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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 |
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Putnam AXIOM Dataset (ICML 2025 Version)
Note: for questions, feedback, bugs, etc. please open a Huggingface discussion here.
Dataset Summary
The Putnam AXIOM dataset is designed for evaluating large language models (LLMs) on advanced mathematical reasoning skills. It is based on challenging problems from the Putnam Mathematical Competition. This version contains 522 original problems prepared for the ICML 2025 submission.
The ICML 2025 paper is available on OpenReview: https://openreview.net/forum?id=kqj2Cn3Sxr
The dataset includes:
- Full Evaluation Set (522 problems): Complete set of original problems
- Originals for Generating Variations (100 problems): A subset of problems used to create variations
- Variations (500 problems): Variations generated from the original problems
Each problem includes:
- Problem statement
- Solution
- Original problem (where applicable)
- Answer type (e.g., numerical, proof)
- Source and type of problem (e.g., Algebra, Calculus, Geometry)
- Year (extracted from problem ID)
- Variation flag (0 for original problems, 1 for variations)
Note About Splits
For experimental purposes, validation and test splits derived from this dataset are available in a separate repository:
- ZIP-FIT experiments splits - Contains validation/test splits used for ZIP-FIT methodology research
Supported Tasks and Leaderboards
- Mathematical Reasoning: Evaluate mathematical reasoning and problem-solving skills.
- Language Model Benchmarking: Use this dataset to benchmark performance of language models on advanced mathematical questions.
Languages
The dataset is presented in English.
Dataset Structure
Data Fields
- year: The year of the competition (extracted from the problem ID).
- id: Unique identifier for each problem.
- problem: The problem statement.
- solution: The solution or explanation for the problem.
- answer_type: The expected type of answer (e.g., numerical, proof).
- source: The origin of the problem (Putnam).
- type: A description of the problem's mathematical topic (e.g., "Algebra Geometry").
- original_problem: Original form of the problem, where applicable.
- original_solution: Original solution to the problem, where applicable.
- variation: Flag for variations (0 for original problems, 1 for generated variations).
Splits
| Split | Description | Number of Problems |
|---|---|---|
full_eval |
Complete set of 522 original problems | 522 |
originals_for_generating_vars |
Original problems used to create variations | 100 |
variations |
Generated variations of the original problems | 500 |
Variations
The variations split contains problems that were algorithmically generated as variations of problems in the originals_for_generating_vars split. These variations maintain the core mathematical concepts of the original problems but present them with different contexts, numbers, or phrasings. The variation field is set to 1 for these problems to distinguish them from the original problems.
Dataset Usage
from datasets import load_dataset
# Load the dataset
dataset = load_dataset("Putnam-AXIOM/putnam-axiom-dataset-ICML-2025-522")
# Access each split
full_eval = dataset["full_eval"] # Original problems
originals = dataset["originals_for_generating_vars"] # Original problems used for variations
variations = dataset["variations"] # Generated variations
# Filter for original problems only (variation = 0)
original_problems = [p for p in full_eval if p["variation"] == 0]
# Filter for variation problems (variation = 1)
variation_problems = [p for p in variations if p["variation"] == 1]
# Example usage: print the first original problem
print(full_eval[0])
Citation
If you use this dataset, please cite it as follows:
@article{putnam_axiom2025,
title={Putnam-AXIOM: A Functional and Static Benchmark for Measuring Higher Level Mathematical Reasoning},
author={Aryan Gulati and Brando Miranda and Eric Chen and Emily Xia and Kai Fronsdal and Bruno de Moraes Dumont and Sanmi Koyejo},
journal={39th International Conference on Machine Learning (ICML 2025)},
year={2025},
note={Preprint available at: https://openreview.net/pdf?id=YXnwlZe0yf, ICML paper: https://openreview.net/forum?id=kqj2Cn3Sxr}
}
License
This dataset is licensed under the Apache 2.0.
Last updated: May 22, 2024
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