4 = 0 \]
has an integral root $r$. Find the value of $4r$. | Since $r$ is a root,
\[ (r - a)(r - b)(r - c)(r - d) = 4, \]
and since $a, b, c, d$ are distinct integers, $r - a, r - b, r - c, r - d$ are distinct integers whose product is 4. But the only set of four distinct integers whose product is 4 is the set \{1, -1, 2, -2\}. Hence
\[ (r - a) + (r - b) + (r - c) + (r - d) = 1 - 1 + 2 - 2 = 0, \]
so
\[ 4r = \boxed{a + b + c + d}. \] | algebraic | putnam (modified boxing) | Algebra Number Theory | a, b, c, d are distinct integers such that
\[ (x - a)(x - b)(x - c)(x - d) - 4 = 0 \]
has an integral root $r$. Show that $4r = a + b + c + d$. | Since $r$ is a root,
\[ (r - a)(r - b)(r - c)(r - d) = 4, \]
and since $a, b, c, d$ are distinct integers, $r - a, r - b, r - c, r - d$ are distinct integers whose product is 4. But the only set of four distinct integers whose product is 4 is the set \{1, -1, 2, -2\}. Hence
\[ (r - a) + (r - b) + (r - c) + (r - d) = 1 - 1 + 2 - 2 = 0, \]
so
\[ 4r = a + b + c + d. \] | 0 |
1947 | 1947_12 | C is a fixed point on $OZ$ and $U, V$ are variable points on $OX, OY$ respectively, where $OX, OY, OZ$ are mutually orthogonal lines. Find the locus of a point $P$ such that $PU, PV, PC$ are mutually orthogonal. | Take $OX, OY, OZ$ as coordinate axes and let $U = (u, 0, 0), V = (0, v, 0), C = (0, 0, c)$. Suppose $P(x, y, z)$ is a point of the locus. Then $(x - u, y, z), (x, y - v, z),$ and $(x, y, z - c)$ must be perpendicular vectors. Therefore
\[
x^2 + y^2 + z^2 = xu + yv \\
x^2 + y^2 + z^2 = xu + zc \\
x^2 + y^2 + z^2 = yv + zc.
\]
Adding the last two equations and subtracting the first, we get
\[
x^2 + y^2 + z^2 = 2zc,
\]
and therefore $P$ lies on the sphere $x^2 + y^2 + (z - c)^2 = c^2$, with center $C$ and radius $|CO|$. Note that, if $c = 0$, i.e., $C = 0$, then $P$ must be 0. But in that case $PC$ is not a line, so there is no locus. We assume henceforth, therefore, that $c \neq 0$.
From (1) we also see that $xu = yv = zc$, so if $z \neq 0$ neither $x$ nor $y$ is 0. If $z = 0$, then $x = y = 0$ from (2).
Conversely, if $P = (x, y, z)$ is any point on the sphere (2) with $xy \neq 0$, then $z \neq 0$ and
\[
u = zc/x, \quad v = zc/y
\]
gives a solution to (1), so the vectors $PU, PV, PC$ are pairwise orthogonal, while if $P = 0$, then for any non-zero choice of $u$ and $v$, $PU, PV, PC$ are pairwise orthogonal. The locus therefore consists of the sphere less two great circles but including the point 0. This makes the final answer \boxed{x^2 + y^2 + (z - c)^2 = c^2}. | algebraic | putnam | Geometry | C is a fixed point on $OZ$ and $U, V$ are variable points on $OX, OY$ respectively, where $OX, OY, OZ$ are mutually orthogonal lines. Find the locus of a point $P$ such that $PU, PV, PC$ are mutually orthogonal. | Take $OX, OY, OZ$ as coordinate axes and let $U = (u, 0, 0), V = (0, v, 0), C = (0, 0, c)$. Suppose $P(x, y, z)$ is a point of the locus. Then $(x - u, y, z), (x, y - v, z),$ and $(x, y, z - c)$ must be perpendicular vectors. Therefore
\[
x^2 + y^2 + z^2 = xu + yv \\
x^2 + y^2 + z^2 = xu + zc \\
x^2 + y^2 + z^2 = yv + zc.
\]
Adding the last two equations and subtracting the first, we get
\[
x^2 + y^2 + z^2 = 2zc,
\]
and therefore $P$ lies on the sphere $\boxed{x^2 + y^2 + (z - c)^2 = c^2}$, with center $C$ and radius $|CO|$. Note that, if $c = 0$, i.e., $C = 0$, then $P$ must be 0. But in that case $PC$ is not a line, so there is no locus. We assume henceforth, therefore, that $c \neq 0$.
From (1) we also see that $xu = yv = zc$, so if $z \neq 0$ neither $x$ nor $y$ is 0. If $z = 0$, then $x = y = 0$ from (2).
Conversely, if $P = (x, y, z)$ is any point on the sphere (2) with $xy \neq 0$, then $z \neq 0$ and
\[
u = zc/x, \quad v = zc/y
\]
gives a solution to (1), so the vectors $PU, PV, PC$ are pairwise orthogonal, while if $P = 0$, then for any non-zero choice of $u$ and $v$, $PU, PV, PC$ are pairwise orthogonal. The locus therefore consists of the sphere less two great circles but including the point 0. | 0 |
1948 | 1948_1 | What is the maximum of $|z^3 - z + 2|$, where $z$ is a complex number with $|z| = 1$? | Let $f(z) = z^3 - z + 2$. We may as well maximize $|f(z)|^2$. If $|z| = 1$, then $z = x + iy$, where $y^2 = 1 - x^2$ and $-1 \leq x \leq 1$, so
\[
|f(z)|^2 = |(x + iy)^3 - (x + iy) + 2|^2 \\
= |x^3 - 3x(1 - x^2) - x + 2 + iy(3x^2 - (1 - x^2) - 1)|^2 \\
= (4x^3 - 4x + 2)^2 + (1 - x^2)(4x^2 - 2)^2 \\
= 16x^3 - 4x^2 - 16x + 8 = L(x).
\]
Hence we seek $\max_{-1 \leq x \leq 1} L(x)$. This maximum must be attained either at a critical point or at an endpoint. The critical points, obtained by solving $L'(x) = 48x^2 - 8x - 16 = 0$, are $x = -\frac{1}{2}, \frac{2}{3}$. Since
\[
L(-1) = 4, \quad L\left(-\frac{1}{2}\right) = 13, \quad L\left(\frac{2}{3}\right) = \frac{8}{27}, \quad L(1) = 4,
\]
the maximum value of $L$ is 13, attained for $x = -\frac{1}{2}$. Hence the maximum value of $|f(z)|$ on the unit circle is $\boxed{\sqrt{13}}$, attained when $\text{Re } z = -\frac{1}{2}$, i.e., when $z = \frac{-1 \pm i\sqrt{3}}{2}$. | numerical | putnam | Complex Numbers Analysis | What is the maximum of $|z^3 - z + 2|$, where $z$ is a complex number with $|z| = 1$? | Let $f(z) = z^3 - z + 2$. We may as well maximize $|f(z)|^2$. If $|z| = 1$, then $z = x + iy$, where $y^2 = 1 - x^2$ and $-1 \leq x \leq 1$, so
\[
|f(z)|^2 = |(x + iy)^3 - (x + iy) + 2|^2 \\
= |x^3 - 3x(1 - x^2) - x + 2 + iy(3x^2 - (1 - x^2) - 1)|^2 \\
= (4x^3 - 4x + 2)^2 + (1 - x^2)(4x^2 - 2)^2 \\
= 16x^3 - 4x^2 - 16x + 8 = L(x).
\]
Hence we seek $\max_{-1 \leq x \leq 1} L(x)$. This maximum must be attained either at a critical point or at an endpoint. The critical points, obtained by solving $L'(x) = 48x^2 - 8x - 16 = 0$, are $x = -\frac{1}{2}, \frac{2}{3}$. Since
\[
L(-1) = 4, \quad L\left(-\frac{1}{2}\right) = 13, \quad L\left(\frac{2}{3}\right) = \frac{8}{27}, \quad L(1) = 4,
\]
the maximum value of $L$ is 13, attained for $x = -\frac{1}{2}$. Hence the maximum value of $|f(z)|$ on the unit circle is $\boxed{\sqrt{13}}$, attained when $\text{Re } z = -\frac{1}{2}$, i.e., when $z = \frac{-1 \pm i\sqrt{3}}{2}$. | 0 |
1948 | 1948_2 | Two spheres in contact have a common tangent cone. These three surfaces divide the space into various parts, only one of which is bounded by all three surfaces; it is "ring-shaped." Being given the radii of the spheres, $r$ and $R$, find the volume of the "ring-shaped" part. (The desired expression is a rational function of $r$ and $R$.) | Let $\alpha$ be the angle of the cone, and place the spheres as shown in the cross-section diagram, with centers at $O_1$ and $O_2$ and origin of the coordinate system at $O$, the point of contact.
The equation of the larger circle is
\[
x^2 + y^2 + 2Rx = 0
\]
and the equation of the smaller is
\[
x^2 + y^2 - 2rx = 0.
\]
From the diagram,
\[
\sin \alpha = \frac{O_1P}{O_1O_2} = \frac{R - r}{R + r},
\]
so
\[
\cos^2 \alpha = \frac{4Rr}{(R + r)^2}.
\]
We compute $V_1$, the volume of the frustum of the cone obtained by revolving area $A_1A_2B_2B_1$ about the $x$-axis.
Volume of larger cone = \[
\frac{1}{3}\pi A_1B_1^2 \cdot B_1T = \frac{1}{3} \pi R^2 \cos^2 \alpha \cdot R \cot \alpha \cos \alpha = \frac{16\pi}{3} R^3 \cdot \frac{r^2R^2}{(R - r)(R + r)^3}.
\]
Volume of smaller cone = \[
\frac{16\pi}{3} r^3 \cdot \frac{r^2R^2}{(R - r)(R + r)^3}.
\]
Their difference is
\[
V_1 = \frac{16\pi}{3} \cdot \frac{r^2R^2}{(R + r)^3} \cdot (R^2 + Rr + r^2).
\]
Let $V_2$ and $V_3$ be, respectively, the volumes of the larger spherical segment and the smaller spherical segment enclosed in the frustum of the cone.
\[
V_2 = \pi \int_{R \sin \alpha}^0 \left(-x^2 - 2Rx\right) dx = \frac{4\pi R^3r^2}{3(R + r)^3}(3R + r),
\]
and
\[
V_3 = \pi \int_0^{r(1 + \sin \alpha)} \left(-x^2 + 2rx\right) dx = \frac{4\pi}{3} \cdot \frac{R^2r^3}{(R + r)^3}(R + 3r).
\]
The desired volume is given by
\[
V_1 - V_2 - V_3 = \frac{4\pi}{3} \cdot \frac{R^2r^2}{(R + r)^3} \cdot \left[4(R^2 + Rr + r^2) - 3R^2 - Rr - 3r^2 - Rr\right] = \boxed{\frac{4\pi}{3} \cdot \frac{R^2r^2}{R + r}}.
\] | algebraic | putnam | Geometry Calculus | Two spheres in contact have a common tangent cone. These three surfaces divide the space into various parts, only one of which is bounded by all three surfaces; it is "ring-shaped." Being given the radii of the spheres, $r$ and $R$, find the volume of the "ring-shaped" part. (The desired expression is a rational function of $r$ and $R$.) | Let $\alpha$ be the angle of the cone, and place the spheres as shown in the cross-section diagram, with centers at $O_1$ and $O_2$ and origin of the coordinate system at $O$, the point of contact.
The equation of the larger circle is
\[
x^2 + y^2 + 2Rx = 0
\]
and the equation of the smaller is
\[
x^2 + y^2 - 2rx = 0.
\]
From the diagram,
\[
\sin \alpha = \frac{O_1P}{O_1O_2} = \frac{R - r}{R + r},
\]
so
\[
\cos^2 \alpha = \frac{4Rr}{(R + r)^2}.
\]
We compute $V_1$, the volume of the frustum of the cone obtained by revolving area $A_1A_2B_2B_1$ about the $x$-axis.
Volume of larger cone = \[
\frac{1}{3}\pi A_1B_1^2 \cdot B_1T = \frac{1}{3} \pi R^2 \cos^2 \alpha \cdot R \cot \alpha \cos \alpha = \frac{16\pi}{3} R^3 \cdot \frac{r^2R^2}{(R - r)(R + r)^3}.
\]
Volume of smaller cone = \[
\frac{16\pi}{3} r^3 \cdot \frac{r^2R^2}{(R - r)(R + r)^3}.
\]
Their difference is
\[
V_1 = \frac{16\pi}{3} \cdot \frac{r^2R^2}{(R + r)^3} \cdot (R^2 + Rr + r^2).
\]
Let $V_2$ and $V_3$ be, respectively, the volumes of the larger spherical segment and the smaller spherical segment enclosed in the frustum of the cone.
\[
V_2 = \pi \int_{R \sin \alpha}^0 \left(-x^2 - 2Rx\right) dx = \frac{4\pi R^3r^2}{3(R + r)^3}(3R + r),
\]
and
\[
V_3 = \pi \int_0^{r(1 + \sin \alpha)} \left(-x^2 + 2rx\right) dx = \frac{4\pi}{3} \cdot \frac{R^2r^3}{(R + r)^3}(R + 3r).
\]
The desired volume is given by
\[
V_1 - V_2 - V_3 = \frac{4\pi}{3} \cdot \frac{R^2r^2}{(R + r)^3} \cdot \left[4(R^2 + Rr + r^2) - 3R^2 - Rr - 3r^2 - Rr\right] = \boxed{\frac{4\pi}{3} \cdot \frac{R^2r^2}{R + r}}.
\] | 0 |
1948 | 1948_3 | Let \( \{a_n\} \) be a decreasing sequence of positive numbers with limit 0 such that
\[ b_n \equiv a_n - 2a_{n+1} + a_{n+2} \geq 0 \]
for all \( n \). Find the value of
\[ \sum_{n=1}^\infty nb_n \] in terms of the sequence given. | Since the \( b_n \)'s are the second differences of the \( a_n \)'s, it is convenient to let \( c_n = a_n - a_{n+1} \); then
\[ c_n - c_{n+1} = a_n - 2a_{n+1} + a_{n+2} \equiv b_n. \]
Since the \( a_n \)'s decrease to zero, \( c_n \geq 0 \) for all \( n \), and \( c_n \to 0 \).
For \( k \geq m \), we have
\[ \sum_{i=m}^k b_i = c_m - c_{k+1}, \]
and therefore
\[ \sum_{i=m}^\infty b_i = c_m = a_m - a_{m+1}. \]
Similarly,
\[ \sum_{m=1}^k (a_m - a_{m+1}) = a_1 - a_{k+1}, \]
so
\[ \sum_{m=1}^\infty (a_m - a_{m+1}) = a_1. \]
Thus
\[ \sum_{m=1}^\infty \left( \sum_{i=m}^\infty b_i \right) = a_1. \]
The \( b_n \)'s are non-negative, and when summing non-negative terms, rearrangement does not affect the value of the sum. For each index \( n \), the term \( b_n \) appears exactly \( n \) times in the preceding double sum, once in each of the sums \( \sum_{i=m}^\infty b_i \) for \( m = 1, 2, \dots, n \). Hence
\[ \sum_{n=1}^\infty nb_n = \sum_{m=1}^\infty \left( \sum_{i=m}^\infty b_i \right) = \boxed{a_1}. \] | algebraic | putnam (modified boxing) | Analysis | Let \( \{a_n\} \) be a decreasing sequence of positive numbers with limit 0 such that
\[ b_n \equiv a_n - 2a_{n+1} + a_{n+2} \geq 0 \]
for all \( n \). Prove that
\[ \sum_{n=1}^\infty nb_n = a_1. \] | Since the \( b_n \)'s are the second differences of the \( a_n \)'s, it is convenient to let \( c_n = a_n - a_{n+1} \); then
\[ c_n - c_{n+1} = a_n - 2a_{n+1} + a_{n+2} \equiv b_n. \]
Since the \( a_n \)'s decrease to zero, \( c_n \geq 0 \) for all \( n \), and \( c_n \to 0 \).
For \( k \geq m \), we have
\[ \sum_{i=m}^k b_i = c_m - c_{k+1}, \]
and therefore
\[ \sum_{i=m}^\infty b_i = c_m = a_m - a_{m+1}. \]
Similarly,
\[ \sum_{m=1}^k (a_m - a_{m+1}) = a_1 - a_{k+1}, \]
so
\[ \sum_{m=1}^\infty (a_m - a_{m+1}) = a_1. \]
Thus
\[ \sum_{m=1}^\infty \left( \sum_{i=m}^\infty b_i \right) = a_1. \]
The \( b_n \)'s are non-negative, and when summing non-negative terms, rearrangement does not affect the value of the sum. For each index \( n \), the term \( b_n \) appears exactly \( n \) times in the preceding double sum, once in each of the sums \( \sum_{i=m}^\infty b_i \) for \( m = 1, 2, \dots, n \). Hence
\[ \sum_{n=1}^\infty nb_n = \sum_{m=1}^\infty \left( \sum_{i=m}^\infty b_i \right) = a_1. \] | 0 |
1948 | 1948_4 | Let \(D\) be a plane region bounded by a circle of radius \(r\). Let \((x, y)\) be a point of \(D\) and consider a circle of radius \(\delta\) and center at \((x, y)\). Denote by \(l(x, y)\) the length of that arc of the circle which is outside \(D\). Find
\[ \lim_{\delta \to 0} \frac{1}{\delta^2} \iint_D l(x, y) \, dx \, dy. \] | First convert the integral to polar coordinates, taking the origin at the center of the given circle. If a point \((x, y)\) has polar coordinates \((\rho, \theta)\), then \(l(x, y) = L(\rho)\), where
\[
L(\rho) = 0 \text{ if } 0 \leq \rho \leq r - \delta
\]
and
\[
L(\rho) = 2\delta \phi = 2\delta \arccos \left( \frac{r^2 - \rho^2 - \delta^2}{2\rho \delta} \right) \text{ if } r - \delta \leq \rho \leq r, \]
as we see by applying the law of cosines to triangle \(OPQ\).
Hence we must find
\[ A = \lim_{\delta \to 0} \frac{1}{\delta^2} \int_0^{2\pi} \int_0^r L(\rho) \rho \, d\rho \, d\theta. \]
Simplifying:
\[
A = \lim_{\delta \to 0} \frac{1}{\delta^2} \int_0^{2\pi} \int_{r-\delta}^r 2\delta \arccos \left( \frac{r^2 - \rho^2 - \delta^2}{2\rho \delta} \right) \rho \, d\rho \, d\theta.
\]
Converting and simplifying further using \( \rho = r - \delta u \):
\[
A = 4\pi \int_0^1 r \arccos u \, du.
\]
Integrating by parts yields:
\[
A = 4\pi \left[ u \arccos u \right]_0^1 + 4\pi \int_0^1 \frac{u}{\sqrt{1 - u^2}} \, du = \boxed{4\pi r}.
\] | numerical | putnam | Analysis | Let \(D\) be a plane region bounded by a circle of radius \(r\). Let \((x, y)\) be a point of \(D\) and consider a circle of radius \(\delta\) and center at \((x, y)\). Denote by \(l(x, y)\) the length of that arc of the circle which is outside \(D\). Find
\[ \lim_{\delta \to 0} \frac{1}{\delta^2} \iint_D l(x, y) \, dx \, dy. \] | First convert the integral to polar coordinates, taking the origin at the center of the given circle. If a point \((x, y)\) has polar coordinates \((\rho, \theta)\), then \(l(x, y) = L(\rho)\), where
\[
L(\rho) = 0 \text{ if } 0 \leq \rho \leq r - \delta
\]
and
\[
L(\rho) = 2\delta \phi = 2\delta \arccos \left( \frac{r^2 - \rho^2 - \delta^2}{2\rho \delta} \right) \text{ if } r - \delta \leq \rho \leq r, \]
as we see by applying the law of cosines to triangle \(OPQ\).
Hence we must find
\[ A = \lim_{\delta \to 0} \frac{1}{\delta^2} \int_0^{2\pi} \int_0^r L(\rho) \rho \, d\rho \, d\theta. \]
Simplifying:
\[
A = \lim_{\delta \to 0} \frac{1}{\delta^2} \int_0^{2\pi} \int_{r-\delta}^r 2\delta \arccos \left( \frac{r^2 - \rho^2 - \delta^2}{2\rho \delta} \right) \rho \, d\rho \, d\theta.
\]
Converting and simplifying further using \( \rho = r - \delta u \):
\[
A = 4\pi \int_0^1 r \arccos u \, du.
\]
Integrating by parts yields:
\[
A = 4\pi \left[ u \arccos u \right]_0^1 + 4\pi \int_0^1 \frac{u}{\sqrt{1 - u^2}} \, du = \boxed{4\pi r}.
\] | 0 |
1948 | 1948_5 | If \(x_1, x_2, \ldots, x_n\) denote the \(n\)-th roots of unity, evaluate:
\[
\prod_{i < j} (x_i - x_j)^2.
\] | Since
\[
t^n - 1 = (t - x_1)(t - x_2) \cdots (t - x_n),
\]
we see that
\[
x_1 x_2 \cdots x_n = (-1)^{n-1}.
\]
Differentiating, we have
\[
n t^{n-1} = \sum_{i=1}^n (t - x_1)(t - x_2) \cdots (t - x_{i-1})(t - x_{i+1}) \cdots (t - x_n).
\]
Evaluating at \(t = x_1, x_2, \ldots, x_n\), we obtain equations such as
\[
nx_1^{n-1} = (x_1 - x_2)(x_1 - x_3) \cdots (x_1 - x_n),
\]
and similarly for all other roots. Multiplying these equations gives
\[
n^n(x_1 x_2 \cdots x_n)^{n-1} = \prod_{i < j} (-(x_i - x_j)^2),
\]
which simplifies to
\[
(-1)^{n(n-1)/2} \prod_{i < j} (x_i - x_j)^2.
\]
Using the earlier expression for the product of the roots, \(x_1 x_2 \cdots x_n = (-1)^{n-1}\), we simplify to
\[
\prod_{i < j} (x_i - x_j)^2 = \boxed{n^n (-1)^{(n-1)(n-2)/2}}.
\] | algebraic | putnam | Algebra | If \(x_1, x_2, \ldots, x_n\) denote the \(n\)-th roots of unity, evaluate:
\[
\prod_{i < j} (x_i - x_j)^2.
\] | Since
\[
t^n - 1 = (t - x_1)(t - x_2) \cdots (t - x_n),
\]
we see that
\[
x_1 x_2 \cdots x_n = (-1)^{n-1}.
\]
Differentiating, we have
\[
n t^{n-1} = \sum_{i=1}^n (t - x_1)(t - x_2) \cdots (t - x_{i-1})(t - x_{i+1}) \cdots (t - x_n).
\]
Evaluating at \(t = x_1, x_2, \ldots, x_n\), we obtain equations such as
\[
nx_1^{n-1} = (x_1 - x_2)(x_1 - x_3) \cdots (x_1 - x_n),
\]
and similarly for all other roots. Multiplying these equations gives
\[
n^n(x_1 x_2 \cdots x_n)^{n-1} = \prod_{i < j} (-(x_i - x_j)^2),
\]
which simplifies to
\[
(-1)^{n(n-1)/2} \prod_{i < j} (x_i - x_j)^2.
\]
Using the earlier expression for the product of the roots, \(x_1 x_2 \cdots x_n = (-1)^{n-1}\), we simplify to
\[
\prod_{i < j} (x_i - x_j)^2 = \boxed{n^n (-1)^{(n-1)(n-2)/2}}.
\] | 0 |
1948 | 1948_6_i | Find the value of
\[
x + \frac{2}{3}x^3 + \frac{2 \cdot 4}{3 \cdot 5}x^5 + \frac{2 \cdot 4 \cdot 6}{3 \cdot 5 \cdot 7}x^7 + \cdots
\] in terms of $x$. | Let
\[
f(x) = x + \frac{2}{3}x^3 + \frac{2 \cdot 4}{3 \cdot 5}x^5 + \frac{2 \cdot 4 \cdot 6}{3 \cdot 5 \cdot 7}x^7 + \cdots.
\]
Then,
\[
f'(x) = 1 + x \left[ 2x + \frac{2}{3} \cdot 4x^3 + \frac{2 \cdot 4}{3 \cdot 5} \cdot 6x^5 + \cdots \right].
\]
Rewriting,
\[
f'(x) = 1 + x \frac{d}{dx}\left[x^2 + \frac{2}{3}x^4 + \frac{2 \cdot 4}{3 \cdot 5}x^6 + \cdots\right],
\]
and factoring further,
\[
f'(x) = 1 + x \frac{d}{dx}[xf(x)] = 1 + xf(x) + x^2f'(x).
\]
Thus, \(f'(x)\) satisfies the differential equation
\[
(1 - x^2)f'(x) = 1 + xf(x),
\]
with the initial condition
\[
f(0) = 0.
\]
Since the series for \(f\) is convergent for \(|x| < 1\), all manipulations are justified. The differential equation is a first-order linear non-singular equation on the interval \((-1, 1)\), and hence it has a unique solution satisfying \(f(0) = 0\).
We observe that \(\frac{\arcsin x}{\sqrt{1 - x^2}}\) satisfies the same differential equation and initial condition. Therefore,
\[
f(x) = \boxed{\frac{\arcsin x}{\sqrt{1 - x^2}}}.
\] | algebraic | putnam (modified boxing) | Differential Equations | Show that
\[
x + \frac{2}{3}x^3 + \frac{2 \cdot 4}{3 \cdot 5}x^5 + \frac{2 \cdot 4 \cdot 6}{3 \cdot 5 \cdot 7}x^7 + \cdots = \frac{\arcsin x}{\sqrt{1 - x^2}}.
\] | Let
\[
f(x) = x + \frac{2}{3}x^3 + \frac{2 \cdot 4}{3 \cdot 5}x^5 + \frac{2 \cdot 4 \cdot 6}{3 \cdot 5 \cdot 7}x^7 + \cdots.
\]
Then,
\[
f'(x) = 1 + x \left[ 2x + \frac{2}{3} \cdot 4x^3 + \frac{2 \cdot 4}{3 \cdot 5} \cdot 6x^5 + \cdots \right].
\]
Rewriting,
\[
f'(x) = 1 + x \frac{d}{dx}\left[x^2 + \frac{2}{3}x^4 + \frac{2 \cdot 4}{3 \cdot 5}x^6 + \cdots\right],
\]
and factoring further,
\[
f'(x) = 1 + x \frac{d}{dx}[xf(x)] = 1 + xf(x) + x^2f'(x).
\]
Thus, \(f'(x)\) satisfies the differential equation
\[
(1 - x^2)f'(x) = 1 + xf(x),
\]
with the initial condition
\[
f(0) = 0.
\]
Since the series for \(f\) is convergent for \(|x| < 1\), all manipulations are justified. The differential equation is a first-order linear non-singular equation on the interval \((-1, 1)\), and hence it has a unique solution satisfying \(f(0) = 0\).
We observe that \(\frac{\arcsin x}{\sqrt{1 - x^2}}\) satisfies the same differential equation and initial condition. Therefore,
\[
f(x) = \frac{\arcsin x}{\sqrt{1 - x^2}}.
\] | 0 |
1948 | 1948_7 | Let \( f(x) \) be a cubic polynomial with roots \( x_1, x_2, \) and \( x_3 \). Assume that \( f(2x) \) is divisible by \( f'(x) \). Compute the ratios \( x_1 : x_2 : x_3 \). | Let \( f(x) = x^3 + ax^2 + bx + c \). Since \( f(2x) \) is divisible by \( f'(x) \), we have
\[
8x^3 + 4ax^2 + 2bx + c = (3x^2 + 2ax + b)(px + q)
\]
for some \( p \) and \( q \). Comparing coefficients, we find
\[
3p = 8, \quad 2ap + 3q = 4a, \quad bp + 2aq = 2b, \quad qb = c,
\]
from which it follows that
\[
p = \frac{8}{3}, \quad q = -\frac{4a}{9}, \quad b = \frac{4a^2}{3}, \quad c = -\frac{16a^3}{27}.
\]
Hence,
\[
f(x) = x^3 + ax^2 + \frac{4}{3}a^2x - \frac{16}{27}a^3.
\]
Now, if \( a = 0 \), all the roots of \( f(x) = 0 \) are zero and their ratios are undefined; so we assume from now on that \( a \neq 0 \). Set \( w = 3x/a \) and consider
\[
F(w) = 27f(x) = 27f\left(\frac{aw}{3}\right) = a^3(w^3 + 3w^2 + 12w - 16).
\]
This simplifies to
\[
F(w) = a^3(w - 1)(w^2 + 4w + 16).
\]
The roots of \( F(w) = 0 \) are \( 1, -2 \pm 2\sqrt{3}i \). The roots of \( f(x) = 0 \) have the same ratios as the roots of \( F(w) = 0 \), so with suitable numbering, we have
\[
x_1 : x_2 : x_3 = \boxed{1 : (-2 + 2\sqrt{3}i) : (-2 - 2\sqrt{3}i)}.
\] | algebraic | putnam | Algebra Complex Numbers | Let \( f(x) \) be a cubic polynomial with roots \( x_1, x_2, \) and \( x_3 \). Assume that \( f(2x) \) is divisible by \( f'(x) \) and compute the ratios \( x_1 : x_2 : x_3 \). | Let \( f(x) = x^3 + ax^2 + bx + c \). Since \( f(2x) \) is divisible by \( f'(x) \), we have
\[
8x^3 + 4ax^2 + 2bx + c = (3x^2 + 2ax + b)(px + q)
\]
for some \( p \) and \( q \). Comparing coefficients, we find
\[
3p = 8, \quad 2ap + 3q = 4a, \quad bp + 2aq = 2b, \quad qb = c,
\]
from which it follows that
\[
p = \frac{8}{3}, \quad q = -\frac{4a}{9}, \quad b = \frac{4a^2}{3}, \quad c = -\frac{16a^3}{27}.
\]
Hence,
\[
f(x) = x^3 + ax^2 + \frac{4}{3}a^2x - \frac{16}{27}a^3.
\]
Now, if \( a = 0 \), all the roots of \( f(x) = 0 \) are zero and their ratios are undefined; so we assume from now on that \( a \neq 0 \). Set \( w = 3x/a \) and consider
\[
F(w) = 27f(x) = 27f\left(\frac{aw}{3}\right) = a^3(w^3 + 3w^2 + 12w - 16).
\]
This simplifies to
\[
F(w) = a^3(w - 1)(w^2 + 4w + 16).
\]
The roots of \( F(w) = 0 \) are \( 1, -2 \pm 2\sqrt{3}i \). The roots of \( f(x) = 0 \) have the same ratios as the roots of \( F(w) = 0 \), so with suitable numbering, we have
\[
x_1 : x_2 : x_3 = \boxed{1 : (-2 + 2\sqrt{3}i) : (-2 - 2\sqrt{3}i)}.
\] | 0 |
1948 | 1948_10 | Let $\min(x,y)$ denote the smaller of the numbers $x$ and $y$. For what $\lambda$'s does the equation \[ \int_0^1 \min(x,y)f(y)\,dy = \lambda f(x) \] have continuous solutions which do not vanish identically in $(0,1)$? What are these solutions? | The given equation can be written as \[ \lambda f(x) = \int_0^x y f(y) dy + x \int_x^1 f(y) dy, \] from which it is clear that, if $\lambda \neq 0$, $f$ is differentiable and hence \[ \lambda f'(x) = x f(x) - x f(x) + \int_x^1 f(y) dy = \int_x^1 f(y) dy. \] Thus $f'$ is also differentiable and \[ \lambda f''(x) = -f(x). \] If $\lambda = 0$, the same steps lead to the equation $0 = -f(x)$. Since we are only interested in functions not identically zero, we shall assume $\lambda \neq 0$ from now on.\nThe general solution of (3) is \[ f(x) = A \cos \mu x + B \sin \mu x, \] where $\mu = \lambda^{-1/2}$ if $\lambda > 0$, or \[ f(x) = A \cosh \nu x + B \sinh \nu x, \] where $\nu = (-\lambda)^{-1/2}$ if $\lambda < 0$.\nIt is evident from (1) that $\lim_{x\to0} f(x) = 0$, so $A = 0$ whether $\lambda$ is positive or negative. From (2) it follows that $\lim_{x\to1} f'(x) = 0$, giving respectively \[ B \mu \cos \mu = 0 \] or \[ B \nu \cosh \nu = 0. \] The last equation can hold only for $B = 0$ and hence $f = 0$; so we conclude that (1) has no non-zero solutions if $\lambda < 0$.\nFor $\lambda > 0$, a non-trivial solution is possible only if $\cos \mu = 0$, that is, $\mu$ is an odd multiple of $\pi/2$. In fact \[ f(x) = B \sin (2k + 1) \frac{\pi}{2} x \] is readily checked to be a solution with \[ \lambda = \mu^{-2} = \boxed{\frac{4}{(2k + 1)^2\pi^2}}, \] where $k$ is any integer and $B$ is arbitrary. Here we need only consider $k = 0, 1, 2, \dots$, since negative values of $k$ produce the same solutions. | algebraic | putnam | Calculus Analysis Differential Equations | Let $\min(x,y)$ denote the smaller of the numbers $x$ and $y$. For what $\lambda$'s does the equation \[ \int_0^1 \min(x,y)f(y)\,dy = \lambda f(x) \] have continuous solutions which do not vanish identically in $(0,1)$? What are these solutions? | The given equation can be written as \[ \lambda f(x) = \int_0^x y f(y) dy + x \int_x^1 f(y) dy, \] from which it is clear that, if $\lambda \neq 0$, $f$ is differentiable and hence \[ \lambda f'(x) = x f(x) - x f(x) + \int_x^1 f(y) dy = \int_x^1 f(y) dy. \] Thus $f'$ is also differentiable and \[ \lambda f''(x) = -f(x). \] If $\lambda = 0$, the same steps lead to the equation $0 = -f(x)$. Since we are only interested in functions not identically zero, we shall assume $\lambda \neq 0$ from now on.\nThe general solution of (3) is \[ f(x) = A \cos \mu x + B \sin \mu x, \] where $\mu = \lambda^{-1/2}$ if $\lambda > 0$, or \[ f(x) = A \cosh \nu x + B \sinh \nu x, \] where $\nu = (-\lambda)^{-1/2}$ if $\lambda < 0$.\nIt is evident from (1) that $\lim_{x\to0} f(x) = 0$, so $A = 0$ whether $\lambda$ is positive or negative. From (2) it follows that $\lim_{x\to1} f'(x) = 0$, giving respectively \[ B \mu \cos \mu = 0 \] or \[ B \nu \cosh \nu = 0. \] The last equation can hold only for $B = 0$ and hence $f = 0$; so we conclude that (1) has no non-zero solutions if $\lambda < 0$.\nFor $\lambda > 0$, a non-trivial solution is possible only if $\cos \mu = 0$, that is, $\mu$ is an odd multiple of $\pi/2$. In fact \[ f(x) = B \sin (2k + 1) \frac{\pi}{2} x \] is readily checked to be a solution with \[ \lambda = \mu^{-2} = \boxed{\frac{4}{(2k + 1)^2\pi^2}}, \] where $k$ is any integer and $B$ is arbitrary. Here we need only consider $k = 0, 1, 2, \dots$, since negative values of $k$ produce the same solutions. | 0 |
1948 | 1948_11 | The pairs of numbers $(a, b)$ such that $|a + bt + t^2| \leq 1$ for $0 \leq t \leq 1$ fill a certain region in the $(a, b)$-plane. What is the area of this region? | Consider \[ f(t) = a + bt + t^2. \] This function has just one critical value, at $t = -b/2$ where its value is $a - b^2/4$. On the interval $[0, 1]$, $f$ can have extreme values only at the endpoints and at the critical point if it should happen to fall in $[0, 1]$ (that is, if $b \in [-2, 0]$). Hence the extreme values of $f$ are \[ f(0) = a \quad \text{and} \quad f(1) = a + b + 1 \] if $b \notin [-2, 0]$, and they are in the set \[ \{a, a + b + 1, a - b^2/4\} \] if $b \in [-2, 0]$.\nHence $|f(t)| \leq 1$ for all $t \in [0, 1]$ if and only if \[ b \notin [-2, 0] \quad \text{and} \quad |a| \leq 1, \ |a + b + 1| \leq 1 \] or \[ b \in [-2, 0], \quad |a| \leq 1, \ |a + b + 1| \leq 1, \ |a - b^2/4| \leq 1. \]\nThe region required is therefore as shown in the diagram where the arc from $A$ to $B$ is part of the parabola $a - b^2/4 = -1$.\nThe area of the required region can be obtained in several ways. The parallelogram $CEDF$ evidently has area 4. We must subtract the area of the piece $AFB$. Since $\overline{AF}$ and $\overline{FB}$ are tangents to the parabola, the area of $AFB$ is $\frac{1}{3}$ that of the triangle $AFB$ (Archimedes' rule). But triangle $AFB$ has base $AF$ of length 1 and altitude 1, so its area is $\frac{1}{2}$ and the curvilinear piece $AFB$ has area $\frac{1}{6}$. Hence the area of $CEDBA$ is $4 - \frac{1}{6} = \boxed{\frac{23}{6}}$. | numerical | putnam | Calculus Geometry | The pairs of numbers $(a, b)$ such that $|a + bt + t^2| \leq 1$ for $0 \leq t \leq 1$ fill a certain region in the $(a, b)$-plane. What is the area of this region? | Consider \[ f(t) = a + bt + t^2. \] This function has just one critical value, at $t = -b/2$ where its value is $a - b^2/4$. On the interval $[0, 1]$, $f$ can have extreme values only at the endpoints and at the critical point if it should happen to fall in $[0, 1]$ (that is, if $b \in [-2, 0]$). Hence the extreme values of $f$ are \[ f(0) = a \quad \text{and} \quad f(1) = a + b + 1 \] if $b \notin [-2, 0]$, and they are in the set \[ \{a, a + b + 1, a - b^2/4\} \] if $b \in [-2, 0]$.\nHence $|f(t)| \leq 1$ for all $t \in [0, 1]$ if and only if \[ b \notin [-2, 0] \quad \text{and} \quad |a| \leq 1, \ |a + b + 1| \leq 1 \] or \[ b \in [-2, 0], \quad |a| \leq 1, \ |a + b + 1| \leq 1, \ |a - b^2/4| \leq 1. \]\nThe region required is therefore as shown in the diagram where the arc from $A$ to $B$ is part of the parabola $a - b^2/4 = -1$.\nThe area of the required region can be obtained in several ways. The parallelogram $CEDF$ evidently has area 4. We must subtract the area of the piece $AFB$. Since $\overline{AF}$ and $\overline{FB}$ are tangents to the parabola, the area of $AFB$ is $\frac{1}{3}$ that of the triangle $AFB$ (Archimedes' rule). But triangle $AFB$ has base $AF$ of length 1 and altitude 1, so its area is $\frac{1}{2}$ and the curvilinear piece $AFB$ has area $\frac{1}{6}$. Hence the area of $CEDBA$ is $4 - \frac{1}{6} = \boxed{\frac{23}{6}}$. | 0 |
1948 | 1948_12 | Let $V_1, V_2, V_3$, and $V$ denote four vertices of a cube. $V_1, V_2$, and $V_3$ are next neighbors of $V$, that is, the lines $VV_1, VV_2$, and $VV_3$ are edges of the cube. Project the cube orthogonally onto a plane (the z-plane, the Gaussian plane) of which the points are marked with complex numbers. Let the projection of $V$ fall in the origin and the projections of $V_1, V_2$, and $V_3$ in points marked with the complex numbers $z_1, z_2$, and $z_3$, respectively. Find the value of $z_1^2 + z_2^2 + z_3^2$. | Take Cartesian coordinates $w, x, y$ in space so that the $x$-axis is the real axis and the $y$-axis is the imaginary axis in the given Gaussian plane. Then the projection maps each point $(w, x, y)$ onto $(0, x, y)$.\n\nSuppose $(w_1, x_1, y_1), (w_2, x_2, y_2)$ and $(w_3, x_3, y_3)$ are mutually orthogonal unit vectors in space. Then the matrix \[ \begin{pmatrix} w_1 & x_1 & y_1 \\ w_2 & x_2 & y_2 \\ w_3 & x_3 & y_3 \end{pmatrix} \] is an orthogonal matrix, so its columns are also mutually orthogonal unit vectors, in particular \[ x_1^2 + x_2^2 + x_3^2 = 1, \quad y_1^2 + y_2^2 + y_3^2 = 1, \quad \text{and} \quad x_1y_1 + x_2y_2 + x_3y_3 = 0. \] The orthogonal projections of the unit vectors into the Gaussian plane are the complex numbers $x_1 + iy_1, x_2 + iy_2$, and $x_3 + iy_3$, and \[ (x_1 + iy_1)^2 + (x_2 + iy_2)^2 + (x_3 + iy_3)^2 = x_1^2 + x_2^2 + x_3^2 - y_1^2 - y_2^2 - y_3^2 + 2i(x_1y_1 + x_2y_2 + x_3y_3) = 0. \]\n\nSuppose the sides of the given cube are of length $a$. Since it is given that the vertex $V$ projects onto the origin, it must be of the form $V = (b, 0, 0)$. There are mutually orthogonal unit vectors $(w_j, x_j, y_j)$ such that \[ V_j = (b, 0, 0) + a(w_j, x_j, y_j) \quad \text{for} \quad j = 1, 2, 3. \] Then the projection of $V_j$ into the Gaussian plane is \[ z_j = a(x_j + iy_j) \] and \[ z_1^2 + z_2^2 + z_3^2 = a^2[(x_1 + iy_1)^2 + (x_2 + iy_2)^2 + (x_3 + iy_3)^2] = \boxed{0}. \] | numerical | putnam (modified boxing) | Geometry Linear Algebra Complex Numbers | Let $V_1, V_2, V_3$, and $V$ denote four vertices of a cube. $V_1, V_2$, and $V_3$ are next neighbors of $V$, that is, the lines $VV_1, VV_2$, and $VV_3$ are edges of the cube. Project the cube orthogonally onto a plane (the z-plane, the Gaussian plane) of which the points are marked with complex numbers. Let the projection of $V$ fall in the origin and the projections of $V_1, V_2$, and $V_3$ in points marked with the complex numbers $z_1, z_2$, and $z_3$, respectively. Show that $z_1^2 + z_2^2 + z_3^2 = 0$. | Take Cartesian coordinates $w, x, y$ in space so that the $x$-axis is the real axis and the $y$-axis is the imaginary axis in the given Gaussian plane. Then the projection maps each point $(w, x, y)$ onto $(0, x, y)$.\n\nSuppose $(w_1, x_1, y_1), (w_2, x_2, y_2)$ and $(w_3, x_3, y_3)$ are mutually orthogonal unit vectors in space. Then the matrix \[ \begin{pmatrix} w_1 & x_1 & y_1 \\ w_2 & x_2 & y_2 \\ w_3 & x_3 & y_3 \end{pmatrix} \] is an orthogonal matrix, so its columns are also mutually orthogonal unit vectors, in particular \[ x_1^2 + x_2^2 + x_3^2 = 1, \quad y_1^2 + y_2^2 + y_3^2 = 1, \quad \text{and} \quad x_1y_1 + x_2y_2 + x_3y_3 = 0. \] The orthogonal projections of the unit vectors into the Gaussian plane are the complex numbers $x_1 + iy_1, x_2 + iy_2$, and $x_3 + iy_3$, and \[ (x_1 + iy_1)^2 + (x_2 + iy_2)^2 + (x_3 + iy_3)^2 = x_1^2 + x_2^2 + x_3^2 - y_1^2 - y_2^2 - y_3^2 + 2i(x_1y_1 + x_2y_2 + x_3y_3) = 0. \]\n\nSuppose the sides of the given cube are of length $a$. Since it is given that the vertex $V$ projects onto the origin, it must be of the form $V = (b, 0, 0)$. There are mutually orthogonal unit vectors $(w_j, x_j, y_j)$ such that \[ V_j = (b, 0, 0) + a(w_j, x_j, y_j) \quad \text{for} \quad j = 1, 2, 3. \] Then the projection of $V_j$ into the Gaussian plane is \[ z_j = a(x_j + iy_j) \] and \[ z_1^2 + z_2^2 + z_3^2 = a^2[(x_1 + iy_1)^2 + (x_2 + iy_2)^2 + (x_3 + iy_3)^2] = 0. \] | 0 |
1949 | 1949_1_i | Three straight lines pass through the three points $(0, -a, a)$, $(a, 0, -a)$, and $(-a, a, 0)$, parallel to the $x$-axis, $y$-axis, and $z$-axis, respectively; $a > 0$. A variable straight line moves so that it has one point in common with each of the three given straight lines. Find the equation of the surface described by the variable line. | Let $L_1, L_2, L_3$ be, respectively, the lines parallel to the $x$-axis through $(0, -a, a)$, parallel to the $y$-axis through $(a, 0, -a)$, and parallel to the $z$-axis through $(-a, a, 0)$. Let $\mathcal{S}$ be the required locus.\n\nLet $P = (p, -a, a)$, $Q = (a, q, -a)$, and $R = (-a, a, r)$ be three collinear points on $L_1, L_2, L_3$, respectively, and let $X = (x, y, z)$ be any point on the same line. Then the vectors $PX, QX, RX$ are proportional, that is, the matrix \[ \begin{pmatrix} x - p & y + a & z - a \\ x - a & y - q & z + a \\ x + a & y - a & z - r \end{pmatrix} \] has rank one. Thus, in particular, \[ (x - p)(y - a) = (x + a)(y + a) \] and \[ (x - p)(z + a) = (x - a)(z - a). \]\n\nTherefore, \[ (x + a)(y + a)(z + a) = (x - p)(y - a)(z + a) = (x - a)(y - a)(z - a), \] so \[ (x + a)(y + a)(z + a) = (x - a)(y - a)(z - a), \] which is equivalent to \[ \boxed{xy + yz + zx + a^2 = 0}. \]\n\nThis equation, then, is satisfied by every point $(x, y, z)$ of $\mathcal{S}$. | algebraic | putnam | Geometry Linear Algebra | Three straight lines pass through the three points $(0, -a, a)$, $(a, 0, -a)$, and $(-a, a, 0)$, parallel to the $x$-axis, $y$-axis, and $z$-axis, respectively; $a > 0$. A variable straight line moves so that it has one point in common with each of the three given straight lines. Find the equation of the surface described by the variable line. | Let $L_1, L_2, L_3$ be, respectively, the lines parallel to the $x$-axis through $(0, -a, a)$, parallel to the $y$-axis through $(a, 0, -a)$, and parallel to the $z$-axis through $(-a, a, 0)$. Let $\mathcal{S}$ be the required locus.\n\nLet $P = (p, -a, a)$, $Q = (a, q, -a)$, and $R = (-a, a, r)$ be three collinear points on $L_1, L_2, L_3$, respectively, and let $X = (x, y, z)$ be any point on the same line. Then the vectors $PX, QX, RX$ are proportional, that is, the matrix \[ \begin{pmatrix} x - p & y + a & z - a \\ x - a & y - q & z + a \\ x + a & y - a & z - r \end{pmatrix} \] has rank one. Thus, in particular, \[ (x - p)(y - a) = (x + a)(y + a) \] and \[ (x - p)(z + a) = (x - a)(z - a). \]\n\nTherefore, \[ (x + a)(y + a)(z + a) = (x - p)(y - a)(z + a) = (x - a)(y - a)(z - a), \] so \[ (x + a)(y + a)(z + a) = (x - a)(y - a)(z - a), \] which is equivalent to \[ \boxed{xy + yz + zx + a^2 = 0}. \]\n\nThis equation, then, is satisfied by every point $(x, y, z)$ of $\mathcal{S}$. | 0 |
1949 | 1949_2 | We consider three vectors drawn from the same initial point $O$, of lengths $a$, $b$, and $c$, respectively. Let $E$ be the parallelepiped with vertex $O$ of which the given vectors are the edges and $H$ the parallelepiped with vertex $O$ of which these vectors are the altitudes. Find the product of the volumes of $E$ and generalize the theorem, with proof, to $n$ dimensions. | We proceed at once to the general case. Let $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n$ be vectors in an $n$-dimensional space. To say that these vectors span a parallelepiped $P$ means that they are linearly independent and that \[ P = \{ \Sigma \lambda_i \mathbf{v}_i : 0 \leq \lambda_i \leq 1 \}. \] The volume of the parallelepiped $P$ is $| \det A |$ where $A$ is the $n \times n$ matrix whose rows are the components of $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n$ in some Cartesian coordinate system.\n\nSince the $\mathbf{v}$'s are linearly independent there are linear functionals $f_1, \ldots, f_n$ such that \[ f_j(\mathbf{v}_i) = \delta_{ij} \| \mathbf{v}_i \|^2 \] for all $i, j$, where $\delta_{ij}$ is the Kronecker delta ($\delta_{ij} = 0$ if $i \neq j$, $\delta_{ij} = 1$). Since every linear functional is realizable with an inner product there are vectors $\mathbf{w}_1, \ldots, \mathbf{w}_n$ such that \[ (\mathbf{v}_i, \mathbf{w}_j) = \delta_{ij} \| \mathbf{v}_i \|^2 \] for all $i, j$.\n\nNow the $\mathbf{w}$'s are linearly independent, for if $\Sigma \alpha_j \mathbf{w}_j = 0$, then \[ (\mathbf{v}_i, \Sigma \alpha_j \mathbf{w}_j) = \alpha_i \| \mathbf{v}_i \|^2 = 0, \quad i = 1, 2, \ldots, n, \] whence $\alpha_1 = \alpha_2 = \cdots = \alpha_n = 0$.\n\nTherefore the vectors $\mathbf{w}_1, \mathbf{w}_2, \ldots, \mathbf{w}_n$ span a parallelepiped $Q$. The vector $\mathbf{v}_i$ is the altitude of $Q$ on the face spanned by all the $\mathbf{w}$'s except $\mathbf{w}_i$, since it is perpendicular to that face, and the projection of $\mathbf{w}_i$ in the direction of $\mathbf{v}_i$ has length $\| \mathbf{v}_i \|$ by (1).\n\nLet $B$ be the $n \times n$ matrix whose rows are $\mathbf{w}_1, \mathbf{w}_2, \ldots, \mathbf{w}_n$. Then \[ \text{vol } Q = | \det B | = | \det B^T |, \] where $B^T$ is the transpose of $B$.\n\nNow equation (1) shows that $AB^T$ is the diagonal matrix \[ \text{diag} ( \| \mathbf{v}_1 \|^2, \| \mathbf{v}_2 \|^2, \ldots, \| \mathbf{v}_n \|^2 ). \] Hence \[ (\text{vol } P)(\text{vol } Q) = | \det A | \cdot | \det B^T | = | \det AB^T | \] \[ = \| \mathbf{v}_1 \|^2 \cdot \| \mathbf{v}_2 \|^2 \cdots \| \mathbf{v}_n \|^2. \]\n\nIn the three-dimensional case, the problem calls the parallelepipeds $E$ and $H$ instead of $P$ and $Q$, and, since $\| \mathbf{v}_1 \| = a$, $\| \mathbf{v}_2 \| = b$, and $\| \mathbf{v}_3 \| = c$, the formula above is the required result for the three-dimensional case. This makes the final answer \boxed{(abc)^2}. | algebraic | putnam (modified boxing) | Geometry Linear Algebra | We consider three vectors drawn from the same initial point $O$, of lengths $a$, $b$, and $c$, respectively. Let $E$ be the parallelepiped with vertex $O$ of which the given vectors are the edges and $H$ the parallelepiped with vertex $O$ of which these vectors are the altitudes. Show that the product of the volumes of $E$ and $H$ equals $(abc)^2$, and generalize the theorem, with proof, to $n$ dimensions. | We proceed at once to the general case. Let $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n$ be vectors in an $n$-dimensional space. To say that these vectors span a parallelepiped $P$ means that they are linearly independent and that \[ P = \{ \Sigma \lambda_i \mathbf{v}_i : 0 \leq \lambda_i \leq 1 \}. \] The volume of the parallelepiped $P$ is $| \det A |$ where $A$ is the $n \times n$ matrix whose rows are the components of $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n$ in some Cartesian coordinate system.\n\nSince the $\mathbf{v}$'s are linearly independent there are linear functionals $f_1, \ldots, f_n$ such that \[ f_j(\mathbf{v}_i) = \delta_{ij} \| \mathbf{v}_i \|^2 \] for all $i, j$, where $\delta_{ij}$ is the Kronecker delta ($\delta_{ij} = 0$ if $i \neq j$, $\delta_{ij} = 1$). Since every linear functional is realizable with an inner product there are vectors $\mathbf{w}_1, \ldots, \mathbf{w}_n$ such that \[ (\mathbf{v}_i, \mathbf{w}_j) = \delta_{ij} \| \mathbf{v}_i \|^2 \] for all $i, j$.\n\nNow the $\mathbf{w}$'s are linearly independent, for if $\Sigma \alpha_j \mathbf{w}_j = 0$, then \[ (\mathbf{v}_i, \Sigma \alpha_j \mathbf{w}_j) = \alpha_i \| \mathbf{v}_i \|^2 = 0, \quad i = 1, 2, \ldots, n, \] whence $\alpha_1 = \alpha_2 = \cdots = \alpha_n = 0$.\n\nTherefore the vectors $\mathbf{w}_1, \mathbf{w}_2, \ldots, \mathbf{w}_n$ span a parallelepiped $Q$. The vector $\mathbf{v}_i$ is the altitude of $Q$ on the face spanned by all the $\mathbf{w}$'s except $\mathbf{w}_i$, since it is perpendicular to that face, and the projection of $\mathbf{w}_i$ in the direction of $\mathbf{v}_i$ has length $\| \mathbf{v}_i \|$ by (1).\n\nLet $B$ be the $n \times n$ matrix whose rows are $\mathbf{w}_1, \mathbf{w}_2, \ldots, \mathbf{w}_n$. Then \[ \text{vol } Q = | \det B | = | \det B^T |, \] where $B^T$ is the transpose of $B$.\n\nNow equation (1) shows that $AB^T$ is the diagonal matrix \[ \text{diag} ( \| \mathbf{v}_1 \|^2, \| \mathbf{v}_2 \|^2, \ldots, \| \mathbf{v}_n \|^2 ). \] Hence \[ (\text{vol } P)(\text{vol } Q) = | \det A | \cdot | \det B^T | = | \det AB^T | \] \[ = \| \mathbf{v}_1 \|^2 \cdot \| \mathbf{v}_2 \|^2 \cdots \| \mathbf{v}_n \|^2. \]\n\nIn the three-dimensional case, the problem calls the parallelepipeds $E$ and $H$ instead of $P$ and $Q$, and, since $\| \mathbf{v}_1 \| = a$, $\| \mathbf{v}_2 \| = b$, and $\| \mathbf{v}_3 \| = c$, the formula above is the required result for the three-dimensional case. | 0 |
1949 | 1949_5 | How many roots of the equation $z^6 + 6z + 10 = 0$ lie in each quadrant of the complex plane? Return the total number of roots in the second and third quadrant. | Let $P(z) = z^6 + 6z + 10$. The minimum value of $P(z)$ for real $z$ is $P(-1) = 5$. Hence the equation has no real roots. There can be no purely imaginary roots since Im $(P(iy)) = 6y \neq 0$ unless $y = 0$, and $P(0) \neq 0$.
The roots sum to zero so they do not all lie in the right half-plane or all in the left half-plane. Since $P$ has real coefficients, the roots occur in conjugate pairs, so the number in the first quadrant is the same as the number in the fourth, and the number in the second is the same as the number in the third. There are, therefore, just two possibilities:
(1) There is one root in each of the first and fourth quadrants, and there are two roots in each of the second and third; or
(2) There are two roots in each of the first and fourth quadrants and there is one root in each of the second and third.
We can decide between these two possibilities through the argument principle of complex variable theory. This states:
If a function $f$ is analytic at each point of a simple closed curve and its interior and does not vanish at any point of the curve, then the number of zeros of $f$ in the interior of the curve (counting multiplicities) is $1/2\pi$ times the variation of $\arg f(z)$ as $z$ describes the curve once in the positive direction.
In this problem take $f = P$ and the simple closed curve formed by the real axis from $O$ to $R$, the arc of the circle $|z| = R$ from $R$ to $iR$, and the imaginary axis from $iR$ to $O$, where $R$ is a large positive number.
The argument variation of $P$ along $[O, R]$ is zero since $P$ remains real and positive. If $R$ is large, then the argument variation of $P(z)$ along the circular arc is approximately the same as that of $z^6$ along that same arc, since $|(P(z)/z^6) - 1|$ is small for all $z$ on this arc. Therefore the argument variation of $P(z)$ along the circular arc is about $6 \cdot \pi/2 = 3\pi$. Finally, as $z$ goes from $iR$ to $O$ along the imaginary axis, $P(z)$ goes back to the value $10$, keeping always in the upper half-plane; hence the argument variation along this part of the path is about $-\pi$.
But the total argument variation, seen to be about $(3\pi - \pi) = 2\pi$, must be an integral multiple of $2\pi$, so it must be exactly $2\pi$ (assuming that $R$ is large enough). Hence the total number of zeros enclosed by the path is one for any large $R$. Therefore the number of zeros in the first quadrant is one. Hence the distribution of zeros by quadrants is given by (1) making the final answer \boxed{4}. | numerical | putnam (modified boxing) | Complex Numbers Algebra | How many roots of the equation $z^6 + 6z + 10 = 0$ lie in each quadrant of the complex plane? | Let $P(z) = z^6 + 6z + 10$. The minimum value of $P(z)$ for real $z$ is $P(-1) = 5$. Hence the equation has no real roots. There can be no purely imaginary roots since Im $(P(iy)) = 6y \neq 0$ unless $y = 0$, and $P(0) \neq 0$.
The roots sum to zero so they do not all lie in the right half-plane or all in the left half-plane. Since $P$ has real coefficients, the roots occur in conjugate pairs, so the number in the first quadrant is the same as the number in the fourth, and the number in the second is the same as the number in the third. There are, therefore, just two possibilities:
(1) There is one root in each of the first and fourth quadrants, and there are two roots in each of the second and third; or
(2) There are two roots in each of the first and fourth quadrants and there is one root in each of the second and third.
We can decide between these two possibilities through the argument principle of complex variable theory. This states:
If a function $f$ is analytic at each point of a simple closed curve and its interior and does not vanish at any point of the curve, then the number of zeros of $f$ in the interior of the curve (counting multiplicities) is $1/2\pi$ times the variation of $\arg f(z)$ as $z$ describes the curve once in the positive direction.
In this problem take $f = P$ and the simple closed curve formed by the real axis from $O$ to $R$, the arc of the circle $|z| = R$ from $R$ to $iR$, and the imaginary axis from $iR$ to $O$, where $R$ is a large positive number.
The argument variation of $P$ along $[O, R]$ is zero since $P$ remains real and positive. If $R$ is large, then the argument variation of $P(z)$ along the circular arc is approximately the same as that of $z^6$ along that same arc, since $|(P(z)/z^6) - 1|$ is small for all $z$ on this arc. Therefore the argument variation of $P(z)$ along the circular arc is about $6 \cdot \pi/2 = 3\pi$. Finally, as $z$ goes from $iR$ to $O$ along the imaginary axis, $P(z)$ goes back to the value $10$, keeping always in the upper half-plane; hence the argument variation along this part of the path is about $-\pi$.
But the total argument variation, seen to be about $(3\pi - \pi) = 2\pi$, must be an integral multiple of $2\pi$, so it must be exactly $2\pi$ (assuming that $R$ is large enough). Hence the total number of zeros enclosed by the path is one for any large $R$. Therefore the number of zeros in the first quadrant is one. Hence the distribution of zeros by quadrants is given by (1). | 0 |
1949 | 1949_11 | Let $a_1, a_2, \ldots, a_n, \ldots$ be an arbitrary sequence of positive numbers. Find the lowerbound of
\[ \limsup_{n \to \infty} \left( \frac{a_1 + a_{n+1}}{a_n} \right)^n. \] | We shall show that there are infinitely many integers $n$ for which
\[ \frac{a_1 + a_{n+1}}{a_n} > 1 + \frac{1}{n}. \]
Our proof is indirect. Suppose it is false. Then for some integer $k$ and for all $n \geq k$
\[ \frac{a_1 + a_{n+1}}{a_n} \leq \frac{n+1}{n}. \]
whence
\[ \frac{a_n}{n} \geq \frac{a_1}{n+1} + \frac{a_{n+1}}{n+1}. \]
Therefore
\[ \frac{a_k}{k} \geq \frac{a_1}{k+1} + \frac{a_{k+1}}{k+1} \geq \frac{a_1}{k+1} + \frac{a_1}{k+1} + \frac{a_{k+2}}{k+2} \geq \frac{a_1}{k+1} + \frac{a_1}{k+2} + \frac{a_1}{k+3} + \frac{a_{k+3}}{k+3}. \]
By induction
\[ \frac{a_k}{k} \geq a_1 \left( \sum_{i=k+1}^p \frac{1}{i} \right) + \frac{a_p}{p} \]
for any $p \geq k$. Then
\[ \sum_{i=k+1}^p \frac{1}{i} \leq \frac{a_k}{k a_1} \]
for $p \geq k$.
But the harmonic series diverges, so the sums $\sum_{i=k+1}^p \frac{1}{i}$ are unbounded. Thus relation (1) cannot hold for all $n \geq k$, and hence there must be infinitely many integers $n$ for which
\[ \frac{a_1 + a_{n+1}}{a_n} > 1 + \frac{1}{n}. \]
Then
\[ \limsup_{n \to \infty} \left( \frac{a_1 + a_{n+1}}{a_n} \right)^n = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n = \boxed{e}. \] | algebraic | putnam (modified boxing) | Analysis Algebra | Let $a_1, a_2, \ldots, a_n, \ldots$ be an arbitrary sequence of positive numbers. Show that
\[ \limsup_{n \to \infty} \left( \frac{a_1 + a_{n+1}}{a_n} \right)^n \geq e. \] | We shall show that there are infinitely many integers $n$ for which
\[ \frac{a_1 + a_{n+1}}{a_n} > 1 + \frac{1}{n}. \]
Our proof is indirect. Suppose it is false. Then for some integer $k$ and for all $n \geq k$
\[ \frac{a_1 + a_{n+1}}{a_n} \leq \frac{n+1}{n}. \]
whence
\[ \frac{a_n}{n} \geq \frac{a_1}{n+1} + \frac{a_{n+1}}{n+1}. \]
Therefore
\[ \frac{a_k}{k} \geq \frac{a_1}{k+1} + \frac{a_{k+1}}{k+1} \geq \frac{a_1}{k+1} + \frac{a_1}{k+1} + \frac{a_{k+2}}{k+2} \geq \frac{a_1}{k+1} + \frac{a_1}{k+2} + \frac{a_1}{k+3} + \frac{a_{k+3}}{k+3}. \]
By induction
\[ \frac{a_k}{k} \geq a_1 \left( \sum_{i=k+1}^p \frac{1}{i} \right) + \frac{a_p}{p} \]
for any $p \geq k$. Then
\[ \sum_{i=k+1}^p \frac{1}{i} \leq \frac{a_k}{k a_1} \]
for $p \geq k$.
But the harmonic series diverges, so the sums $\sum_{i=k+1}^p \frac{1}{i}$ are unbounded. Thus relation (1) cannot hold for all $n \geq k$, and hence there must be infinitely many integers $n$ for which
\[ \frac{a_1 + a_{n+1}}{a_n} > 1 + \frac{1}{n}. \]
Then
\[ \limsup_{n \to \infty} \left( \frac{a_1 + a_{n+1}}{a_n} \right)^n = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n = e. \] | 0 |
1950 | 1950_1 | For what values of the ratio $a/b$ is the limaçon $r = a - b\cos\theta$ a convex curve? ($a > b > 0$) Return the lower bound of the values. | The graph of $r = f(\theta)$ in polar coordinates is a simple closed curve surrounding the origin if $f$ is periodic with period $2\pi$ and everywhere positive. This is the case in the present problem, since by hypothesis $a > b > 0$. Such a curve is nonsingularly parametrized by $\theta$ if $r^2 + (r')^2 > 0$, again true in the present problem. The curvature is given by
\[ \kappa = \frac{r^2 + 2(r')^2 - rr''}{[r^2 + (r')^2]^{3/2}} \]
(whenever $f$ is of class $C^2$).
The curve is convex if and only if the curvature is everywhere non-negative, i.e., if and only if
\[ r^2 + 2(r')^2 - rr'' \geq 0. \]
For $r = a - b\cos\theta$, we have $r' = b\sin\theta$, $r'' = b\cos\theta$ and
\[ r^2 + 2(r')^2 - rr'' = a^2 + 2b^2 - 3ab\cos\theta. \]
This last expression is always non-negative if and only if
\[ a^2 + 2b^2 - 3ab \geq 0. \]
(Since $a$ and $b$ are positive, the least value occurs for $\theta = 0$.) This is equivalent to
\[ (a - 2b)(a - b) \geq 0, \]
and since $a - b > 0$ by hypothesis, to
\[ a \geq 2b. \]
Thus the limaçon is convex if and only if $a \geq 2b$ which makes the lower bound \boxed{2}. | numerical | putnam (modified boxing) | Geometry Analysis | For what values of the ratio $a/b$ is the limaçon $r = a - b\cos\theta$ a convex curve? ($a > b > 0$) | The graph of $r = f(\theta)$ in polar coordinates is a simple closed curve surrounding the origin if $f$ is periodic with period $2\pi$ and everywhere positive. This is the case in the present problem, since by hypothesis $a > b > 0$. Such a curve is nonsingularly parametrized by $\theta$ if $r^2 + (r')^2 > 0$, again true in the present problem. The curvature is given by
\[ \kappa = \frac{r^2 + 2(r')^2 - rr''}{[r^2 + (r')^2]^{3/2}} \]
(whenever $f$ is of class $C^2$).
The curve is convex if and only if the curvature is everywhere non-negative, i.e., if and only if
\[ r^2 + 2(r')^2 - rr'' \geq 0. \]
For $r = a - b\cos\theta$, we have $r' = b\sin\theta$, $r'' = b\cos\theta$ and
\[ r^2 + 2(r')^2 - rr'' = a^2 + 2b^2 - 3ab\cos\theta. \]
This last expression is always non-negative if and only if
\[ a^2 + 2b^2 - 3ab \geq 0. \]
(Since $a$ and $b$ are positive, the least value occurs for $\theta = 0$.) This is equivalent to
\[ (a - 2b)(a - b) \geq 0, \]
and since $a - b > 0$ by hypothesis, to
\[ a \geq 2b. \]
Thus the limaçon is convex if and only if $\boxed{a \geq 2b}$. | 0 |
1950 | 1950_3 | The sequence $x_0, x_1, x_2, \dots$ is defined by the conditions
\[ x_0 = a, \quad x_1 = b, \quad x_{n+1} = \frac{x_{n-1} + (2n-1)x_n}{2n} \quad \text{for } n \geq 1, \]
where $a$ and $b$ are given numbers. Express $\lim_{n \to \infty} x_n$ concisely in terms of $a$ and $b$. | The recursion can be rearranged as
\[ x_{n+1} - x_n = -\frac{1}{2n} (x_n - x_{n-1}) \]
whence it follows that
\[ x_{n+1} - x_n = \left(-\frac{1}{2}\right)^n \frac{1}{n!} (x_1 - x_0) = \left(-\frac{1}{2}\right)^n \frac{1}{n!} (b - a). \]
But
\[ x_{n+1} = x_0 + \sum_{i=0}^n (x_{i+1} - x_i) = a + (b - a) \sum_{i=0}^n \left(-\frac{1}{2}\right)^i \frac{1}{i!}. \]
The sum on the right is the partial sum of the power series for $e^{-1/2}$. Therefore
\[ \lim_{n \to \infty} x_n = \boxed{a + (b - a)e^{-1/2}}. \] | algebraic | putnam | Analysis Algebra | The sequence $x_0, x_1, x_2, \dots$ is defined by the conditions
\[ x_0 = a, \quad x_1 = b, \quad x_{n+1} = \frac{x_{n-1} + (2n-1)x_n}{2n} \quad \text{for } n \geq 1, \]
where $a$ and $b$ are given numbers. Express $\lim_{n \to \infty} x_n$ concisely in terms of $a$ and $b$. | The recursion can be rearranged as
\[ x_{n+1} - x_n = -\frac{1}{2n} (x_n - x_{n-1}) \]
whence it follows that
\[ x_{n+1} - x_n = \left(-\frac{1}{2}\right)^n \frac{1}{n!} (x_1 - x_0) = \left(-\frac{1}{2}\right)^n \frac{1}{n!} (b - a). \]
But
\[ x_{n+1} = x_0 + \sum_{i=0}^n (x_{i+1} - x_i) = a + (b - a) \sum_{i=0}^n \left(-\frac{1}{2}\right)^i \frac{1}{i!}. \]
The sum on the right is the partial sum of the power series for $e^{-1/2}$. Therefore
\[ \lim_{n \to \infty} x_n = \boxed{a + (b - a)e^{-1/2}}. \] | 0 |
1950 | 1950_5 | A function $D(n)$ of the positive integral variable $n$ is defined by the following properties: $D(1) = 0, D(p) = 1$ if $p$ is a prime, $D(uv) = uD(v) + vD(u)$ for any two positive integers $u$ and $v$. Answer all three parts below.
(i) Show that these properties are compatible and determine uniquely $D(n)$. (Derive a formula for $D(n)/n$, assuming that $n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$ where $p_1, p_2, \ldots, p_k$ are different primes.)
(ii) For what values of $n$ is $D(n) = n$?
(iii) Define $D^2(n) = D(D(n))$, etc., and find the limit of $D^m(63)$ as $m$ tends to $\infty$. Return (iii) as your final answer. | (i) Suppose there is a function $D$ with the required properties. We have
\[ \frac{D(uv)}{uv} = \frac{D(u)}{u} + \frac{D(v)}{v} \]
and by induction
\[ \frac{D(u_1u_2\cdots u_k)}{u_1u_2\cdots u_k} = \sum_i \frac{D(u_i)}{u_i}. \]
Hence
\[ \frac{D(p^\alpha)}{p^\alpha} = \alpha \frac{D(p)}{p} = \frac{\alpha}{p} \]
if $p$ is a prime, and for any integer $n$ with prime factorization $p_1^{\alpha_1} \cdots p_k^{\alpha_k}$ we have
\[ \frac{D(n)}{n} = \sum_i \frac{\alpha_i}{p_i}. \]
This equation shows that there is at most one function with the given properties.
On the other hand, since every integer $n > 1$ has a factorization into primes that is unique apart from order, and since the order in which the primes are numbered does not affect the sum in (2), we can define $D(1) = 0$ and use (2) to define $D(n)$ for $n > 1$. So defined, $D(p) = 1$ for $p$ a prime, and $D$ clearly satisfies (1), which is equivalent to $D(uv) = uD(v) + vD(u)$. Thus there is a unique function with the prescribed properties.
We note for future reference that $D(n) > 0$ for $n > 1$.
(ii) The equation $D(n) = n$ is equivalent to
\[ \frac{\alpha_1}{p_1} + \frac{\alpha_2}{p_2} + \cdots + \frac{\alpha_k}{p_k} = 1, \]
where $n = p_1^{\alpha_1} \cdots p_k^{\alpha_k}$ is the prime factorization of $n$. If (3) is multiplied through by $p_1p_2 \cdots p_k$, we see that $p_1p_2 \cdots p_k - \alpha_k/p_k$ is an integer. Since the $p$'s are all different, we conclude that $p_k$ divides $\alpha_k$. So $\alpha_k/p_k$ is an integer, and it is clear from (3) that $k = 1$ and $\alpha_k = p_k$. Thus any solution of $D(n) = n$ has the form $n = p^p$ where $p$ is prime. Conversely, any such $n$ is a solution.
(iii) $D(63) = 51,$
\[ D^2(63) = D(51) = 20, \]
\[ D^3(63) = D(20) = 24, \]
\[ D^4(63) = D(24) = 44, \]
\[ D^5(63) = D(44) = 48. \]
It appears that $D^m(63)$ has started to increase with $m$.
Suppose $n = 4k$ where $k > 1$. Then $D(n) = D(4)k + 4D(k) = 4k + D(k) > 4k = n$. Thus, if $n > 4$ and $n$ is divisible by 4, then $D(n) > n$ and $D(n)$ is divisible by 4. This implies that the sequence
\[ D(n), D^2(n), D^3(n), \ldots, D^m(n), \ldots \]
is strictly increasing. Since $D$ takes integral values, $D^m(n) \to \infty$ as $m \to \infty$ whenever $n = 4k, k > 1$.
Applying this result to the case above, we see that $D^m(63) = D^{m-2}(20) \to \boxed{\infty}$ as $m \to \infty$. | numerical | putnam (modified boxing) | Number Theory Algebra | A function $D(n)$ of the positive integral variable $n$ is defined by the following properties: $D(1) = 0, D(p) = 1$ if $p$ is a prime, $D(uv) = uD(v) + vD(u)$ for any two positive integers $u$ and $v$. Answer all three parts below.
(i) Show that these properties are compatible and determine uniquely $D(n)$. (Derive a formula for $D(n)/n$, assuming that $n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$ where $p_1, p_2, \ldots, p_k$ are different primes.)
(ii) For what values of $n$ is $D(n) = n$?
(iii) Define $D^2(n) = D(D(n))$, etc., and find the limit of $D^m(63)$ as $m$ tends to $\infty$. | (i) Suppose there is a function $D$ with the required properties. We have
\[ \frac{D(uv)}{uv} = \frac{D(u)}{u} + \frac{D(v)}{v} \]
and by induction
\[ \frac{D(u_1u_2\cdots u_k)}{u_1u_2\cdots u_k} = \sum_i \frac{D(u_i)}{u_i}. \]
Hence
\[ \frac{D(p^\alpha)}{p^\alpha} = \alpha \frac{D(p)}{p} = \frac{\alpha}{p} \]
if $p$ is a prime, and for any integer $n$ with prime factorization $p_1^{\alpha_1} \cdots p_k^{\alpha_k}$ we have
\[ \frac{D(n)}{n} = \sum_i \frac{\alpha_i}{p_i}. \]
This equation shows that there is at most one function with the given properties.
On the other hand, since every integer $n > 1$ has a factorization into primes that is unique apart from order, and since the order in which the primes are numbered does not affect the sum in (2), we can define $D(1) = 0$ and use (2) to define $D(n)$ for $n > 1$. So defined, $D(p) = 1$ for $p$ a prime, and $D$ clearly satisfies (1), which is equivalent to $D(uv) = uD(v) + vD(u)$. Thus there is a unique function with the prescribed properties.
We note for future reference that $D(n) > 0$ for $n > 1$.
(ii) The equation $D(n) = n$ is equivalent to
\[ \frac{\alpha_1}{p_1} + \frac{\alpha_2}{p_2} + \cdots + \frac{\alpha_k}{p_k} = 1, \]
where $n = p_1^{\alpha_1} \cdots p_k^{\alpha_k}$ is the prime factorization of $n$. If (3) is multiplied through by $p_1p_2 \cdots p_k$, we see that $p_1p_2 \cdots p_k - \alpha_k/p_k$ is an integer. Since the $p$'s are all different, we conclude that $p_k$ divides $\alpha_k$. So $\alpha_k/p_k$ is an integer, and it is clear from (3) that $k = 1$ and $\alpha_k = p_k$. Thus any solution of $D(n) = n$ has the form $n = p^p$ where $p$ is prime. Conversely, any such $n$ is a solution.
(iii) $D(63) = 51,$
\[ D^2(63) = D(51) = 20, \]
\[ D^3(63) = D(20) = 24, \]
\[ D^4(63) = D(24) = 44, \]
\[ D^5(63) = D(44) = 48. \]
It appears that $D^m(63)$ has started to increase with $m$.
Suppose $n = 4k$ where $k > 1$. Then $D(n) = D(4)k + 4D(k) = 4k + D(k) > 4k = n$. Thus, if $n > 4$ and $n$ is divisible by 4, then $D(n) > n$ and $D(n)$ is divisible by 4. This implies that the sequence
\[ D(n), D^2(n), D^3(n), \ldots, D^m(n), \ldots \]
is strictly increasing. Since $D$ takes integral values, $D^m(n) \to \infty$ as $m \to \infty$ whenever $n = 4k, k > 1$.
Applying this result to the case above, we see that $D^m(63) = D^{m-2}(20) \to \infty$ as $m \to \infty$. | 0 |
1950 | 1950_8 | Two obvious approximations to the length of the perimeter of the ellipse with semi-axes $a$ and $b$ are $\pi(a + b)$ and $2\pi\sqrt{ab}$. Which one comes nearer to the truth when the ratio $b/a$ is very close to 1? | Let the ellipse be taken in the parametric form $x = a \cos t, \; y = b \sin t$. Then the length $L$ is given by \[ L = \int_0^{2\pi} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt = \int_0^{2\pi} \sqrt{a^2 \sin^2 t + b^2 \cos^2 t} dt, \] a well-known elliptic integral. Since we are asked to consider $L$ when $b/a$ is nearly 1 we put $b = (1 + \lambda)a$ and consider \[ L(\lambda) = a \int_0^{2\pi} \sqrt{1 + (2\lambda + \lambda^2) \cos^2 t} dt. \] This is evidently an analytic function of $\lambda$, and we calculate its power series to terms of degree 2 \[ L(\lambda) = a \int_0^{2\pi} \left( 1 + \frac{1}{2} (2\lambda + \lambda^2) \cos^2 t - \frac{1}{8} (2\lambda + \lambda^2)^2 \cos^4 t + \cdots \right) dt \] \[ = 2\pi a \left[ 1 + \frac{1}{4} (2\lambda + \lambda^2) - \frac{3}{64} (2\lambda + \lambda^2)^2 + \cdots \right] \] \[ = 2\pi a \left[ 1 + \frac{1}{2} \lambda + \frac{1}{16} \lambda^2 + \cdots \right]. \] The first expression was obtained using the binomial expansion of $(1 + z)^{1/2}$. All the terms omitted are of degree at least three in $\lambda$. The formal manipulation is justified because the series in question converges absolutely for small values of $\lambda$ (in fact if $|2\lambda| + \lambda^2 < 1$). (We could also find the power series by differentiating under the integral sign.)
The proposed approximations to the perimeter are \[ \pi(a + b) = 2\pi a \left( 1 + \frac{1}{2} \lambda \right) \] and \[ 2\pi \sqrt{ab} = 2\pi a \sqrt{1 + \lambda} = 2\pi a \left( 1 + \frac{1}{2} \lambda - \frac{1}{8} \lambda^2 + \cdots \right). \] Since the three functions have the same constant and first degree terms their differences are controlled by the second degree terms for small $\lambda$. We have \[ L(\lambda) > 2\pi a \left( 1 + \frac{1}{2} \lambda \right) > 2\pi a \left( 1 + \frac{1}{2} \lambda - \frac{1}{8} \lambda^2 + \cdots \right). \] For small $\lambda$. Thus $\boxed{\pi(a + b)}$ is a better approximation to the length of an ellipse than $2\pi\sqrt{ab}$; in fact it is roughly three times better since \[ L(\lambda) - \pi(a + b) \approx 3 \left(L(\lambda) - 2\pi \sqrt{ab}\right). \] | algebraic | putnam | Analysis Calculus | Two obvious approximations to the length of the perimeter of the ellipse with semi-axes $a$ and $b$ are $\pi(a + b)$ and $2\pi\sqrt{ab}$. Which one comes nearer to the truth when the ratio $b/a$ is very close to 1? | Let the ellipse be taken in the parametric form $x = a \cos t, \; y = b \sin t$. Then the length $L$ is given by \[ L = \int_0^{2\pi} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt = \int_0^{2\pi} \sqrt{a^2 \sin^2 t + b^2 \cos^2 t} dt, \] a well-known elliptic integral. Since we are asked to consider $L$ when $b/a$ is nearly 1 we put $b = (1 + \lambda)a$ and consider \[ L(\lambda) = a \int_0^{2\pi} \sqrt{1 + (2\lambda + \lambda^2) \cos^2 t} dt. \] This is evidently an analytic function of $\lambda$, and we calculate its power series to terms of degree 2 \[ L(\lambda) = a \int_0^{2\pi} \left( 1 + \frac{1}{2} (2\lambda + \lambda^2) \cos^2 t - \frac{1}{8} (2\lambda + \lambda^2)^2 \cos^4 t + \cdots \right) dt \] \[ = 2\pi a \left[ 1 + \frac{1}{4} (2\lambda + \lambda^2) - \frac{3}{64} (2\lambda + \lambda^2)^2 + \cdots \right] \] \[ = 2\pi a \left[ 1 + \frac{1}{2} \lambda + \frac{1}{16} \lambda^2 + \cdots \right]. \] The first expression was obtained using the binomial expansion of $(1 + z)^{1/2}$. All the terms omitted are of degree at least three in $\lambda$. The formal manipulation is justified because the series in question converges absolutely for small values of $\lambda$ (in fact if $|2\lambda| + \lambda^2 < 1$). (We could also find the power series by differentiating under the integral sign.)
The proposed approximations to the perimeter are \[ \pi(a + b) = 2\pi a \left( 1 + \frac{1}{2} \lambda \right) \] and \[ 2\pi \sqrt{ab} = 2\pi a \sqrt{1 + \lambda} = 2\pi a \left( 1 + \frac{1}{2} \lambda - \frac{1}{8} \lambda^2 + \cdots \right). \] Since the three functions have the same constant and first degree terms their differences are controlled by the second degree terms for small $\lambda$. We have \[ L(\lambda) > 2\pi a \left( 1 + \frac{1}{2} \lambda \right) > 2\pi a \left( 1 + \frac{1}{2} \lambda - \frac{1}{8} \lambda^2 + \cdots \right). \] For small $\lambda$. Thus $\boxed{\pi(a + b)}$ is a better approximation to the length of an ellipse than $2\pi\sqrt{ab}$; in fact it is roughly three times better since \[ L(\lambda) - \pi(a + b) \approx 3 \left(L(\lambda) - 2\pi \sqrt{ab}\right). \] | 0 |
1950 | 1950_9 | In the Gregorian calendar: \begin{enumerate} \item[(i)] years not divisible by 4 are common years; \item[(ii)] years divisible by 4 but not by 100 are leap years; \item[(iii)] years divisible by 100 but not by 400 are common years; \item[(iv)] years divisible by 400 are leap years; \item[(v)] a leap year contains 366 days; a common year 365 days. \end{enumerate} Find the probability that Christmas falls on a Wednesday given its calculated across $N$ years. | According to the rules given, any 400 consecutive Gregorian years will involve 303 ordinary years and 97 leap years making a total of $400 \cdot 365 + 97$ days. Since this number is divisible by 7 ($400 \equiv 1, \; 365 \equiv 1, \; 97 \equiv -1 \; \text{(mod 7)}$), there is an integral number of weeks in 400 years (in fact, 20,871 weeks) and therefore the day of the week on which Christmas (or any other calendar date) falls repeats in cycles of 400. If there are $N$ years in such a period on which Christmas falls on Wednesday, then the probability that Christmas falls on Wednesday is $\boxed{N/400}$. But $N/400 \neq 1/7$ for any integer $N$. | algebraic | putnam (modified boxing) | Probability Number Theory | In the Gregorian calendar: \begin{enumerate} \item[(i)] years not divisible by 4 are common years; \item[(ii)] years divisible by 4 but not by 100 are leap years; \item[(iii)] years divisible by 100 but not by 400 are common years; \item[(iv)] years divisible by 400 are leap years; \item[(v)] a leap year contains 366 days; a common year 365 days. \end{enumerate} Prove that the probability that Christmas falls on a Wednesday is not $1/7$. | According to the rules given, any 400 consecutive Gregorian years will involve 303 ordinary years and 97 leap years making a total of $400 \cdot 365 + 97$ days. Since this number is divisible by 7 ($400 \equiv 1, \; 365 \equiv 1, \; 97 \equiv -1 \; \text{(mod 7)}$), there is an integral number of weeks in 400 years (in fact, 20,871 weeks) and therefore the day of the week on which Christmas (or any other calendar date) falls repeats in cycles of 400. If there are $N$ years in such a period on which Christmas falls on Wednesday, then the probability that Christmas falls on Wednesday is $N/400$. But $N/400 \neq 1/7$ for any integer $N$. | 0 |
1950 | 1950_10 | The cross-section of a right cylinder is an ellipse, with semi-axes $a$ and $b$, where $a > b$. The cylinder is very long, made of very light homogeneous material. The cylinder rests on the horizontal ground which it touches along the straight line joining the lower endpoints of the minor axes of its several cross-sections. Along the upper endpoints of these minor axes lies a very heavy homogeneous wire, straight and just as long as the cylinder. The wire and the cylinder are rigidly connected. We neglect the weight of the cylinder, the breadth of the wire, and the friction of the ground. \newline The system described is in equilibrium, because of its symmetry. This equilibrium seems to be stable when the ratio $b/a$ is very small, but unstable when this ratio comes close to 1. Examine this assertion and find the value of the ratio $b/a$ which separates the cases of stable and unstable equilibrium. | Because the cylinder is long, the only displacements we need to consider are the rolling motions of the cylinder. Hence we may confine our attention to a plane perpendicular to the axis of the cylinder, and the problem becomes effectively two-dimensional. It is equivalent to the following. An ellipse, whose equation we take to be \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \] is restricted to the plane but is free to roll along the line $y = -b$. The weight of the ellipse is negligible but there is a heavy particle attached to the ellipse at $P = (0, b)$. The force of gravity acts parallel to the $y$-axis in the negative direction. We are to determine, in terms of $a/b$, the condition for stability. Suppose the ellipse rolls a little way from the starting position. If this motion raises $P$ it will be opposed by the force of gravity and the ellipse will be in stable equilibrium when resting on $Q$. On the other hand, if a slight motion lowers $P$ the equilibrium at $Q$ will be unstable. \newline Thus, to decide whether the equilibrium is stable or not, we consider the distance $|PZ|$ from $P$ to a variable point $Z$ on the ellipse. If this function has a strict local minimum at $Q$ then the equilibrium is stable. On the other hand, if $|PZ|$ has a strict local maximum for $Z = Q$, the equilibrium is unstable. (For a sharper form of this criterion, see the solution to Problem A.M. 7 of the Fourth Competition.) We can equally well consider the function $|PZ|^2$. \newline Suppose $Z$ is the point $(a \sin t, -b \cos t)$ (which is on the ellipse). Then $t = 0$ corresponds to $Z = Q$, and \[ |PZ|^2 = a^2 \sin^2 t + b^2 (1 + \cos t)^2. \] If we put $a = rb$ this becomes \[ b^2 [(r^2 - 1) \sin^2 t + 2 + 2 \cos t]. \] The first derivative (with respect to $t$) is \[ b^2 [(r^2 - 1) \sin 2t - 2 \sin t], \] which vanishes for $t = 0$; and the second derivative is \[ b^2 [2(r^2 - 1) \cos 2t - 2 \cos t] \] which is $2b^2(r^2 - 2)$ for $t = 0$. Hence we have a strict local maximum at $t = 0$ (and therefore instability) if $r^2 < 2$ and a strict local minimum (and therefore stability) if $r^2 > 2$. The critical value of $b/a$ (=$1/r$) is therefore $\boxed{\frac{1}{\sqrt{2}}}$. | numerical | putnam | Geometry | The cross-section of a right cylinder is an ellipse, with semi-axes $a$ and $b$, where $a > b$. The cylinder is very long, made of very light homogeneous material. The cylinder rests on the horizontal ground which it touches along the straight line joining the lower endpoints of the minor axes of its several cross-sections. Along the upper endpoints of these minor axes lies a very heavy homogeneous wire, straight and just as long as the cylinder. The wire and the cylinder are rigidly connected. We neglect the weight of the cylinder, the breadth of the wire, and the friction of the ground. \newline The system described is in equilibrium, because of its symmetry. This equilibrium seems to be stable when the ratio $b/a$ is very small, but unstable when this ratio comes close to 1. Examine this assertion and find the value of the ratio $b/a$ which separates the cases of stable and unstable equilibrium. | Because the cylinder is long, the only displacements we need to consider are the rolling motions of the cylinder. Hence we may confine our attention to a plane perpendicular to the axis of the cylinder, and the problem becomes effectively two-dimensional. It is equivalent to the following. An ellipse, whose equation we take to be \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \] is restricted to the plane but is free to roll along the line $y = -b$. The weight of the ellipse is negligible but there is a heavy particle attached to the ellipse at $P = (0, b)$. The force of gravity acts parallel to the $y$-axis in the negative direction. We are to determine, in terms of $a/b$, the condition for stability. Suppose the ellipse rolls a little way from the starting position. If this motion raises $P$ it will be opposed by the force of gravity and the ellipse will be in stable equilibrium when resting on $Q$. On the other hand, if a slight motion lowers $P$ the equilibrium at $Q$ will be unstable. \newline Thus, to decide whether the equilibrium is stable or not, we consider the distance $|PZ|$ from $P$ to a variable point $Z$ on the ellipse. If this function has a strict local minimum at $Q$ then the equilibrium is stable. On the other hand, if $|PZ|$ has a strict local maximum for $Z = Q$, the equilibrium is unstable. (For a sharper form of this criterion, see the solution to Problem A.M. 7 of the Fourth Competition.) We can equally well consider the function $|PZ|^2$. \newline Suppose $Z$ is the point $(a \sin t, -b \cos t)$ (which is on the ellipse). Then $t = 0$ corresponds to $Z = Q$, and \[ |PZ|^2 = a^2 \sin^2 t + b^2 (1 + \cos t)^2. \] If we put $a = rb$ this becomes \[ b^2 [(r^2 - 1) \sin^2 t + 2 + 2 \cos t]. \] The first derivative (with respect to $t$) is \[ b^2 [(r^2 - 1) \sin 2t - 2 \sin t], \] which vanishes for $t = 0$; and the second derivative is \[ b^2 [2(r^2 - 1) \cos 2t - 2 \cos t] \] which is $2b^2(r^2 - 2)$ for $t = 0$. Hence we have a strict local maximum at $t = 0$ (and therefore instability) if $r^2 < 2$ and a strict local minimum (and therefore stability) if $r^2 > 2$. The critical value of $b/a$ (=$1/r$) is therefore $\boxed{\frac{1}{\sqrt{2}}}$. | 0 |
1950 | 1950_11_ii | A plane varies so that it includes a cone of constant volume equal to $\pi a^3 / 3$ with the surface the equation of which in rectangular coordinates is $2xy = z$. Find the equation of the envelope of the various positions of this plane. \newline State the result so that it applies to a general cone (that is, conic surface) of the second order. | Make the change of coordinates \[ x = \frac{1}{2} \sqrt{2} (u + v), \quad y = \frac{1}{2} \sqrt{2} (u - v). \] Then the $uv$-axes are orthogonal and rotated by $\pi/4$ radians from the $xy$-axes. The equation of the given surface in the new coordinates is \[ z^2 + v^2 = uz, \] which is a right circular cone with the $u$-axis. \newline Next we find the volume of the conical region cut off from the solid cone by a plane. Because of rotational symmetry we need only consider planes of the form $v = mu + b$. In order that the plane should cut off a bounded region it is necessary that $|m| < 1$. The region cut off is then a cone with an elliptical base. We will find the altitude of that cone and the area of its base. \newline The altitude is the distance from the origin to the plane, namely \[ \frac{|b|}{\sqrt{1 + m^2}}. \] The area of the base is $\sqrt{1 + m^2}$ times the area of the ellipse obtained by projecting it orthogonally onto the $vz$-plane. To find the equation of the projected ellipse we eliminate $u$ between the equations of the cone and the plane, getting \[ z^2 + v^2 = (mv + b)^2. \] Collecting the $v$ terms and completing the square we get \[ z^2 + (1 - m^2) \left( v - \frac{mb}{1 - m^2} \right)^2 = b^2 \frac{1}{1 - m^2}. \] [Note that this is indeed an ellipse because $|m| < 1$.] The area of this ellipse is \[ A = \pi b^2 \frac{1}{(1 - m^2)}. \] The volume of the conical region is therefore \[ \frac{1}{3} \text{ base } \times \text{ altitude } = \frac{1}{3} \left( \sqrt{1 + m^2} \pi b^2 \frac{1}{(1 - m^2)} \right) \left( \frac{|b|}{\sqrt{1 + m^2}} \right) = \frac{1}{3} \pi |b| \frac{1}{(1 - m^2)^2}. \] The problem restricts consideration to those planes which cut off a volume $\pi a^3 / 3$, that is, to planes for which \[ |b| = a \sqrt{1 - m^2}, \] where $m$ is the tangent of the angle between the plane and the $uv$-plane. \newline We are to find the envelope $E$ of all such planes. Clearly $E$ must share the rotational symmetry of the entire configuration, so it suffices to find the intersection of $E$ with the $uv$-plane. \newline Consider the plane $P$ tangent to $E$ at a point of $I$. Reflection in the $uv$-plane preserves $P$, since it preserves $E$ and fixes each point of $I$. Therefore $P$ is perpendicular to the $uv$-plane, so it has an equation of the form $v = mu + b$ where, as we have seen, $|b| = a \sqrt{1 - m^2}$. The line in which $P$ meets the $uv$-plane is clearly tangent to $I$. Hence $I$ is the envelope of all lines in the $uv$-plane having equations of the form \[ (1) \quad v = mu \pm a \sqrt{1 - m^2}. \] The envelope problem is thus reduced to two dimensions. \newline To find the envelope, we first take the positive sign in (1), and eliminate $m$ between the equation \[ (2) \quad v = mu + \sqrt{1 - m^2} \] and the equation obtained by differentiating (2) with respect to $m$, namely \[ (3) \quad 0 = v - \frac{m}{\sqrt{1 - m^2}} u + a \sqrt{1 - m^2}. \] From (3) we find \[ m^2 = \frac{v^2}{v^2 + a^2}, \quad \text{so } 1 - m^2 = \frac{a^2}{v^2 + a^2}. \] Equation (2) now becomes \[ u = \sqrt{1 - m^2} \left( \frac{m}{\sqrt{1 - m^2}} v + a \right) = \sqrt{1 - m^2} \sqrt{1 - m^2} \left( \frac{v^2}{a^2} + 1 \right), \] giving finally \[ (4) \quad u^2 = v^2 + a^2 \] as the equation of the envelope $I$. The same equation results if we start with the negative sign. \newline [Since the algebraic work could introduce extraneous points not in the locus, we should check to see that the given family of lines is exactly the family of tangent lines to the hyperbola (4). This examination reveals that the tangents to the upper branch of the hyperbola involve the plus sign in (1), while the tangents to the lower branch involve the minus sign. Thus the lower branch is indeed extraneous to the problem of finding the envelope of the family of lines (2).] \newline The three-dimensional envelope $E$ is obtained by revolving $I$ around the $u$ axis; hence its equation is \[ u^2 = v^2 + z^2 + a^2. \] Transforming back to the original coordinates its equation is \[ \boxed{2xy = z^2 + a^2}. \] | algebraic | putnam | Geometry Analysis | (ii) A plane varies so that it includes a cone of constant volume equal to $\pi a^3 / 3$ with the surface the equation of which in rectangular coordinates is $2xy = z$. Find the equation of the envelope of the various positions of this plane. \newline State the result so that it applies to a general cone (that is, conic surface) of the second order. | Make the change of coordinates \[ x = \frac{1}{2} \sqrt{2} (u + v), \quad y = \frac{1}{2} \sqrt{2} (u - v). \] Then the $uv$-axes are orthogonal and rotated by $\pi/4$ radians from the $xy$-axes. The equation of the given surface in the new coordinates is \[ z^2 + v^2 = uz, \] which is a right circular cone with the $u$-axis. \newline Next we find the volume of the conical region cut off from the solid cone by a plane. Because of rotational symmetry we need only consider planes of the form $v = mu + b$. In order that the plane should cut off a bounded region it is necessary that $|m| < 1$. The region cut off is then a cone with an elliptical base. We will find the altitude of that cone and the area of its base. \newline The altitude is the distance from the origin to the plane, namely \[ \frac{|b|}{\sqrt{1 + m^2}}. \] The area of the base is $\sqrt{1 + m^2}$ times the area of the ellipse obtained by projecting it orthogonally onto the $vz$-plane. To find the equation of the projected ellipse we eliminate $u$ between the equations of the cone and the plane, getting \[ z^2 + v^2 = (mv + b)^2. \] Collecting the $v$ terms and completing the square we get \[ z^2 + (1 - m^2) \left( v - \frac{mb}{1 - m^2} \right)^2 = b^2 \frac{1}{1 - m^2}. \] [Note that this is indeed an ellipse because $|m| < 1$.] The area of this ellipse is \[ A = \pi b^2 \frac{1}{(1 - m^2)}. \] The volume of the conical region is therefore \[ \frac{1}{3} \text{ base } \times \text{ altitude } = \frac{1}{3} \left( \sqrt{1 + m^2} \pi b^2 \frac{1}{(1 - m^2)} \right) \left( \frac{|b|}{\sqrt{1 + m^2}} \right) = \frac{1}{3} \pi |b| \frac{1}{(1 - m^2)^2}. \] The problem restricts consideration to those planes which cut off a volume $\pi a^3 / 3$, that is, to planes for which \[ |b| = a \sqrt{1 - m^2}, \] where $m$ is the tangent of the angle between the plane and the $uv$-plane. \newline We are to find the envelope $E$ of all such planes. Clearly $E$ must share the rotational symmetry of the entire configuration, so it suffices to find the intersection of $E$ with the $uv$-plane. \newline Consider the plane $P$ tangent to $E$ at a point of $I$. Reflection in the $uv$-plane preserves $P$, since it preserves $E$ and fixes each point of $I$. Therefore $P$ is perpendicular to the $uv$-plane, so it has an equation of the form $v = mu + b$ where, as we have seen, $|b| = a \sqrt{1 - m^2}$. The line in which $P$ meets the $uv$-plane is clearly tangent to $I$. Hence $I$ is the envelope of all lines in the $uv$-plane having equations of the form \[ (1) \quad v = mu \pm a \sqrt{1 - m^2}. \] The envelope problem is thus reduced to two dimensions. \newline To find the envelope, we first take the positive sign in (1), and eliminate $m$ between the equation \[ (2) \quad v = mu + \sqrt{1 - m^2} \] and the equation obtained by differentiating (2) with respect to $m$, namely \[ (3) \quad 0 = v - \frac{m}{\sqrt{1 - m^2}} u + a \sqrt{1 - m^2}. \] From (3) we find \[ m^2 = \frac{v^2}{v^2 + a^2}, \quad \text{so } 1 - m^2 = \frac{a^2}{v^2 + a^2}. \] Equation (2) now becomes \[ u = \sqrt{1 - m^2} \left( \frac{m}{\sqrt{1 - m^2}} v + a \right) = \sqrt{1 - m^2} \sqrt{1 - m^2} \left( \frac{v^2}{a^2} + 1 \right), \] giving finally \[ (4) \quad u^2 = v^2 + a^2 \] as the equation of the envelope $I$. The same equation results if we start with the negative sign. \newline [Since the algebraic work could introduce extraneous points not in the locus, we should check to see that the given family of lines is exactly the family of tangent lines to the hyperbola (4). This examination reveals that the tangents to the upper branch of the hyperbola involve the plus sign in (1), while the tangents to the lower branch involve the minus sign. Thus the lower branch is indeed extraneous to the problem of finding the envelope of the family of lines (2).] \newline The three-dimensional envelope $E$ is obtained by revolving $I$ around the $u$ axis; hence its equation is \[ u^2 = v^2 + z^2 + a^2. \] Transforming back to the original coordinates its equation is \[ \boxed{2xy = z^2 + a^2}. \] | 0 |
1951 | 1951_3 | Find the sum to infinity of the series: \[ 1 - \frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \cdots + \frac{(-1)^{n+1}}{3n - 2} + \cdots. \] | We first show that \[ 1 - \frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \cdots + \frac{(-1)^{n+1}}{3n - 2} + \cdots = \int_0^1 \frac{dt}{1 + t^3}. \] \newline Note that \[ \frac{1}{1 + t^3} = \sum_{n=1}^k (-1)^{n+1} t^{3n-3} + \frac{(-1)^k t^{3k}}{1 + t^3} \] for any $t \neq -1$. \newline Integrate from 0 to 1 to obtain \[ \int_0^1 \frac{dt}{1 + t^3} - \sum_{n=1}^k (-1)^{n+1} \int_0^1 t^{3n-3} dt = (-1)^k \int_0^1 \frac{t^{3k} dt}{1 + t^3}. \] \newline Hence \[ \int_0^1 \frac{dt}{1 + t^3} - \sum_{n=1}^k \frac{(-1)^{n+1}}{3n - 2} = \left| \int_0^1 \frac{t^{3k} dt}{1 + t^3} \right| \leq \int_0^1 t^{3k} dt = \frac{1}{3k + 1}. \] \newline Letting $k \to \infty$, we get \[ \sum_{n=1}^\infty \frac{(-1)^{n+1}}{3n - 2} = \int_0^1 \frac{dt}{1 + t^3}. \] \newline This integral can be evaluated by partial fractions: \[ \int_0^1 \frac{dt}{1 + t^3} = \frac{1}{3} \int_0^1 \left[ \frac{1}{1 + t} + \frac{2 - t}{1 - t + t^2} \right] dt \] \[ = \frac{1}{3} \left[ \log(1 + t) - \frac{1}{2} \log(1 - t + t^2) + \sqrt{3} \arctan \frac{2t - 1}{\sqrt{3}} \right]_0^1 \] \[ = \frac{1}{3} \left[ \log 2 + \sqrt{3} \left( \frac{\pi}{6} + \frac{\pi}{6} \right) \right] = \boxed{\frac{1}{3} \left( \log 2 + \frac{\pi}{\sqrt{3}} \right)}. \] | algebraic | putnam | Analysis | Find the sum to infinity of the series: \[ 1 - \frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \cdots + \frac{(-1)^{n+1}}{3n - 2} + \cdots. \] | We first show that \[ 1 - \frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \cdots + \frac{(-1)^{n+1}}{3n - 2} + \cdots = \int_0^1 \frac{dt}{1 + t^3}. \] \newline Note that \[ \frac{1}{1 + t^3} = \sum_{n=1}^k (-1)^{n+1} t^{3n-3} + \frac{(-1)^k t^{3k}}{1 + t^3} \] for any $t \neq -1$. \newline Integrate from 0 to 1 to obtain \[ \int_0^1 \frac{dt}{1 + t^3} - \sum_{n=1}^k (-1)^{n+1} \int_0^1 t^{3n-3} dt = (-1)^k \int_0^1 \frac{t^{3k} dt}{1 + t^3}. \] \newline Hence \[ \int_0^1 \frac{dt}{1 + t^3} - \sum_{n=1}^k \frac{(-1)^{n+1}}{3n - 2} = \left| \int_0^1 \frac{t^{3k} dt}{1 + t^3} \right| \leq \int_0^1 t^{3k} dt = \frac{1}{3k + 1}. \] \newline Letting $k \to \infty$, we get \[ \sum_{n=1}^\infty \frac{(-1)^{n+1}}{3n - 2} = \int_0^1 \frac{dt}{1 + t^3}. \] \newline This integral can be evaluated by partial fractions: \[ \int_0^1 \frac{dt}{1 + t^3} = \frac{1}{3} \int_0^1 \left[ \frac{1}{1 + t} + \frac{2 - t}{1 - t + t^2} \right] dt \] \[ = \frac{1}{3} \left[ \log(1 + t) - \frac{1}{2} \log(1 - t + t^2) + \sqrt{3} \arctan \frac{2t - 1}{\sqrt{3}} \right]_0^1 \] \[ = \frac{1}{3} \left[ \log 2 + \sqrt{3} \left( \frac{\pi}{6} + \frac{\pi}{6} \right) \right] = \boxed{\frac{1}{3} \left( \log 2 + \frac{\pi}{\sqrt{3}} \right)}. \] | 0 |
1951 | 1951_6 | Determine the position of a normal chord of a parabola such that it cuts off of the parabola a segment of minimum area. | Choose coordinates so that the equation of the parabola is \( 4ay = x^2, \ a > 0 \). \newline The chord connecting the point \( P(2as, as^2) \) to the point \( Q(2at, at^2) \) has the equation \[ y = \frac{1}{2}(t + s)x - ast \] \newline and the tangent line at \( (2at, at^2) \) has slope \( t \). Hence the line (1) will be normal to the parabola at \( Q \) if and only if \[ \frac{1}{2}t(t + s) = -1, \] \newline which may be written as \[ s = -\frac{2}{t} - t. \] \newline We see, therefore, that \( s \) and \( t \) have opposite signs. Take \( s < 0 \) and \( t > 0 \). Then the area cut off by the chord is \[ \int_{2as}^{2at} \left[ \frac{1}{2}(t + s)x - ast - \frac{1}{4a}x^2 \right] dx = \frac{1}{3}a^2(t - s)^3. \] \newline This area will be minimal when \( t - s \) is minimal. But \[ t - s = 2t + \frac{2}{t} = 2\left( \sqrt{t} - \frac{1}{\sqrt{t}} \right)^2 + 4 \geq 4. \] \newline Equality is attained only when \( \sqrt{t} = 1 \) and hence \( t = 1 \). \newline Thus, of all normals to the parabola at points to the right of the axis, the normal at \( (2a, a) \) cuts off the least area. The area cut off is \( 64a^2/3 \). \newline By symmetry, the normal at \( \boxed{(-2a, a)} \) cuts off the least area among normals at points to the left of the axis. \newline The critical normals can be characterized as those which meet the axis at an angle of \( 45^\circ \). | algebraic | putnam | Geometry | Determine the position of a normal chord of a parabola such that it cuts off of the parabola a segment of minimum area. | Choose coordinates so that the equation of the parabola is \( 4ay = x^2, \ a > 0 \). \newline The chord connecting the point \( P(2as, as^2) \) to the point \( Q(2at, at^2) \) has the equation \[ y = \frac{1}{2}(t + s)x - ast \] \newline and the tangent line at \( (2at, at^2) \) has slope \( t \). Hence the line (1) will be normal to the parabola at \( Q \) if and only if \[ \frac{1}{2}t(t + s) = -1, \] \newline which may be written as \[ s = -\frac{2}{t} - t. \] \newline We see, therefore, that \( s \) and \( t \) have opposite signs. Take \( s < 0 \) and \( t > 0 \). Then the area cut off by the chord is \[ \int_{2as}^{2at} \left[ \frac{1}{2}(t + s)x - ast - \frac{1}{4a}x^2 \right] dx = \frac{1}{3}a^2(t - s)^3. \] \newline This area will be minimal when \( t - s \) is minimal. But \[ t - s = 2t + \frac{2}{t} = 2\left( \sqrt{t} - \frac{1}{\sqrt{t}} \right)^2 + 4 \geq 4. \] \newline Equality is attained only when \( \sqrt{t} = 1 \) and hence \( t = 1 \). \newline Thus, of all normals to the parabola at points to the right of the axis, the normal at \( (2a, a) \) cuts off the least area. The area cut off is \( 64a^2/3 \). \newline By symmetry, the normal at \( \boxed{(-2a, a)} \) cuts off the least area among normals at points to the left of the axis. \newline The critical normals can be characterized as those which meet the axis at an angle of \( 45^\circ \). | 0 |
1951 | 1951_13 | Assuming that all the roots of the cubic equation $x^3 + ax^2 + bx + c = 0$ are real, find the upper bound of the difference between the greatest and the least roots. | Let the roots be ordered so that $r_1 \leq r_2 \leq r_3$ and let $r_2 = r_1 + u$ and $r_3 = r_2 + v$ where $u$ and $v$ are non-negative. Then \[ a = -(r_1 + r_2 + r_3), \] \[ b = r_1r_2 + r_2r_3 + r_3r_1; \] \[ a^2 - 3b = \frac{1}{2} [(r_1 - r_2)^2 + (r_2 - r_3)^2 + (r_3 - r_1)^2] = \frac{1}{2} [u^2 + v^2 + (u + v)^2] = (u + v)^2 - uv. \] Since $u$ and $v$ are non-negative, this gives \[ a^2 - 3b \leq (u + v)^2. \] Since the difference between the greatest and the least roots is $u + v$, this difference is at least $(a^2 - 3b)^{1/2}$. \[ \text{On the other hand,} \] \[ u^2 + v^2 = \frac{1}{2} [(u + v)^2 + (u - v)^2] \geq \frac{1}{2} (u + v)^2, \] \[ \text{so} \ a^2 - 3b \geq \frac{3}{4} (u + v)^2 \] \[ \text{and the difference } u + v \text{ is at most } \boxed{\frac{2}{\sqrt{3}} (a^2 - 3b)^{1/2}}. \] | algebraic | putnam (modified boxing) | Algebra | Assuming that all the roots of the cubic equation $x^3 + ax^2 + bx + c = 0$ are real, show that the difference between the greatest and the least roots is not less than $(a^2 - 3b)^{1/2}$ or greater than $2(a^2 - 3b)^{1/2}/\sqrt{3}$. | Let the roots be ordered so that $r_1 \leq r_2 \leq r_3$ and let $r_2 = r_1 + u$ and $r_3 = r_2 + v$ where $u$ and $v$ are non-negative. Then \[ a = -(r_1 + r_2 + r_3), \] \[ b = r_1r_2 + r_2r_3 + r_3r_1; \] \[ a^2 - 3b = \frac{1}{2} [(r_1 - r_2)^2 + (r_2 - r_3)^2 + (r_3 - r_1)^2] = \frac{1}{2} [u^2 + v^2 + (u + v)^2] = (u + v)^2 - uv. \] Since $u$ and $v$ are non-negative, this gives \[ a^2 - 3b \leq (u + v)^2. \] Since the difference between the greatest and the least roots is $u + v$, this difference is at least $(a^2 - 3b)^{1/2}$. \[ \text{On the other hand,} \] \[ u^2 + v^2 = \frac{1}{2} [(u + v)^2 + (u - v)^2] \geq \frac{1}{2} (u + v)^2, \] \[ \text{so} \ a^2 - 3b \geq \frac{3}{4} (u + v)^2 \] \[ \text{and the difference } u + v \text{ is at most } \frac{2}{\sqrt{3}} (a^2 - 3b)^{1/2}. \] | 0 |
1951 | 1951_14 | Find the volume of the four-dimensional hypersphere $x^2 + y^2 + z^2 + t^2 = r^2$, and also the hypervolume of its interior $x^2 + y^2 + z^2 + t^2 < r^2$. Return the hypervolume as the final answer. | Let $V_4(r)$ be the hypervolume of the interior of the hypersphere, i.e., the four-dimensional ball of radius $r$. If the hypervolume is "sliced" perpendicular to the $x$-axis, one gets \[ V_4(r) = \int_{-r}^r V_3(\sqrt{r^2 - x^2}) dx \] where \[ V_3(\rho) = (4/3) \pi \rho^3 \] is the ordinary volume of the three-dimensional ball of radius $\rho$. Hence \[ V_4(r) = (4/3) \pi \int_{-r}^r (r^2 - x^2)^{3/2} dx. \] Making the substitution $x = r \sin \theta$ \[ V_4(r) = \frac{4 \pi r^4}{3} \int_{-\pi/2}^{\pi/2} \cos^4 \theta \, d \theta \] \[ = \frac{\pi r^4}{3} \int_{-\pi/2}^{\pi/2} \left( \frac{3}{8} + \frac{1}{2} \cos 2\theta + \frac{1}{8} \cos 4\theta \right) d\theta, \] where we have used the double-angle formula $2 \cos^2 \theta = 1 + \cos 2\theta$ twice to simplify the integrand. Hence \[ V_4(r) = \frac{\pi r^4}{3} \cdot \frac{3}{2} \pi = \boxed{\frac{\pi^2 r^4}{2}}| \] is the hypervolume of the four-dimensional ball. Let $A_4(r)$ be the three-dimensional volume of the hyperspherical surface of radius $r$. We can relate $A_4$ to $V_4$ as follows. If we compute $V_4(r)$ by considering concentric spherical shells we see that \[ V_4(r) = \int_0^r A_4(\rho) d \rho. \] Hence by the fundamental theorem of calculus \[ \frac{d}{dr} V_4(r) = A_4(r), \] so that \[ A_4(r) = 2 \pi^2 r^3. \] | algebraic | putnam | Geometry Analysis | Find the volume of the four-dimensional hypersphere $x^2 + y^2 + z^2 + t^2 = r^2$, and also the hypervolume of its interior $x^2 + y^2 + z^2 + t^2 < r^2$. | Let $V_4(r)$ be the hypervolume of the interior of the hypersphere, i.e., the four-dimensional ball of radius $r$. If the hypervolume is "sliced" perpendicular to the $x$-axis, one gets \[ V_4(r) = \int_{-r}^r V_3(\sqrt{r^2 - x^2}) dx \] where \[ V_3(\rho) = (4/3) \pi \rho^3 \] is the ordinary volume of the three-dimensional ball of radius $\rho$. Hence \[ V_4(r) = (4/3) \pi \int_{-r}^r (r^2 - x^2)^{3/2} dx. \] Making the substitution $x = r \sin \theta$ \[ V_4(r) = \frac{4 \pi r^4}{3} \int_{-\pi/2}^{\pi/2} \cos^4 \theta \, d \theta \] \[ = \frac{\pi r^4}{3} \int_{-\pi/2}^{\pi/2} \left( \frac{3}{8} + \frac{1}{2} \cos 2\theta + \frac{1}{8} \cos 4\theta \right) d\theta, \] where we have used the double-angle formula $2 \cos^2 \theta = 1 + \cos 2\theta$ twice to simplify the integrand. Hence \[ V_4(r) = \frac{\pi r^4}{3} \cdot \frac{3}{2} \pi = \frac{\pi^2 r^4}{2} \] is the hypervolume of the four-dimensional ball. Let $A_4(r)$ be the three-dimensional volume of the hyperspherical surface of radius $r$. We can relate $A_4$ to $V_4$ as follows. If we compute $V_4(r)$ by considering concentric spherical shells we see that \[ V_4(r) = \int_0^r A_4(\rho) d \rho. \] Hence by the fundamental theorem of calculus \[ \frac{d}{dr} V_4(r) = A_4(r), \] so that \[ A_4(r) = 2 \pi^2 r^3. \] | 0 |
1952 | 1952_1 | Let \[ f(x) = \sum_{i=0}^n a_i x^{n-i} \] be a polynomial of degree $n$ with integral coefficients. If $a_0$, $a_n$, and $f(1)$ are odd, find the number of rational roots of $f(x) = 0$. | Suppose $f(p/q) = 0$ where $p$ and $q$ are integers having no common factor. Then \[ q^n f\left(\frac{p}{q}\right) = a_0 p^n + q a_1 p^{n-1} + \cdots + q^{n-1} a_{n-1} p + q^n a_n = 0. \] It follows that $q$ divides $a_0$ and $p$ divides $a_n$. Therefore $p$ and $q$ are both odd. Hence \[ a_0 p^n + q a_1 p^{n-1} + \cdots + q^{n-1} a_{n-1} p + q^n a_n \equiv a_0 + a_1 + \cdots + a_{n-1} + a_n \equiv f(1) \equiv 1 \pmod{2}, \] contradicting $(1)$. This makes the answer \boxed{0}. | numerical | putnam (modified boxing) | Algebra Number Theory | Let \[ f(x) = \sum_{i=0}^n a_i x^{n-i} \] be a polynomial of degree $n$ with integral coefficients. If $a_0$, $a_n$, and $f(1)$ are odd, prove that $f(x) = 0$ has no rational roots. | Suppose $f(p/q) = 0$ where $p$ and $q$ are integers having no common factor. Then \[ q^n f\left(\frac{p}{q}\right) = a_0 p^n + q a_1 p^{n-1} + \cdots + q^{n-1} a_{n-1} p + q^n a_n = 0. \] It follows that $q$ divides $a_0$ and $p$ divides $a_n$. Therefore $p$ and $q$ are both odd. Hence \[ a_0 p^n + q a_1 p^{n-1} + \cdots + q^{n-1} a_{n-1} p + q^n a_n \equiv a_0 + a_1 + \cdots + a_{n-1} + a_n \equiv f(1) \equiv 1 \pmod{2}, \] contradicting $(1)$. | 0 |
1952 | 1952_4 | The flag of the United Nations consists of a polar map of the world, with the North Pole as center, extending approximately to 45° South Latitude. The parallels of latitude are concentric circles with radii proportional to their co-latitudes. Australia is near the periphery of the map and is intersected by the parallel of latitude 30° S. In the very close vicinity of this parallel how much are East and West distances exaggerated as compared to North and South distances? | Let a point on the earth with co-latitude $\alpha$ and longitude $\theta$ be represented by a point on the flag map with polar coordinates $k\alpha$ and $\theta$ where $k$ is a constant of proportionality. \n\nLet $R$ be the radius of the earth. A meridian on the earth has length $\pi R$ and on the (complete) flag map is represented by a segment of length $\pi k$. Hence a short distance $d$ in the North-South direction is represented on the flag map by the distance $d_1 = kd/R$.\n\nOn the earth the small circle of latitude 30° S (= co-latitude $2\pi/3$) is represented on the flag map by a circle of radius $2\pi k/3$ and of circumference $4\pi^2 k/3$. On the earth that circle has length $2\pi R \sin (2\pi/3) = \pi R\sqrt{3}$.\n\nHence a short distance $d$ in the East-West direction is represented on the flag map by a distance \[ \frac{4\pi^2 k/3}{\pi R\sqrt{3}} d = \frac{4\pi kd}{3\sqrt{3}R} = \frac{4\pi}{3\sqrt{3}} d_1. \] \n\nTherefore distances in the East-West direction at latitude 30° S are magnified relative to distances in the North-South direction by the factor $\boxed{\frac{4\pi}{3\sqrt{3}}} \approx 2.42$. | numerical | putnam | Geometry Trigonometry | The flag of the United Nations consists of a polar map of the world, with the North Pole as center, extending approximately to 45° South Latitude. The parallels of latitude are concentric circles with radii proportional to their co-latitudes. Australia is near the periphery of the map and is intersected by the parallel of latitude 30° S. In the very close vicinity of this parallel how much are East and West distances exaggerated as compared to North and South distances? | Let a point on the earth with co-latitude $\alpha$ and longitude $\theta$ be represented by a point on the flag map with polar coordinates $k\alpha$ and $\theta$ where $k$ is a constant of proportionality. \n\nLet $R$ be the radius of the earth. A meridian on the earth has length $\pi R$ and on the (complete) flag map is represented by a segment of length $\pi k$. Hence a short distance $d$ in the North-South direction is represented on the flag map by the distance $d_1 = kd/R$.\n\nOn the earth the small circle of latitude 30° S (= co-latitude $2\pi/3$) is represented on the flag map by a circle of radius $2\pi k/3$ and of circumference $4\pi^2 k/3$. On the earth that circle has length $2\pi R \sin (2\pi/3) = \pi R\sqrt{3}$.\n\nHence a short distance $d$ in the East-West direction is represented on the flag map by a distance \[ \frac{4\pi^2 k/3}{\pi R\sqrt{3}} d = \frac{4\pi kd}{3\sqrt{3}R} = \frac{4\pi}{3\sqrt{3}} d_1. \] \n\nTherefore distances in the East-West direction at latitude 30° S are magnified relative to distances in the North-South direction by the factor $\boxed{\frac{4\pi}{3\sqrt{3}}} \approx 2.42$. | 0 |
1952 | 1952_7 | Directed lines are drawn from the center of a circle, making angles of $0, \pm 1, \pm 2, \pm 3, \dots$ (measured in radians from a prime direction). If these lines meet the circle in points $P_0, P_1, P_{-1}, P_2, P_{-2}, \dots$, find the number of intervals on the circumference of the circle which does not contain some $P_j$. (You may assume that $\pi$ is irrational.) | The various points $P_0, P_{\pm 1}, P_{\pm 2}, \dots$ must all be distinct, else $\pi$ would be a rational number. (For $P_i = P_j$ implies that $j - i$ is an integral multiple of $2\pi$.)\n\nLet $I$ be a given interval on the circumference, say of length $\epsilon$. Choose $N$ so that $\epsilon > 2\pi/N$. The points $P_0, P_1, P_2, \dots, P_N$ are all different and divide the circle into $N$ arcs, of which one must have length at most $2\pi/N$. Hence there are indices $m$ and $m + k$ ($k > 0$) such that $P_m$ and $P_{m+k}$ bound an arc of length $\delta < \epsilon$. If $q$ is the greatest integer in $2\pi/\delta$, then\n\n\[ P_m, P_{m+k}, P_{m+2k}, \dots, P_{m+qk} \]\n\ndivide the circle into $q + 1$ arcs each of length $\delta$ or less. Since $I$ is longer than any of these arcs, it contains one of the points listed in (1). Thus the final answer becomes \boxed{0}. | numerical | putnam (modified boxing) | Geometry Number Theory | Directed lines are drawn from the center of a circle, making angles of $0, \pm 1, \pm 2, \pm 3, \dots$ (measured in radians from a prime direction). If these lines meet the circle in points $P_0, P_1, P_{-1}, P_2, P_{-2}, \dots$, show that there is no interval on the circumference of the circle which does not contain some $P_j$. (You may assume that $\pi$ is irrational.) | The various points $P_0, P_{\pm 1}, P_{\pm 2}, \dots$ must all be distinct, else $\pi$ would be a rational number. (For $P_i = P_j$ implies that $j - i$ is an integral multiple of $2\pi$.)\n\nLet $I$ be a given interval on the circumference, say of length $\epsilon$. Choose $N$ so that $\epsilon > 2\pi/N$. The points $P_0, P_1, P_2, \dots, P_N$ are all different and divide the circle into $N$ arcs, of which one must have length at most $2\pi/N$. Hence there are indices $m$ and $m + k$ ($k > 0$) such that $P_m$ and $P_{m+k}$ bound an arc of length $\delta < \epsilon$. If $q$ is the greatest integer in $2\pi/\delta$, then\n\n\[ P_m, P_{m+k}, P_{m+2k}, \dots, P_{m+qk} \]\n\ndivide the circle into $q + 1$ arcs each of length $\delta$ or less. Since $I$ is longer than any of these arcs, it contains one of the points listed in (1). | 0 |
1952 | 1952_11 | A homogeneous solid body is made by joining a base of a circular cylinder of height $h$ and radius $r$, and the base of a hemisphere of radius $r$. This body is placed with the hemispherical end on a horizontal table, with the axis of the cylinder in a vertical position, and then slightly oscillated. It is intuitively evident that if $r$ is large as compared to $h$, the equilibrium will be stable; but if $r$ is small as compared to $h$, the equilibrium will be unstable. What is the critical value of the ratio $r/h$ which enables the body to rest in neutral equilibrium in any position? | Since the body is homogeneous, the center of gravity and the centroid are the same. Assume a coordinate system such that $z = 0$ is the equation of the plane where the hemisphere is joined to the cylinder (when the body is in a vertical position); the equation of the table is $z = -r$.\n\nBy cylindrical symmetry, it suffices to consider a section perpendicular to the plane of the table through the contact point. If the solid is slightly tilted, the normal force $N$ at the point of contact will act through the point $O$.\n\nIf the centroid of the solid is above $O$, then the couple formed by $N$ and the weight $W$ will produce a toppling rotation. If the centroid of the solid is below $O$ then the couple will produce a restoring rotation. Hence the problem amounts to determining the ratio of $r$ to $h$ which will give a centroid at $O$.\n\nBy symmetry we need only consider the $z$-coordinate of the centroid.\n\nThe $z$-coordinate of the centroid of the cylinder is $h/2$, and the volume is $\pi r^2 h$. For the hemisphere the volume is $(2/3)\pi r^3$, and the $z$-coordinate of the centroid is given by\n\n\[ \bar{z} = \frac{\int_{-r}^0 z\pi(r^2 - z^2) dz}{(2/3)\pi r^3} = -(3/8)r. \]\n\nThe centroid of the whole body will be at the origin if and only if\n\n\[ (\pi r^2 h) \frac{h}{2} - \left( \frac{2}{3} \pi r^3 \right) \left( \frac{3r}{8} \right) = 0, \]\n\nthat is $r^2 = 2h^2$, whence $r/h = \boxed{\sqrt{2}}$ is the critical value. For this shape, the center of gravity will be over the point of support whenever the body rests on a point of the hemispherical surface, so there will be neutral equilibrium. | numerical | putnam | Geometry Analysis | A homogeneous solid body is made by joining a base of a circular cylinder of height $h$ and radius $r$, and the base of a hemisphere of radius $r$. This body is placed with the hemispherical end on a horizontal table, with the axis of the cylinder in a vertical position, and then slightly oscillated. It is intuitively evident that if $r$ is large as compared to $h$, the equilibrium will be stable; but if $r$ is small as compared to $h$, the equilibrium will be unstable. What is the critical value of the ratio $r/h$ which enables the body to rest in neutral equilibrium in any position? | Since the body is homogeneous, the center of gravity and the centroid are the same. Assume a coordinate system such that $z = 0$ is the equation of the plane where the hemisphere is joined to the cylinder (when the body is in a vertical position); the equation of the table is $z = -r$.\n\nBy cylindrical symmetry, it suffices to consider a section perpendicular to the plane of the table through the contact point. If the solid is slightly tilted, the normal force $N$ at the point of contact will act through the point $O$.\n\nIf the centroid of the solid is above $O$, then the couple formed by $N$ and the weight $W$ will produce a toppling rotation. If the centroid of the solid is below $O$ then the couple will produce a restoring rotation. Hence the problem amounts to determining the ratio of $r$ to $h$ which will give a centroid at $O$.\n\nBy symmetry we need only consider the $z$-coordinate of the centroid.\n\nThe $z$-coordinate of the centroid of the cylinder is $h/2$, and the volume is $\pi r^2 h$. For the hemisphere the volume is $(2/3)\pi r^3$, and the $z$-coordinate of the centroid is given by\n\n\[ \bar{z} = \frac{\int_{-r}^0 z\pi(r^2 - z^2) dz}{(2/3)\pi r^3} = -(3/8)r. \]\n\nThe centroid of the whole body will be at the origin if and only if\n\n\[ (\pi r^2 h) \frac{h}{2} - \left( \frac{2}{3} \pi r^3 \right) \left( \frac{3r}{8} \right) = 0, \]\n\nthat is $r^2 = 2h^2$, whence $r/h = \boxed{\sqrt{2}}$ is the critical value. For this shape, the center of gravity will be over the point of support whenever the body rests on a point of the hemispherical surface, so there will be neutral equilibrium. | 0 |
1953 | 1953_4 | From the identity \[ \int_0^{\pi/2} \log \sin 2x \, dx = \int_0^{\pi/2} \log \sin x \, dx + \int_0^{\pi/2} \log \cos x \, dx + \int_0^{\pi/2} \log 2 \, dx, \] deduce the value of \[ \int_0^{\pi/2} \log \sin x \, dx. \] | Let \[ I = \int_0^{\pi/2} \log \sin x \, dx. \] [This is, of course, an improper integral which we assume for the moment to exist.] Making the substitutions $x = \pi/2 - u$ and $x = \pi - v$ we see that\[ I = \int_0^{\pi/2} \log \cos u \, du = \int_{\pi/2}^{\pi} \log \sin v \, dv. \] Therefore \[ \int_0^{\pi} \log \sin v \, dv = \int_0^{\pi/2} \log \sin v \, dv + \int_{\pi/2}^{\pi} \log \sin v \, dv = 2I. \] Making the substitution $v = 2w$ we have also\[ \int_0^{\pi} \log \sin v \, dv = 2 \int_0^{\pi/2} \log \sin 2w \, dw. \] Then \[ I = \int_0^{\pi/2} \log \sin 2w \, dw = \int_0^{\pi/2} (\log 2) dw + \int_0^{\pi/2} \log \sin w \, dw + \int_0^{\pi/2} \log \cos w \, dw \] \[ = \frac{\pi}{2} \log 2 + 2I. \] Hence \[ I = -\frac{\pi}{2} \log 2. \] To justify these manipulations, note that\[ \frac{2}{\pi} x < \sin x \leq 1 \text{ for } 0 < x \leq \frac{\pi}{2} \] and hence \[ \log x - \log \frac{\pi}{2} < \log \sin x \leq 0. \] Now \[ \int_\epsilon^1 \log x \, dx = -1 - \epsilon \log \epsilon + \epsilon. \] So \[ \lim_{\epsilon \to 0} \int_\epsilon^1 \log x \, dx = -1. \] It follows that the improper integral \[ \int_0^{\pi/2} \log \sin x \, dx = \lim_{\epsilon \to 0} \int_\epsilon^{\pi/2} \log \sin x \, dx \] exists. The several substitutions employed above can now be justified routinely. Thus the solution is \boxed{-\frac{\pi}{2} \log 2}. | numerical | putnam | Calculus Analysis | From the identity \[ \int_0^{\pi/2} \log \sin 2x \, dx = \int_0^{\pi/2} \log \sin x \, dx + \int_0^{\pi/2} \log \cos x \, dx + \int_0^{\pi/2} \log 2 \, dx, \] deduce the value of \[ \int_0^{\pi/2} \log \sin x \, dx. \] | Let \[ I = \int_0^{\pi/2} \log \sin x \, dx. \] [This is, of course, an improper integral which we assume for the moment to exist.] Making the substitutions $x = \pi/2 - u$ and $x = \pi - v$ we see that\[ I = \int_0^{\pi/2} \log \cos u \, du = \int_{\pi/2}^{\pi} \log \sin v \, dv. \] Therefore \[ \int_0^{\pi} \log \sin v \, dv = \int_0^{\pi/2} \log \sin v \, dv + \int_{\pi/2}^{\pi} \log \sin v \, dv = 2I. \] Making the substitution $v = 2w$ we have also\[ \int_0^{\pi} \log \sin v \, dv = 2 \int_0^{\pi/2} \log \sin 2w \, dw. \] Then \[ I = \int_0^{\pi/2} \log \sin 2w \, dw = \int_0^{\pi/2} (\log 2) dw + \int_0^{\pi/2} \log \sin w \, dw + \int_0^{\pi/2} \log \cos w \, dw \] \[ = \frac{\pi}{2} \log 2 + 2I. \] Hence \[ I = \boxed{-\frac{\pi}{2} \log 2}. \] To justify these manipulations, note that\[ \frac{2}{\pi} x < \sin x \leq 1 \text{ for } 0 < x \leq \frac{\pi}{2} \] and hence \[ \log x - \log \frac{\pi}{2} < \log \sin x \leq 0. \] Now \[ \int_\epsilon^1 \log x \, dx = -1 - \epsilon \log \epsilon + \epsilon. \] So \[ \lim_{\epsilon \to 0} \int_\epsilon^1 \log x \, dx = -1. \] It follows that the improper integral \[ \int_0^{\pi/2} \log \sin x \, dx = \lim_{\epsilon \to 0} \int_\epsilon^{\pi/2} \log \sin x \, dx \] exists. The several substitutions employed above can now be justified routinely. | 0 |
1953 | 1953_6 | Show that the sequence \[ \sqrt{7}, \sqrt{7 - \sqrt{7}}, \sqrt{7 - \sqrt{7 + \sqrt{7}}}, \sqrt{7 - \sqrt{7 + \sqrt{7 - \sqrt{7}}}}, \dots \] converges, and evaluate the limit. | Let $x_0 = \sqrt{7}, x_1 = \sqrt{7 - \sqrt{7}}, x_2 = \sqrt{7 - \sqrt{7 + \sqrt{7}}}, \dots.$ The later terms of the intended sequence are given by the recursion \[ x_{n+2} = \sqrt{7 - \sqrt{7 + x_n}} \quad \text{for } n \geq 0. \]
We have therefore two interlocked recursions of the form \[ a_{n+1} = f(a_n), \] where \[ f(x) = \sqrt{7 - \sqrt{7 + x}}. \]
The diagram (explained on p. 223) indicates that for any start in the domain of $f$ the recursion converges to 2, which is the unique fixed point of $f$. We shall prove this analytically.
By the mean value theorem, \[ |f(x) - 2| = |f(x) - f(2)| = |f'(\xi)||x - 2| \] for some number $\xi$ between $2$ and $x$. If $0 \leq x \leq 7$, then surely $0 \leq \xi \leq 7$ and \[ |f'(\xi)| = \frac{-1}{4\sqrt{7 - \sqrt{7 + \xi}}\sqrt{7 + \xi}} \leq \frac{1}{4\sqrt{7 - \sqrt{14}}} = \alpha < 1. \]
Hence \[ |x_{n+2} - 2| = |f(x_n) - 2| \leq \alpha|x_n - 2| \] if $0 \leq x_n \leq 7$. Since $x_0$ and $x_1$ are both in this interval it follows that $0 \leq x_n \leq 7$ for all $n$ and that \[ |x_n - 2| \leq \alpha^k |x_0 - 2| \] and \[ |x_{n+1} - 2| \leq \alpha^k |x_1 - 2| \] for all positive integers $k$. Since $\alpha < 1$, these inequalities imply \[ x_n \to \boxed{2} \] as $n \to \infty. | numerical | putnam | Analysis Calculus | Show that the sequence \[ \sqrt{7}, \sqrt{7 - \sqrt{7}}, \sqrt{7 - \sqrt{7 + \sqrt{7}}}, \sqrt{7 - \sqrt{7 + \sqrt{7 - \sqrt{7}}}}, \dots \] converges, and evaluate the limit. | Let $x_0 = \sqrt{7}, x_1 = \sqrt{7 - \sqrt{7}}, x_2 = \sqrt{7 - \sqrt{7 + \sqrt{7}}}, \dots.$ The later terms of the intended sequence are given by the recursion \[ x_{n+2} = \sqrt{7 - \sqrt{7 + x_n}} \quad \text{for } n \geq 0. \]
We have therefore two interlocked recursions of the form \[ a_{n+1} = f(a_n), \] where \[ f(x) = \sqrt{7 - \sqrt{7 + x}}. \]
The diagram (explained on p. 223) indicates that for any start in the domain of $f$ the recursion converges to 2, which is the unique fixed point of $f$. We shall prove this analytically.
By the mean value theorem, \[ |f(x) - 2| = |f(x) - f(2)| = |f'(\xi)||x - 2| \] for some number $\xi$ between $2$ and $x$. If $0 \leq x \leq 7$, then surely $0 \leq \xi \leq 7$ and \[ |f'(\xi)| = \frac{-1}{4\sqrt{7 - \sqrt{7 + \xi}}\sqrt{7 + \xi}} \leq \frac{1}{4\sqrt{7 - \sqrt{14}}} = \alpha < 1. \]
Hence \[ |x_{n+2} - 2| = |f(x_n) - 2| \leq \alpha|x_n - 2| \] if $0 \leq x_n \leq 7$. Since $x_0$ and $x_1$ are both in this interval it follows that $0 \leq x_n \leq 7$ for all $n$ and that \[ |x_n - 2| \leq \alpha^k |x_0 - 2| \] and \[ |x_{n+1} - 2| \leq \alpha^k |x_1 - 2| \] for all positive integers $k$. Since $\alpha < 1$, these inequalities imply \[ x_n \to \boxed{2} \] as $n \to \infty. | 0 |
1953 | 1953_7 | Assuming that the roots of \[ x^3 + px^2 + qx + r = 0 \] are all real and positive, find the relation between $p, q$, and $r$ which is a necessary and sufficient condition that the roots may be the cosines of the angles of a triangle. | For any triangle $ABC$, we have \[ a = b \cos C + c \cos B \] \[ b = c \cos A + a \cos C \] \[ c = a \cos B + b \cos A. \] Regarding these as three homogeneous linear equations for $a, b$, and $c$ having a non-trivial solution we see that \[ \det \begin{bmatrix} 1 & -\cos C & -\cos B \\ -\cos C & 1 & -\cos A \\ -\cos B & -\cos A & 1 \end{bmatrix} = 0, \] that is \[ \cos^2 A + \cos^2 B + \cos^2 C + 2 \cos A \cos B \cos C = 1. \] If the roots of the equation \[ x^3 + px^2 + qx + r = 0 \] are $\cos A, \cos B, \cos C$, then \[ -p = \cos A + \cos B + \cos C \] \[ q = \cos A \cos B + \cos B \cos C + \cos C \cos A \] \[ -r = \cos A \cos B \cos C \] and (1) becomes \[ p^2 - 2q - 2r = 1, \] which is thus a necessary condition.
Now suppose that (3) holds and that the roots, say $x_1, x_2, x_3$, of (2) are all real and positive. Then \[ x_1^2 + x_2^2 + x_3^2 + 2x_1x_2x_3 = 1. \] From this it is clear that each root lies between $0$ and $1$; hence there are unique acute angles $A, B, C$ such that $x_1 = \cos A, x_2 = \cos B, x_3 = \cos C$. To prove that these are the angles of a triangle, it is sufficient to show that $A + B + C = \pi$. Substituting in (4), we get \[ \cos^2 C + 2 \cos A \cos B \cos C = 1 - \cos^2 A - \cos^2 B. \] Completing the square on the left, we obtain \[ (\cos C + \cos A \cos B)^2 = \sin^2 A \sin^2 B. \] Since the angles are all acute, taking the positive square root gives \[ \cos C + \cos A \cos B = \sin A \sin B \] and therefore \[ \cos C = - (\cos A \cos B - \sin A \sin B) = -\cos (A + B) = \cos (\pi - A - B). \] Since both $C$ and $\pi - A - B$ are in $(0, \pi)$, we have $C = \pi - A - B$, as required.
Thus, if the roots of (2) are all real and positive, then (3) is necessary and sufficient that the roots be the cosines of the angles of some triangle. Thus the final answer is \boxed{p^2 - 2q - 2r = 1}. | algebraic | putnam | Algebra Trigonometry | Assuming that the roots of \[ x^3 + px^2 + qx + r = 0 \] are all real and positive, find the relation between $p, q$, and $r$ which is a necessary and sufficient condition that the roots may be the cosines of the angles of a triangle. | For any triangle $ABC$, we have \[ a = b \cos C + c \cos B \] \[ b = c \cos A + a \cos C \] \[ c = a \cos B + b \cos A. \] Regarding these as three homogeneous linear equations for $a, b$, and $c$ having a non-trivial solution we see that \[ \det \begin{bmatrix} 1 & -\cos C & -\cos B \\ -\cos C & 1 & -\cos A \\ -\cos B & -\cos A & 1 \end{bmatrix} = 0, \] that is \[ \cos^2 A + \cos^2 B + \cos^2 C + 2 \cos A \cos B \cos C = 1. \] If the roots of the equation \[ x^3 + px^2 + qx + r = 0 \] are $\cos A, \cos B, \cos C$, then \[ -p = \cos A + \cos B + \cos C \] \[ q = \cos A \cos B + \cos B \cos C + \cos C \cos A \] \[ -r = \cos A \cos B \cos C \] and (1) becomes \[ \boxed{p^2 - 2q - 2r = 1}, \] which is thus a necessary condition.
Now suppose that (3) holds and that the roots, say $x_1, x_2, x_3$, of (2) are all real and positive. Then \[ x_1^2 + x_2^2 + x_3^2 + 2x_1x_2x_3 = 1. \] From this it is clear that each root lies between $0$ and $1$; hence there are unique acute angles $A, B, C$ such that $x_1 = \cos A, x_2 = \cos B, x_3 = \cos C$. To prove that these are the angles of a triangle, it is sufficient to show that $A + B + C = \pi$. Substituting in (4), we get \[ \cos^2 C + 2 \cos A \cos B \cos C = 1 - \cos^2 A - \cos^2 B. \] Completing the square on the left, we obtain \[ (\cos C + \cos A \cos B)^2 = \sin^2 A \sin^2 B. \] Since the angles are all acute, taking the positive square root gives \[ \cos C + \cos A \cos B = \sin A \sin B \] and therefore \[ \cos C = - (\cos A \cos B - \sin A \sin B) = -\cos (A + B) = \cos (\pi - A - B). \] Since both $C$ and $\pi - A - B$ are in $(0, \pi)$, we have $C = \pi - A - B$, as required.
Thus, if the roots of (2) are all real and positive, then (3) is necessary and sufficient that the roots be the cosines of the angles of some triangle. | 0 |
1953 | 1953_10 | Solve the equations \[ \frac{dy}{dx} = z(y + z)^n, \quad \frac{dz}{dx} = y(y + z)^n, \] given the initial conditions $y = 1$ and $z = 0$ when $x = 0$ and give the value of y+z. | Let $u = y + z$. Adding the two given equations we get \[ \frac{du}{dx} = u^{n+1} \] where $u = 1$ when $x = 0$. We temporarily assume $n \neq 0$. Then we can solve (1) by separating the variables. We have \[ -n \frac{du}{u^{n-1}} = -n dx \] and therefore \[ \frac{1}{u^n} = a - nx. \] Using the initial condition we see that $u = 1$ and \[ u^n = \frac{1}{1 - nx} \] for $-\infty < x < 1/n$ if $n > 0$ and for $1/n < x < \infty$ if $n < 0$. Now let $v = y - z$ and subtract the original equations, \[ \frac{dv}{dx} = -vu^n = \frac{-v}{1 - nx} \] where $v = 1$ when $x = 0$. Again the variables separate: \[ n \ln v = b + \ln(1 - nx); \] and using the initial condition, \[ v^n = 1 - nx. \] Therefore, \[ y = \frac{1}{2}(u + v) = \frac{1}{2}[(1 - nx)^{-1/n} + (1 - nx)^{1/n}] \] \[ z = \frac{1}{2}(u - v) = \frac{1}{2}[(1 - nx)^{-1/n} - (1 - nx)^{1/n}] \] for $-\infty < x < 1/n$ if $n > 0$ and for $1/n < x < \infty$ if $n < 0$. The case $n = 0$ can be solved by the same method. In this case $du/dx = u, dv/dx = -v$, so $u = e^x, v = e^{-x}$, and \[ y = \frac{1}{2}(e^x + e^{-x}) = \cosh x \] \[ z = \frac{1}{2}(e^x - e^{-x}) = \sinh x, \] for all $x$. Thus the final answer is \boxed{sinh x + cosh x}. | algebraic | putnam (modified boxing) | Differential Equations Algebra | Solve the equations \[ \frac{dy}{dx} = z(y + z)^n, \quad \frac{dz}{dx} = y(y + z)^n, \] given the initial conditions $y = 1$ and $z = 0$ when $x = 0$. | Let $u = y + z$. Adding the two given equations we get \[ \frac{du}{dx} = u^{n+1} \] where $u = 1$ when $x = 0$. We temporarily assume $n \neq 0$. Then we can solve (1) by separating the variables. We have \[ -n \frac{du}{u^{n-1}} = -n dx \] and therefore \[ \frac{1}{u^n} = a - nx. \] Using the initial condition we see that $u = 1$ and \[ u^n = \frac{1}{1 - nx} \] for $-\infty < x < 1/n$ if $n > 0$ and for $1/n < x < \infty$ if $n < 0$. Now let $v = y - z$ and subtract the original equations, \[ \frac{dv}{dx} = -vu^n = \frac{-v}{1 - nx} \] where $v = 1$ when $x = 0$. Again the variables separate: \[ n \ln v = b + \ln(1 - nx); \] and using the initial condition, \[ v^n = 1 - nx. \] Therefore, \[ y = \frac{1}{2}(u + v) = \frac{1}{2}[(1 - nx)^{-1/n} + (1 - nx)^{1/n}] \] \[ z = \frac{1}{2}(u - v) = \frac{1}{2}[(1 - nx)^{-1/n} - (1 - nx)^{1/n}] \] for $-\infty < x < 1/n$ if $n > 0$ and for $1/n < x < \infty$ if $n < 0$. The case $n = 0$ can be solved by the same method. In this case $du/dx = u, dv/dx = -v$, so $u = e^x, v = e^{-x}$, and \[ \boxed{y = \frac{1}{2}(e^x + e^{-x}) = \cosh x \] \[ z = \frac{1}{2}(e^x - e^{-x}) = \sinh x}, \] for all $x$. | 0 |
1953 | 1953_11 | Determine the equation of a surface in three-dimensional Cartesian space which has the following properties: (a) it passes through the point $(1, 1, 1)$; and (b) if the tangent plane be drawn at any point $P$, and $A, B$ and $C$ are the intersections of this plane with the $x, y$ and $z$ axes respectively, then $P$ is the orthocenter (intersection of the altitudes) of the triangle $ABC$. | Consider any plane $\Pi$ that crosses the coordinate axes at three distinct points $A, B,$ and $C$, and let $P$ be the orthocenter of the triangle $ABC$. Treat the points as vectors from the origin, use $(\,\,)$ to denote the inner product of two vectors, and recall that $(A, B) = (B, C) = (C, A) = 0$ since the axes are orthogonal. Then the orthocenter property becomes \[ (P - A, B - C) = (P - B, C - A) = (P - C, A - B) = 0. \] Using the linearity properties of the inner product, we get \[ (P, B - C) = (P, C - A) = (P, A - B) = 0. \] Thus $P$, as a vector, is orthogonal to the plane $\Pi$. This means that $P$ is the foot of the perpendicular on $\Pi$ from the origin $O$. The condition on the surface is, then, that every normal passes through the origin. Obviously the sphere \[ x^2 + y^2 + z^2 = 3 \] has this property and passes through the required point $(1, 1, 1)$. A surface that satisfies the original conditions, however, cannot contain a point of a coordinate plane, since it is impossible (by the above analysis) that such a point be the orthocenter of a triangle whose vertices are on the coordinate axes. We shall show that the largest connected surface satisfying the conditions is that portion of the sphere $(1)$ satisfying $x > 0, y > 0, z > 0$. [We consider only connected surfaces, since without this restriction we can adjoin portions of other spheres at will.] The condition on the surface is that the differential form \[ xdx + ydy + zdz = \frac{1}{2} d(x^2 + y^2 + z^2) \] vanish on the surface. We conclude that $x^2 + y^2 + z^2$ is constant. Since the surface is required to contain $(1, 1, 1)$ it must be a part of the sphere $(1)$. Since, as we have seen, it can contain no point of the coordinate planes, it must lie entirely in the first octant. Thus the final answer is \boxed{x^2 + y^2 + z^2 = 3}. | algebraic | putnam | Geometry Linear Algebra | Determine the equation of a surface in three-dimensional Cartesian space which has the following properties: (a) it passes through the point $(1, 1, 1)$; and (b) if the tangent plane be drawn at any point $P$, and $A, B$ and $C$ are the intersections of this plane with the $x, y$ and $z$ axes respectively, then $P$ is the orthocenter (intersection of the altitudes) of the triangle $ABC$. | Consider any plane $\Pi$ that crosses the coordinate axes at three distinct points $A, B,$ and $C$, and let $P$ be the orthocenter of the triangle $ABC$. Treat the points as vectors from the origin, use $(\,\,)$ to denote the inner product of two vectors, and recall that $(A, B) = (B, C) = (C, A) = 0$ since the axes are orthogonal. Then the orthocenter property becomes \[ (P - A, B - C) = (P - B, C - A) = (P - C, A - B) = 0. \] Using the linearity properties of the inner product, we get \[ (P, B - C) = (P, C - A) = (P, A - B) = 0. \] Thus $P$, as a vector, is orthogonal to the plane $\Pi$. This means that $P$ is the foot of the perpendicular on $\Pi$ from the origin $O$. The condition on the surface is, then, that every normal passes through the origin. Obviously the sphere \[ \boxed{x^2 + y^2 + z^2 = 3} \] has this property and passes through the required point $(1, 1, 1)$. A surface that satisfies the original conditions, however, cannot contain a point of a coordinate plane, since it is impossible (by the above analysis) that such a point be the orthocenter of a triangle whose vertices are on the coordinate axes. We shall show that the largest connected surface satisfying the conditions is that portion of the sphere $(1)$ satisfying $x > 0, y > 0, z > 0$. [We consider only connected surfaces, since without this restriction we can adjoin portions of other spheres at will.] The condition on the surface is that the differential form \[ xdx + ydy + zdz = \frac{1}{2} d(x^2 + y^2 + z^2) \] vanish on the surface. We conclude that $x^2 + y^2 + z^2$ is constant. Since the surface is required to contain $(1, 1, 1)$ it must be a part of the sphere $(1)$. Since, as we have seen, it can contain no point of the coordinate planes, it must lie entirely in the first octant. | 0 |
1953 | 1953_14 | Let $w$ be an irrational number with $0 < w < 1$. Prove that $w$ has a unique convergent expansion of the form \[ w = \frac{1}{p_0} - \frac{1}{p_0p_1} + \frac{1}{p_0p_1p_2} - \frac{1}{p_0p_1p_2p_3} + \cdots, \] where $p_0, p_1, p_2, \ldots$ are integers and $1 \leq p_0 < p_1 < p_2 < \cdots$. If $w = \frac{1}{2} \sqrt{2}$, find the sum of $p_0, p_1, p_2$. | Before starting the construction we make an observation. \[(1) \text{If } \alpha \text{ is an irrational number such that } 0 < \alpha < 1, \text{ there is a unique integer } p \text{ such that } \frac{1}{p + 1} < \alpha < \frac{1}{p}. \] Moreover, $p$ is positive, $1 - p\alpha$ is irrational, and $0 < 1 - p\alpha < 1/(p + 1)$.\n\nNow define a sequence $w_0, w_1, w_2, \ldots$ of irrational numbers between $0$ and $1$ and a sequence of positive integers by induction as follows:\nLet $w_0 = w$ and let $p_0$ be the integer such that \[ \frac{1}{p_0 + 1} < w_0 < \frac{1}{p_0}. \] Let \[ w_1 = 1 - p_0w_0 \] (which is irrational and between $0$ and $1$, by $(1)$) and let $p_1$ be the integer such that \[ \frac{1}{p_1 + 1} < w_1 < \frac{1}{p_1}. \] After $w_k$ and $p_k$ have been defined so that \[ \frac{1}{p_k + 1} < w_k < \frac{1}{p_k}, \] let \[ w_{k+1} = 1 - p_kw_k, \] which is irrational and between $0$ and $1$, and let $p_{k+1}$ be the integer such that \[ \frac{1}{p_{k+1} + 1} < w_{k+1} < \frac{1}{p_{k+1}}. \] From $(1)$ and $(2)$ it follows that \[ w_k < \frac{1}{p_k + 1}. \]\n\nSo $p_{k+1} \geq p_k + 1$. Hence the $p$'s increase strictly. Now \[ w_k = \frac{1}{p_k}w_{k+1}, \] so \[ w = w_0 = \frac{1}{p_0} - \frac{w_1}{p_0p_1} = \frac{1}{p_0} - \frac{1}{p_0p_1} + \frac{w_2}{p_0p_1p_2} = \sum_{k=0}^\infty \frac{(-1)^k}{p_0p_1 \cdots p_k}, \] as required.\n\nNext we show uniqueness. Suppose \[ w = \sum_{n=0}^\infty \frac{(-1)^n}{q_0q_1 \cdots q_n}, \] where $q_0, q_1, q_2, \ldots$ is a strictly increasing sequence of positive integers. We shall prove that, if the preceding construction is carried out, $p_n = q_n$ for all $n$.\n\nSince the series appearing in $(4)$ is alternating with terms strictly decreasing in absolute value, we have \[ \frac{1}{q_0} > w_0 > \frac{1}{q_0} - \frac{1}{q_0q_1} \geq \frac{1}{q_0} - \frac{1}{q_0(q_0 + 1)} = \frac{1}{q_0 + 1}. \] Therefore $p_0 = q_0$, and \[ w_1 = 1 - p_0w_0 = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{q_1q_2 \cdots q_n}. \]\n\nRepeating this argument inductively we find $p_{k-1} = q_{k-1}$ and \[ w_k = \sum_{n=k}^\infty \frac{(-1)^{n-k}}{q_kq_{k+1} \cdots q_n}, \] for $k = 1, 2, 3, \ldots$. This proves the uniqueness of the expansion.\n\nIf $w = \frac{1}{2}\sqrt{2}$, then $p_0 = 1$. \[ w_1 = 1 - \frac{1}{2}\sqrt{2} = \frac{2 - \sqrt{2}}{2}, \text{ so } p_1 = \Big[\frac{2 + \sqrt{2}}{2}\Big] = 3, \] \[ w_2 = 1 - 3\Big(\frac{2 - \sqrt{2}}{2}\Big) = \frac{4 + 3\sqrt{2}}{8}, \text{ so } p_2 = \Big[\frac{4 + 3\sqrt{2}}{8}\Big] = 8. \] The final sum is \boxed{12}. | numerical | putnam (modified boxing) | Number Theory Analysis | Let $w$ be an irrational number with $0 < w < 1$. Prove that $w$ has a unique convergent expansion of the form \[ w = \frac{1}{p_0} - \frac{1}{p_0p_1} + \frac{1}{p_0p_1p_2} - \frac{1}{p_0p_1p_2p_3} + \cdots, \] where $p_0, p_1, p_2, \ldots$ are integers and $1 \leq p_0 < p_1 < p_2 < \cdots$. If $w = \frac{1}{2} \sqrt{2}$, find $p_0, p_1, p_2$. | Before starting the construction we make an observation. \[(1) \text{If } \alpha \text{ is an irrational number such that } 0 < \alpha < 1, \text{ there is a unique integer } p \text{ such that } \frac{1}{p + 1} < \alpha < \frac{1}{p}. \] Moreover, $p$ is positive, $1 - p\alpha$ is irrational, and $0 < 1 - p\alpha < 1/(p + 1)$.\n\nNow define a sequence $w_0, w_1, w_2, \ldots$ of irrational numbers between $0$ and $1$ and a sequence of positive integers by induction as follows:\nLet $w_0 = w$ and let $p_0$ be the integer such that \[ \frac{1}{p_0 + 1} < w_0 < \frac{1}{p_0}. \] Let \[ w_1 = 1 - p_0w_0 \] (which is irrational and between $0$ and $1$, by $(1)$) and let $p_1$ be the integer such that \[ \frac{1}{p_1 + 1} < w_1 < \frac{1}{p_1}. \] After $w_k$ and $p_k$ have been defined so that \[ \frac{1}{p_k + 1} < w_k < \frac{1}{p_k}, \] let \[ w_{k+1} = 1 - p_kw_k, \] which is irrational and between $0$ and $1$, and let $p_{k+1}$ be the integer such that \[ \frac{1}{p_{k+1} + 1} < w_{k+1} < \frac{1}{p_{k+1}}. \] From $(1)$ and $(2)$ it follows that \[ w_k < \frac{1}{p_k + 1}. \]\n\nSo $p_{k+1} \geq p_k + 1$. Hence the $p$'s increase strictly. Now \[ w_k = \frac{1}{p_k}w_{k+1}, \] so \[ w = w_0 = \frac{1}{p_0} - \frac{w_1}{p_0p_1} = \frac{1}{p_0} - \frac{1}{p_0p_1} + \frac{w_2}{p_0p_1p_2} = \sum_{k=0}^\infty \frac{(-1)^k}{p_0p_1 \cdots p_k}, \] as required.\n\nNext we show uniqueness. Suppose \[ w = \sum_{n=0}^\infty \frac{(-1)^n}{q_0q_1 \cdots q_n}, \] where $q_0, q_1, q_2, \ldots$ is a strictly increasing sequence of positive integers. We shall prove that, if the preceding construction is carried out, $p_n = q_n$ for all $n$.\n\nSince the series appearing in $(4)$ is alternating with terms strictly decreasing in absolute value, we have \[ \frac{1}{q_0} > w_0 > \frac{1}{q_0} - \frac{1}{q_0q_1} \geq \frac{1}{q_0} - \frac{1}{q_0(q_0 + 1)} = \frac{1}{q_0 + 1}. \] Therefore $p_0 = q_0$, and \[ w_1 = 1 - p_0w_0 = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{q_1q_2 \cdots q_n}. \]\n\nRepeating this argument inductively we find $p_{k-1} = q_{k-1}$ and \[ w_k = \sum_{n=k}^\infty \frac{(-1)^{n-k}}{q_kq_{k+1} \cdots q_n}, \] for $k = 1, 2, 3, \ldots$. This proves the uniqueness of the expansion.\n\nIf $w = \frac{1}{2}\sqrt{2}$, then $p_0 = 1$. \[ w_1 = 1 - \frac{1}{2}\sqrt{2} = \frac{2 - \sqrt{2}}{2}, \text{ so } p_1 = \Big[\frac{2 + \sqrt{2}}{2}\Big] = 3, \] \[ w_2 = 1 - 3\Big(\frac{2 - \sqrt{2}}{2}\Big) = \frac{4 + 3\sqrt{2}}{8}, \text{ so } p_2 = \Big[\frac{4 + 3\sqrt{2}}{8}\Big] = 8. \] | 0 |
1954 | 1954_4 | A uniform rod of length $2k$ and weight $w$ rests with the end $A$ against a smooth vertical wall, while to the lower end $B$ is fastened a string $BC$ of length $2b$ coming from a point $C$ in the wall directly above $A$. If the system is in equilibrium, determine the angle $\angle ABC$. | Obviously the system will be in equilibrium if the rod and the string are both vertical, i.e., if $B$ is on the wall and $\angle ABC = 0$.
\[ \text{(Diagram showing the setup)} \]
Assume that the rod is in equilibrium in some non-vertical position, as pictured. Let $D$ be the midpoint of $BC$. The force of tension in the string and the force of gravity on the rod both act through the point $D$ (since the latter force acts vertically through the midpoint of the rod). Hence for equilibrium the reaction of the wall must also act through $D$. Since the wall is smooth this force is perpendicular to the wall. Hence we have $AD$ perpendicular to $AC$ and
\[ |AC|^2 + |AD|^2 = |CD|^2 = b^2. \]
By the law of cosines
\[ |AC|^2 = 4k^2 + 4b^2 - 8bk \cos \phi, \]
\[ |AD|^2 = 4k^2 + b^2 - 4bk \cos \phi. \]
Hence
\[ \cos \phi = \frac{2k^2 + b^2}{3bk} = \frac{2}{3}\frac{k}{b} + \frac{1}{3}\frac{b}{k}. \]
Thus
\[ \angle ABC = \boxed{\arccos \left( \frac{2}{3} \frac{k}{b} + \frac{1}{3} \frac{b}{k} \right)}. \]
Since angles $ADB$ and $CAB$ are obtuse, we must have $b < 2k < 2b$. Since $2x/3 + 1/3x < 1$ for $1/2 < x < 1$, equation $(1)$ determines a unique acute angle whenever $1/2 < k/b < 1$. Otherwise there is no equilibrium position except with $B$ on the wall. | algebraic | putnam | Geometry Physics | A uniform rod of length $2k$ and weight $w$ rests with the end $A$ against a smooth vertical wall, while to the lower end $B$ is fastened a string $BC$ of length $2b$ coming from a point $C$ in the wall directly above $A$. If the system is in equilibrium, determine the angle $\angle ABC$. | Obviously the system will be in equilibrium if the rod and the string are both vertical, i.e., if $B$ is on the wall and $\angle ABC = 0$.
\[ \text{(Diagram showing the setup)} \]
Assume that the rod is in equilibrium in some non-vertical position, as pictured. Let $D$ be the midpoint of $BC$. The force of tension in the string and the force of gravity on the rod both act through the point $D$ (since the latter force acts vertically through the midpoint of the rod). Hence for equilibrium the reaction of the wall must also act through $D$. Since the wall is smooth this force is perpendicular to the wall. Hence we have $AD$ perpendicular to $AC$ and
\[ |AC|^2 + |AD|^2 = |CD|^2 = b^2. \]
By the law of cosines
\[ |AC|^2 = 4k^2 + 4b^2 - 8bk \cos \phi, \]
\[ |AD|^2 = 4k^2 + b^2 - 4bk \cos \phi. \]
Hence
\[ \cos \phi = \frac{2k^2 + b^2}{3bk} = \frac{2}{3}\frac{k}{b} + \frac{1}{3}\frac{b}{k}. \]
Thus
\[ \angle ABC = \boxed{\arccos \left( \frac{2}{3} \frac{k}{b} + \frac{1}{3} \frac{b}{k} \right)}. \]
Since angles $ADB$ and $CAB$ are obtuse, we must have $b < 2k < 2b$. Since $2x/3 + 1/3x < 1$ for $1/2 < x < 1$, equation $(1)$ determines a unique acute angle whenever $1/2 < k/b < 1$. Otherwise there is no equilibrium position except with $B$ on the wall. | 0 |
1954 | 1954_6 | Suppose that $u_0, u_1, u_2, \dots$ is a sequence of real numbers such that
\[ u_n = \sum_{k=1}^\infty u_{n+k}^2 \text{ for } n = 0, 1, 2, \dots. \]
Ff $\sum u_n$ converges, find the maximum value value of $u_k$ for all $k$. | It is obvious from $(1)$ that the terms $u_n$ are non-increasing and non-negative. Thus, if any one term is zero, so are all its successors, and by induction using $(1)$ so are all its predecessors.
Suppose $\sum u_n$ converges. Let $p$ be chosen so that $\sum_{n > p} u_n < 1$. Then, from $(1)$,
\[ u_p = \sum_{n > p} u_n^2 \leq \sum_{n > p} u_p u_n = u_p \sum_{n > p} u_n \leq u_p \]
with equality at the last step only if $u_p = 0$. But, looking at the first member and the last member, we see that equality must hold throughout. Thus $u_p = \boxed{0}$ and hence all the $u$'s are zero, as we have shown above. | numerical | putnam (modified boxing) | Analysis | Suppose that $u_0, u_1, u_2, \dots$ is a sequence of real numbers such that
\[ u_n = \sum_{k=1}^\infty u_{n+k}^2 \text{ for } n = 0, 1, 2, \dots. \]
Prove that if $\sum u_n$ converges, then $u_k = 0$ for all $k$. | It is obvious from $(1)$ that the terms $u_n$ are non-increasing and non-negative. Thus, if any one term is zero, so are all its successors, and by induction using $(1)$ so are all its predecessors.
Suppose $\sum u_n$ converges. Let $p$ be chosen so that $\sum_{n > p} u_n < 1$. Then, from $(1)$,
\[ u_p = \sum_{n > p} u_n^2 \leq \sum_{n > p} u_p u_n = u_p \sum_{n > p} u_n \leq u_p \]
with equality at the last step only if $u_p = 0$. But, looking at the first member and the last member, we see that equality must hold throughout. Thus $u_p = 0$ and hence all the $u$'s are zero, as we have shown above. | 0 |
1954 | 1954_8 | Given the equation $x^2 - y^2 = a^3$ find the least value of $a$ for which it always has integral solutions for $x$ and $y$. | Let $x + y = a^2$, and $x - y = a$. Then $x^2 - y^2 = a^3$, and
\[ x = \frac{1}{2}(a^2 + a) \text{ and } y = \frac{1}{2}(a^2 - a). \]
Since $a^2$ and $a$ are both even or both odd, $x$ and $y$ are both integers and a solution exists for every integer $a$ thus making the least value \boxed{0}. | numerical | putnam (modified boxing) | Algebra Number Theory | Show that the equation $x^2 - y^2 = a^3$ has always integral solutions for $x$ and $y$ whenever $a$ is a positive integer. | Let $x + y = a^2$, and $x - y = a$. Then $x^2 - y^2 = a^3$, and
\[ x = \frac{1}{2}(a^2 + a) \text{ and } y = \frac{1}{2}(a^2 - a). \]
Since $a^2$ and $a$ are both even or both odd, $x$ and $y$ are both integers and a solution exists for every integer $a$. | 0 |
1954 | 1954_12 | Let $f(x)$ be a real-valued function, defined for $-1 < x < 1$, such that $f'(0)$ exists. Let $a_n, b_n$ be two sequences such that
\[-1 < a_n < 0 < b_n < 1, \quad \lim_{n \to \infty} a_n = 0, \quad \lim_{n \to \infty} b_n = 0. \]
Find the value of
\[ \lim_{n \to \infty} \frac{f(b_n) - f(a_n)}{b_n - a_n} in terms of $f'(0)$. \] | Let $\epsilon > 0$ be given. Then we can choose $\delta$, $0 < \delta < 1$, so that for all $x$ with $0 < |x| < \delta$,
\[
\left|\frac{f(x) - f(0)}{x} - f'(0)\right| < \epsilon.
\]
Let $k$ be chosen so that, for all $n \geq k$,
\[ |a_n| < \delta, \quad \text{and} \quad |b_n| < \delta. \]
Then for $m \geq k$ we have
\[ \left|\frac{f(a_m) - f(0)}{a_m} - f'(0)\right| < \epsilon, \quad \left|\frac{f(b_m) - f(0)}{b_m} - f'(0)\right| < \epsilon. \]
But
\[ |f(b_m) - f(a_m) - (b_m - a_m)f'(0)| \\
= |(f(b_m) - f(0) - b_m f'(0)) - (f(a_m) - f(0) - a_m f'(0))| \\
\leq |f(b_m) - f(0) - b_m f'(0)| + |f(a_m) - f(0) - a_m f'(0)| \\
\leq \epsilon |b_m| + \epsilon |a_m| = \epsilon (b_m - a_m).
\]
The penultimate step follows from (1) and the last from $a_m < 0 < b_m$. Therefore
\[ \left|\frac{f(b_m) - f(a_m)}{b_m - a_m} - f'(0)\right| < \epsilon \]
for all $m > k$. Since this is true for any $\epsilon > 0$, we have proved that
\[ \lim_{n \to \infty} \frac{f(b_n) - f(a_n)}{b_n - a_n} = \boxed{f'(0)}. \] | algebraic | putnam (modified boxing) | Analysis | Let $f(x)$ be a real-valued function, defined for $-1 < x < 1$, such that $f'(0)$ exists. Let $a_n, b_n$ be two sequences such that
\[-1 < a_n < 0 < b_n < 1, \quad \lim_{n \to \infty} a_n = 0, \quad \lim_{n \to \infty} b_n = 0. \]
Prove that
\[ \lim_{n \to \infty} \frac{f(b_n) - f(a_n)}{b_n - a_n} = f'(0). \] | Let $\epsilon > 0$ be given. Then we can choose $\delta$, $0 < \delta < 1$, so that for all $x$ with $0 < |x| < \delta$,
\[
\left|\frac{f(x) - f(0)}{x} - f'(0)\right| < \epsilon.
\]
Let $k$ be chosen so that, for all $n \geq k$,
\[ |a_n| < \delta, \quad \text{and} \quad |b_n| < \delta. \]
Then for $m \geq k$ we have
\[ \left|\frac{f(a_m) - f(0)}{a_m} - f'(0)\right| < \epsilon, \quad \left|\frac{f(b_m) - f(0)}{b_m} - f'(0)\right| < \epsilon. \]
But
\[ |f(b_m) - f(a_m) - (b_m - a_m)f'(0)| \\
= |(f(b_m) - f(0) - b_m f'(0)) - (f(a_m) - f(0) - a_m f'(0))| \\
\leq |f(b_m) - f(0) - b_m f'(0)| + |f(a_m) - f(0) - a_m f'(0)| \\
\leq \epsilon |b_m| + \epsilon |a_m| = \epsilon (b_m - a_m).
\]
The penultimate step follows from (1) and the last from $a_m < 0 < b_m$. Therefore
\[ \left|\frac{f(b_m) - f(a_m)}{b_m - a_m} - f'(0)\right| < \epsilon \]
for all $m > k$. Since this is true for any $\epsilon > 0$, we have proved that
\[ \lim_{n \to \infty} \frac{f(b_n) - f(a_n)}{b_n - a_n} = f'(0). \] | 0 |
1954 | 1954_14 | Find the sum of the lower and upper bound of the following:
\[
\lim_{n \to \infty} \sum_{s=1}^n \left( \frac{a + s}{n} \right)^n \quad (a > 0)
\]. | We shall evaluate the limit exactly using the facts:
\[
\left( 1 + \frac{x}{n} \right)^n \leq e^x
\]
if $n > 0$ and $1 + x/n > 0$, and
\[
\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n = e^x
\]
for all real $x$.
Let
\[
S_n = \sum_{s=1}^n \left( \frac{a + s}{n} \right)^n = \sum_{r=0}^{n-1} \left( 1 + \frac{a - r}{n} \right)^n.
\]
Recalling that $a > 0$ and using $(1)$, we have
\[
S_n \leq \sum_{r=0}^{n-1} \exp(a - r) < \sum_{r=0}^{\infty} \exp(a - r) = \frac{e^{a+1}}{e - 1}.
\]
Therefore,
\[
\limsup_{n \to \infty} S_n \leq \frac{e^{a+1}}{e - 1}.
\]
Also, for fixed $k$ and $n > k$,
\[
S_n \geq \sum_{r=0}^k \left( 1 + \frac{a - r}{n} \right)^n.
\]
The limit on the right exists as $n \to \infty$ by $(2)$, so
\[
\liminf_{n \to \infty} S_n \geq \sum_{r=0}^k \exp(a - r).
\]
Since $k$ is arbitrary, we get
\[
\liminf_{n \to \infty} S_n \geq \sum_{r=0}^{\infty} \exp(a - r) = \frac{e^{a+1}}{e - 1}.
\]
Comparing $(3)$ and $(4)$, we obtain
\[
\lim_{n \to \infty} S_n = \frac{e^{a+1}}{e - 1}.
\]
Evidently,
\[
1 < \frac{e}{e - 1} < e, \quad \text{so} \quad e^a < \lim S_n < e^{a+1},
\]
as required. Thus the final answer is $\boxed{e^a + e^{a+1}}$. | algebraic | putnam (modified boxing) | Analysis | Show that
\[
\lim_{n \to \infty} \sum_{s=1}^n \left( \frac{a + s}{n} \right)^n \quad (a > 0)
\]
lies between $e^a$ and $e^{a+1}$. | We shall evaluate the limit exactly using the facts:
\[
\left( 1 + \frac{x}{n} \right)^n \leq e^x
\]
if $n > 0$ and $1 + x/n > 0$, and
\[
\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n = e^x
\]
for all real $x$.
Let
\[
S_n = \sum_{s=1}^n \left( \frac{a + s}{n} \right)^n = \sum_{r=0}^{n-1} \left( 1 + \frac{a - r}{n} \right)^n.
\]
Recalling that $a > 0$ and using $(1)$, we have
\[
S_n \leq \sum_{r=0}^{n-1} \exp(a - r) < \sum_{r=0}^{\infty} \exp(a - r) = \frac{e^{a+1}}{e - 1}.
\]
Therefore,
\[
\limsup_{n \to \infty} S_n \leq \frac{e^{a+1}}{e - 1}.
\]
Also, for fixed $k$ and $n > k$,
\[
S_n \geq \sum_{r=0}^k \left( 1 + \frac{a - r}{n} \right)^n.
\]
The limit on the right exists as $n \to \infty$ by $(2)$, so
\[
\liminf_{n \to \infty} S_n \geq \sum_{r=0}^k \exp(a - r).
\]
Since $k$ is arbitrary, we get
\[
\liminf_{n \to \infty} S_n \geq \sum_{r=0}^{\infty} \exp(a - r) = \frac{e^{a+1}}{e - 1}.
\]
Comparing $(3)$ and $(4)$, we obtain
\[
\lim_{n \to \infty} S_n = \frac{e^{a+1}}{e - 1}.
\]
Evidently,
\[
1 < \frac{e}{e - 1} < e, \quad \text{so} \quad e^a < \lim S_n < e^{a+1},
\]
as required. | 0 |
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