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1994 | 1994_A2 | Let $A$ be the area of the region in the first quadrant bounded by the line $y = \frac{1}{2} x$, the $x$-axis, and the ellipse $\frac{1}{9} x^2 + y^2 = 1$. Find the positive number $m$ such that $A$ is equal to the area of the region in the first quadrant bounded by the line $y = mx$, the $y$-axis, and the ellipse $\frac{1}{9} x^2 + y^2 = 1$. | The linear transformation given by $x_1 = \frac{1}{3}x$, $y_1 = y$ transforms the region $R$ bounded by $y = \frac{1}{2}x$, the $x$-axis, and the ellipse $\frac{1}{9}x^2 + y^2 = 1$ into the region $R'$ bounded by $y_1 = \frac{3}{2}x_1$, the $x_1$-axis, and the circle $x_1^2 + y_1^2 = 1$; it also transforms the region $S$ bounded by $y = mx$, the $y$-axis, and $\frac{1}{9}x^2 + y^2 = 1$ into the region $S'$ bounded by $y_1 = 3mx_1$, the $y_1$-axis, and the circle. Since all areas are multiplied by the same (nonzero) factor under the transformation, $R$ and $S$ have the same area if and only if $R'$ and $S'$ have the same area. However, we can see by symmetry about the line $y_1 = x_1$ that this happens if and only if $3m = \frac{2}{3}$, that is, $m = \boxed{\frac{2}{9}}$. | numerical | putnam | Geometry Linear Algebra | Let $A$ be the area of the region in the first quadrant bounded by the line $y = \frac{1}{2} x$, the $x$-axis, and the ellipse $\frac{1}{9} x^2 + y^2 = 1$. Find the positive number $m$ such that $A$ is equal to the area of the region in the first quadrant bounded by the line $y = mx$, the $y$-axis, and the ellipse $\frac{1}{9} x^2 + y^2 = 1$. | 0 |
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2008 | 2008_B2 | Let $F_0(x) = \ln x$. For $n \geq 0$ and $x > 0$, let $F_{n+1}(x) = \int_0^x F_n(t)\,dt$. Evaluate \[ \lim_{n \to \infty} \frac{n! F_n(1)}{\ln n}. \] | We claim that $F_n(x) = (\ln x-a_n)x^n/n!$, where $a_n = \sum_{k=1}^n 1/k$. Indeed, temporarily write $G_n(x) = (\ln x-a_n)x^n/n!$ for $x>0$ and $n\geq 1$; then $\lim_{x\to 0} G_n(x) = 0$ and $G_n'(x) = (\ln x-a_n+1/n) x^{n-1}/(n-1)! = G_{n-1}(x)$, and the claim follows by the Fundamental Theorem of Calculus and induction on $n$. Given the claim, we have $F_n(1) = -a_n/n!$ and so we need to evaluate $-\lim_{n\to\infty} \frac{a_n}{\ln n}$. But since the function $1/x$ is strictly decreasing for $x$ positive, $\sum_{k=2}^n 1/k = a_n-1$ is bounded below by $\int_2^n dx/x = \ln n-\ln 2$ and above by $\int_1^n dx/x=\ln n$. It follows that $\lim_{n\to\infty} \frac{a_n}{\ln n} = 1$, and the desired limit is $\boxed{-1}$. | numerical | putnam | Analysis Calculus | Let $F_0(x) = \ln x$. For $n \geq 0$ and $x > 0$, let $F_{n+1}(x) = \int_0^x F_n(t)\,dt$. Evaluate \[ \lim_{n \to \infty} \frac{n! F_n(1)}{\ln n}. \] | We claim that $F_n(x) = (\ln x-a_n)x^n/n!$, where $a_n = \sum_{k=1}^n 1/k$. Indeed, temporarily write $G_n(x) = (\ln x-a_n)x^n/n!$ for $x>0$ and $n\geq 1$; then $\lim_{x\to 0} G_n(x) = 0$ and $G_n'(x) = (\ln x-a_n+1/n) x^{n-1}/(n-1)! = G_{n-1}(x)$, and the claim follows by the Fundamental Theorem of Calculus and induction on $n$. Given the claim, we have $F_n(1) = -a_n/n!$ and so we need to evaluate $-\lim_{n\to\infty} \frac{a_n}{\ln n}$. But since the function $1/x$ is strictly decreasing for $x$ positive, $\sum_{k=2}^n 1/k = a_n-1$ is bounded below by $\int_2^n dx/x = \ln n-\ln 2$ and above by $\int_1^n dx/x=\ln n$. It follows that $\lim_{n\to\infty} \frac{a_n}{\ln n} = 1$, and the desired limit is $\boxed{-1}$. | 0 |
1986 | 1986_B2 | Find all the ordered triples $T=(x-y,y-z,z-x)$, where $x,y,z$ are complex numbers satisfying the simultaneous equations \[ x(x-1)+2yz = y(y-1)+2zx = z(z-1)+2xy, \] and sum the maximum values of $x-y$, $y-z$ and $z-x$ across all possible triples you find as the final answer. | The system is equivalent to \[ 0 = (x-y)(x+y-1-2z) = (y-z)(y+z-1-2x) = (z-x)(z+x-1-2y). \] If no two of $x$, $y$, $z$ are equal, then \[ x+y-1-2z = y+z-1-2x = z+x-1-2y = 0. \] Adding these gives the contradiction $-3 = 0$. So at least two of $x$, $y$, $z$ are equal. If $x = y$ and $y \neq z$, then $z = 2x+1-y = x+1$. In this case we find that $x-y = 0$, $y-z = -1$, and $z-x = 1$. A similar result follows when $y = z$ and $z \neq x$, and when $z = x$ and $x \neq y$. Thus, the only possibilities for $(x-y, y-z, z-x)$ are $(0,0,0)$, $(0,-1,1)$, $(1,0,-1)$, and $(-1,1,0)$. One shows easily that each of these actually occurs. The final answer thus becomes \boxed{3}. | numerical | putnam (modified boxing) | Complex Numbers Algebra | Prove that there are only a finite number of possibilities for the ordered triple $T=(x-y,y-z,z-x)$, where $x,y,z$ are complex numbers satisfying the simultaneous equations \[ x(x-1)+2yz = y(y-1)+2zx = z(z-1)+2xy, \] and list all such triples $T$. | 0 |
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2011 | 2011_A4 | Find the sum of the first $k$ positive integers $n$ for which there is an $n \times n$ matrix with integer entries such that every dot product of a row with itself is even, while every dot product of two different rows is odd? | Let $I$ denote the $n\times n$ identity matrix, and let $A$ denote the $n\times n$ matrix all of whose entries are $1$. If $n$ is odd, then the matrix $A-I$ satisfies the conditions of the problem: the dot product of any row with itself is $n-1$, and the dot product of any two distinct rows is $n-2$. Conversely, suppose $n$ is even, and suppose that the matrix $M$ satisfied the conditions of the problem. Consider all matrices and vectors mod $2$. Since the dot product of a row with itself is equal mod $2$ to the sum of the entries of the row, we have $M v = 0$ where $v$ is the vector $(1,1,\ldots,1)$, and so $M$ is singular. On the other hand, $M M^T = A-I$; since \[ (A-I)^2 = A^2-2A+I = (n-2)A+I = I, \] we have $(\det M)^2 = \det(A-I) = 1$ and $\det M = 1$, contradicting the fact that $M$ is singular. This is true for n odd and hence the answer is \boxed{k^2}. | algebraic | putnam (modified boxing) | Linear Algebra Number Theory | For which positive integers $n$ is there an $n \times n$ matrix with integer entries such that every dot product of a row with itself is even, while every dot product of two different rows is odd? | The answer is \boxed{$n$ odd}. Let $I$ denote the $n\times n$ identity matrix, and let $A$ denote the $n\times n$ matrix all of whose entries are $1$. If $n$ is odd, then the matrix $A-I$ satisfies the conditions of the problem: the dot product of any row with itself is $n-1$, and the dot product of any two distinct rows is $n-2$. Conversely, suppose $n$ is even, and suppose that the matrix $M$ satisfied the conditions of the problem. Consider all matrices and vectors mod $2$. Since the dot product of a row with itself is equal mod $2$ to the sum of the entries of the row, we have $M v = 0$ where $v$ is the vector $(1,1,\ldots,1)$, and so $M$ is singular. On the other hand, $M M^T = A-I$; since \[ (A-I)^2 = A^2-2A+I = (n-2)A+I = I, \] we have $(\det M)^2 = \det(A-I) = 1$ and $\det M = 1$, contradicting the fact that $M$ is singular. | 0 |
1962 | 1962_3 | In a triangle $ABC$ in the Euclidean plane, let $A'$ be a point on the segment from $B$ to $C$, $B'$ a point on the segment from $C$ to $A$, and $C'$ a point on the segment from $A$ to $B$ such that
\[\frac{AB'}{B'C} = \frac{BC'}{C'A} = \frac{CA'}{A'B} = k,\]
where $k$ is a positive constant. Let $\Delta$ be the triangle formed by parts of the segments obtained by joining $A$ and $A'$, $B$ and $B'$, and $C$ and $C'$. Find the ratio of the areas of the triangles $\Delta$ and $\triangle ABC$ in terms of $k$.\] | Use barycentric coordinates for the triangle $ABC$.
Set \( A = (1, 0, 0) \), \( B = (0, 1, 0) \), and \( C = (0, 0, 1) \).
The points $A'$, $B'$, and $C'$ divide their respective sides in the ratio $k:1$. Using the barycentric coordinate formula for such division, we have:
\[A' = \frac{1}{k + 1}(0, k, 1), \quad B' = \frac{1}{k + 1}(1, 0, k), \quad C' = \frac{1}{k + 1}(k, 1, 0).\]
The equations of $BB'$ and $CC'$ are $z = kx$ and $x = ky$, respectively. Their intersection $P$ is given by $(\lambda, \mu, \nu)$, where $\lambda = k\mu$, $\nu = k\lambda$, and $\lambda + \mu + \nu = 1$. Solving gives:
\[P = \frac{1}{k^2 + k + 1}(k, 1, k^2).\]
Similarly, by symmetry:
\[Q = \frac{1}{k^2 + k + 1}(k^2, k, 1), \quad R = \frac{1}{k^2 + k + 1}(1, k^2, k).\]
The area of $\triangle PQR$ relative to $\triangle ABC$ can be calculated using the determinant formula for the area of a triangle in barycentric coordinates:
\[\text{area ratio} = \frac{1}{(k^2 + k + 1)^3} \begin{vmatrix} k & 1 & k^2 \\ k^2 & k & 1 \\ 1 & k^2 & k \ \end{vmatrix}.\]
Expanding the determinant gives:
\[k^3 - 2k^3 + 1 = (k - 1)^2.\]
Thus:
\[\text{area ratio} = \boxed{\frac{(k - 1)^2}{k^2 + k + 1}}.\] | algebraic | putnam (modified boxing) | Geometry | In a triangle $ABC$ in the Euclidean plane, let $A'$ be a point on the segment from $B$ to $C$, $B'$ a point on the segment from $C$ to $A$, and $C'$ a point on the segment from $A$ to $B$ such that
\[\frac{AB'}{B'C} = \frac{BC'}{C'A} = \frac{CA'}{A'B} = k,\]
where $k$ is a positive constant. Let $\Delta$ be the triangle formed by parts of the segments obtained by joining $A$ and $A'$, $B$ and $B'$, and $C$ and $C'$. Prove that the areas of the triangles $\Delta$ and $\triangle ABC$ are in the ratio:
\[\frac{(k - 1)^2}{k^2 + k + 1}.\] | Use barycentric coordinates for the triangle $ABC$.
Set \( A = (1, 0, 0) \), \( B = (0, 1, 0) \), and \( C = (0, 0, 1) \).
The points $A'$, $B'$, and $C'$ divide their respective sides in the ratio $k:1$. Using the barycentric coordinate formula for such division, we have:
\[A' = \frac{1}{k + 1}(0, k, 1), \quad B' = \frac{1}{k + 1}(1, 0, k), \quad C' = \frac{1}{k + 1}(k, 1, 0).\]
The equations of $BB'$ and $CC'$ are $z = kx$ and $x = ky$, respectively. Their intersection $P$ is given by $(\lambda, \mu, \nu)$, where $\lambda = k\mu$, $\nu = k\lambda$, and $\lambda + \mu + \nu = 1$. Solving gives:
\[P = \frac{1}{k^2 + k + 1}(k, 1, k^2).\]
Similarly, by symmetry:
\[Q = \frac{1}{k^2 + k + 1}(k^2, k, 1), \quad R = \frac{1}{k^2 + k + 1}(1, k^2, k).\]
The area of $\triangle PQR$ relative to $\triangle ABC$ can be calculated using the determinant formula for the area of a triangle in barycentric coordinates:
\[\text{area ratio} = \frac{1}{(k^2 + k + 1)^3} \begin{vmatrix} k & 1 & k^2 \\ k^2 & k & 1 \\ 1 & k^2 & k \ \end{vmatrix}.\]
Expanding the determinant gives:
\[k^3 - 2k^3 + 1 = (k - 1)^2.\]
Thus:
\[\text{area ratio} = \frac{(k - 1)^2}{k^2 + k + 1}.\] | 0 |
1974 | 1974_B6 | For a set with $n$ elements, how many subsets are there whose cardinality (the number of elements in the subset) is respectively $\equiv 0 \pmod{3}, \equiv 1 \pmod{3}, \equiv 2 \pmod{3}$? In other words, calculate \[ s_{i,n} = \sum_{k \equiv i \pmod{3}} \binom{n}{k} \quad \text{for } i = 0, 1, 2 and any given $n$. \] | Let $n \equiv r \pmod{6}$ with $r \in \{0,1,2,3,4,5\}$. Then the pattern is \[ \begin{aligned} r & \quad 0 \quad 1 \quad 2 \quad 3 \quad 4 \quad 5 \\ s_{0,n} & \quad a+1 \quad b \quad c \quad d-1 \quad e \quad f \\ s_{1,n} & \quad a \quad b \quad c+1 \quad d \quad e \quad f-1 \\ s_{2,n} & \quad a \quad b-1 \quad c \quad d \quad e+1 \quad f \end{aligned} \] This is easily proved by mathematical induction using the formulas \[ s_{i,n} = s_{i-1,n-1} + s_{i,n-1}. \quad \text{[Here } 0-1 = 2 \pmod{3}\text{.]} \] These formulas follow immediately from the rule \[ \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}. \] The sums may be computed readily using the above patterns and \[ s_{0,n} + s_{1,n} + s_{2,n} = 2^n. \] For $n = 1000$, $r = 4$ and \[ s_{0,1000} = s_{1,1000} = s_{2,1000} - 1 = 2^{1000} - 1}{3}. Thus making the sum for any $n$, \boxed{2^n}. \] | numerical | putnam (modified boxing) | Combinatorics | For a set with $n$ elements, how many subsets are there whose cardinality (the number of elements in the subset) is respectively $\equiv 0 \pmod{3}, \equiv 1 \pmod{3}, \equiv 2 \pmod{3}$? In other words, calculate \[ s_{i,n} = \sum_{k \equiv i \pmod{3}} \binom{n}{k} \quad \text{for } i = 0, 1, 2. \] | Let $n \equiv r \pmod{6}$ with $r \in \{0,1,2,3,4,5\}$. Then the pattern is \[ \begin{aligned} r & \quad 0 \quad 1 \quad 2 \quad 3 \quad 4 \quad 5 \\ s_{0,n} & \quad a+1 \quad b \quad c \quad d-1 \quad e \quad f \\ s_{1,n} & \quad a \quad b \quad c+1 \quad d \quad e \quad f-1 \\ s_{2,n} & \quad a \quad b-1 \quad c \quad d \quad e+1 \quad f \end{aligned} \] This is easily proved by mathematical induction using the formulas \[ s_{i,n} = s_{i-1,n-1} + s_{i,n-1}. \quad \text{[Here } 0-1 = 2 \pmod{3}\text{.]} \] These formulas follow immediately from the rule \[ \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}. \] The sums may be computed readily using the above patterns and \[ s_{0,n} + s_{1,n} + s_{2,n} = 2^n. \] For $n = 1000$, $r = 4$ and \[ s_{0,1000} = s_{1,1000} = s_{2,1000} - 1 = \boxed{\frac{2^{1000} - 1}{3}}. \] | 0 |
1948 | 1948_12 | Let $V_1, V_2, V_3$, and $V$ denote four vertices of a cube. $V_1, V_2$, and $V_3$ are next neighbors of $V$, that is, the lines $VV_1, VV_2$, and $VV_3$ are edges of the cube. Project the cube orthogonally onto a plane (the z-plane, the Gaussian plane) of which the points are marked with complex numbers. Let the projection of $V$ fall in the origin and the projections of $V_1, V_2$, and $V_3$ in points marked with the complex numbers $z_1, z_2$, and $z_3$, respectively. Find the value of $z_1^2 + z_2^2 + z_3^2$. | Take Cartesian coordinates $w, x, y$ in space so that the $x$-axis is the real axis and the $y$-axis is the imaginary axis in the given Gaussian plane. Then the projection maps each point $(w, x, y)$ onto $(0, x, y)$.\n\nSuppose $(w_1, x_1, y_1), (w_2, x_2, y_2)$ and $(w_3, x_3, y_3)$ are mutually orthogonal unit vectors in space. Then the matrix \[ \begin{pmatrix} w_1 & x_1 & y_1 \\ w_2 & x_2 & y_2 \\ w_3 & x_3 & y_3 \end{pmatrix} \] is an orthogonal matrix, so its columns are also mutually orthogonal unit vectors, in particular \[ x_1^2 + x_2^2 + x_3^2 = 1, \quad y_1^2 + y_2^2 + y_3^2 = 1, \quad \text{and} \quad x_1y_1 + x_2y_2 + x_3y_3 = 0. \] The orthogonal projections of the unit vectors into the Gaussian plane are the complex numbers $x_1 + iy_1, x_2 + iy_2$, and $x_3 + iy_3$, and \[ (x_1 + iy_1)^2 + (x_2 + iy_2)^2 + (x_3 + iy_3)^2 = x_1^2 + x_2^2 + x_3^2 - y_1^2 - y_2^2 - y_3^2 + 2i(x_1y_1 + x_2y_2 + x_3y_3) = 0. \]\n\nSuppose the sides of the given cube are of length $a$. Since it is given that the vertex $V$ projects onto the origin, it must be of the form $V = (b, 0, 0)$. There are mutually orthogonal unit vectors $(w_j, x_j, y_j)$ such that \[ V_j = (b, 0, 0) + a(w_j, x_j, y_j) \quad \text{for} \quad j = 1, 2, 3. \] Then the projection of $V_j$ into the Gaussian plane is \[ z_j = a(x_j + iy_j) \] and \[ z_1^2 + z_2^2 + z_3^2 = a^2[(x_1 + iy_1)^2 + (x_2 + iy_2)^2 + (x_3 + iy_3)^2] = \boxed{0}. \] | numerical | putnam (modified boxing) | Geometry Linear Algebra Complex Numbers | Let $V_1, V_2, V_3$, and $V$ denote four vertices of a cube. $V_1, V_2$, and $V_3$ are next neighbors of $V$, that is, the lines $VV_1, VV_2$, and $VV_3$ are edges of the cube. Project the cube orthogonally onto a plane (the z-plane, the Gaussian plane) of which the points are marked with complex numbers. Let the projection of $V$ fall in the origin and the projections of $V_1, V_2$, and $V_3$ in points marked with the complex numbers $z_1, z_2$, and $z_3$, respectively. Show that $z_1^2 + z_2^2 + z_3^2 = 0$. | Take Cartesian coordinates $w, x, y$ in space so that the $x$-axis is the real axis and the $y$-axis is the imaginary axis in the given Gaussian plane. Then the projection maps each point $(w, x, y)$ onto $(0, x, y)$.\n\nSuppose $(w_1, x_1, y_1), (w_2, x_2, y_2)$ and $(w_3, x_3, y_3)$ are mutually orthogonal unit vectors in space. Then the matrix \[ \begin{pmatrix} w_1 & x_1 & y_1 \\ w_2 & x_2 & y_2 \\ w_3 & x_3 & y_3 \end{pmatrix} \] is an orthogonal matrix, so its columns are also mutually orthogonal unit vectors, in particular \[ x_1^2 + x_2^2 + x_3^2 = 1, \quad y_1^2 + y_2^2 + y_3^2 = 1, \quad \text{and} \quad x_1y_1 + x_2y_2 + x_3y_3 = 0. \] The orthogonal projections of the unit vectors into the Gaussian plane are the complex numbers $x_1 + iy_1, x_2 + iy_2$, and $x_3 + iy_3$, and \[ (x_1 + iy_1)^2 + (x_2 + iy_2)^2 + (x_3 + iy_3)^2 = x_1^2 + x_2^2 + x_3^2 - y_1^2 - y_2^2 - y_3^2 + 2i(x_1y_1 + x_2y_2 + x_3y_3) = 0. \]\n\nSuppose the sides of the given cube are of length $a$. Since it is given that the vertex $V$ projects onto the origin, it must be of the form $V = (b, 0, 0)$. There are mutually orthogonal unit vectors $(w_j, x_j, y_j)$ such that \[ V_j = (b, 0, 0) + a(w_j, x_j, y_j) \quad \text{for} \quad j = 1, 2, 3. \] Then the projection of $V_j$ into the Gaussian plane is \[ z_j = a(x_j + iy_j) \] and \[ z_1^2 + z_2^2 + z_3^2 = a^2[(x_1 + iy_1)^2 + (x_2 + iy_2)^2 + (x_3 + iy_3)^2] = 0. \] | 0 |
1999 | 1999_B5 | For an integer $n\geq 3$, let $\theta=2\pi/n$. Evaluate the determinant of the $n\times n$ matrix $I+A$, where $I$ is the $n\times n$ identity matrix and $A=(a_{jk})$ has entries $a_{jk}=\cos(j\theta+k\theta)$ for all $j,k$. | We claim that the eigenvalues of $A$ are $0$ with multiplicity $n-2$, and $n/2$ and $-n/2$, each with multiplicity $1$. To prove this claim, define vectors $v^{(m)}$, $0\leq m\leq n-1$, componentwise by $(v^{(m)})_k = e^{ikm\theta}$, and note that the $v^{(m)}$ form a basis for $\CC^n$. (If we arrange the $v^{(m)}$ into an $n\times n$ matrix, then the determinant of this matrix is a Vandermonde product which is nonzero.) Now note that \begin{align*} (Av^{(m)})_j &= \sum_{k=1}^n \cos(j\theta+k\theta) e^{ikm\theta} \\ &= \frac{e^{ij\theta}}{2} \sum_{k=1}^n e^{ik(m+1)\theta} + \frac{e^{-ij\theta}}{2} \sum_{k=1}^n e^{ik(m-1)\theta}. \end{align*} Since $\sum_{k=1}^n e^{ik\ell\theta} = 0$ for integer $\ell$ unless $n\,|\,\ell$, we conclude that $Av^{(m)}=0$ for $m=0$ or for $2 \leq m \leq n-1$. In addition, we find that $(Av^{(1)})_j = \frac{n}{2} e^{-ij\theta} = \frac{n}{2}(v^{(n-1)})_j$ and $(Av^{(n-1)})_j = \frac{n}{2} e^{ij\theta} = \frac{n}{2}(v^{(1)})_j$, so that $A(v^{(1)} \pm v^{(n-1)}) = \pm \frac{n}{2} (v^{(1)} \pm v^{(n-1)})$. Thus $\{v^{(0)},v^{(2)},v^{(3)},\ldots,v^{(n-2)}, v^{(1)}+v^{(n-1)},v^{(1)}-v^{(n-1)}\}$ is a basis for $\CC^n$ of eigenvectors of $A$ with the claimed eigenvalues. Finally, the determinant of $I+A$ is the product of $(1+\lambda)$ over all eigenvalues $\lambda$ of $A$; in this case, $\det (I+A) = (1+n/2)(1-n/2) = \boxed{1-n^2/4}$. | algebraic | putnam | Complex Numbers Linear Algebra Number Theory | For an integer $n\geq 3$, let $\theta=2\pi/n$. Evaluate the determinant of the $n\times n$ matrix $I+A$, where $I$ is the $n\times n$ identity matrix and $A=(a_{jk})$ has entries $a_{jk}=\cos(j\theta+k\theta)$ for all $j,k$. | We claim that the eigenvalues of $A$ are $0$ with multiplicity $n-2$, and $n/2$ and $-n/2$, each with multiplicity $1$. To prove this claim, define vectors $v^{(m)}$, $0\leq m\leq n-1$, componentwise by $(v^{(m)})_k = e^{ikm\theta}$, and note that the $v^{(m)}$ form a basis for $\CC^n$. (If we arrange the $v^{(m)}$ into an $n\times n$ matrix, then the determinant of this matrix is a Vandermonde product which is nonzero.) Now note that \begin{align*} (Av^{(m)})_j &= \sum_{k=1}^n \cos(j\theta+k\theta) e^{ikm\theta} \\ &= \frac{e^{ij\theta}}{2} \sum_{k=1}^n e^{ik(m+1)\theta} + \frac{e^{-ij\theta}}{2} \sum_{k=1}^n e^{ik(m-1)\theta}. \end{align*} Since $\sum_{k=1}^n e^{ik\ell\theta} = 0$ for integer $\ell$ unless $n\,|\,\ell$, we conclude that $Av^{(m)}=0$ for $m=0$ or for $2 \leq m \leq n-1$. In addition, we find that $(Av^{(1)})_j = \frac{n}{2} e^{-ij\theta} = \frac{n}{2}(v^{(n-1)})_j$ and $(Av^{(n-1)})_j = \frac{n}{2} e^{ij\theta} = \frac{n}{2}(v^{(1)})_j$, so that $A(v^{(1)} \pm v^{(n-1)}) = \pm \frac{n}{2} (v^{(1)} \pm v^{(n-1)})$. Thus $\{v^{(0)},v^{(2)},v^{(3)},\ldots,v^{(n-2)}, v^{(1)}+v^{(n-1)},v^{(1)}-v^{(n-1)}\}$ is a basis for $\CC^n$ of eigenvectors of $A$ with the claimed eigenvalues. Finally, the determinant of $I+A$ is the product of $(1+\lambda)$ over all eigenvalues $\lambda$ of $A$; in this case, $\det (I+A) = (1+n/2)(1-n/2) = \boxed{1-n^2/4}$. | 0 |
1987 | 1987_B1 | Evaluate \[ \int_2^4 \frac{\sqrt{\ln(9-x)}\,dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}. \] | Let $I$ denote the value of the integral. The substitution $9 - x = y + 3$ gives \[ I = \int_2^4 \frac{\sqrt{\ln(y+3)}\,dy}{\sqrt{\ln(y+3)}+\sqrt{\ln(9-y)}}. \] So, \[ 2I = \int_2^4 \frac{\sqrt{\ln(x+3)} + \sqrt{\ln(9-x)}}{\sqrt{\ln(x+3)} + \sqrt{\ln(9-x)}}\,dx = 2, \] and $I = \boxed{1}$. | numerical | putnam | Calculus | Evaluate \[ \int_2^4 \frac{\sqrt{\ln(9-x)}\,dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}. \] | 0 |
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1965 | 1965_B3 | Find the sum of all sides of all the right-angled triangles whose sides are integers while the area is numerically equal to twice the perimeter. | All Pythagorean triples can be obtained from $x = \lambda(p^2 - q^2)$, $y = 2\lambda pq$, $z = \lambda(p^2 + q^2)$ where $0 < q < p$, $(p, q) = 1$ and $p \not\equiv q \pmod{2}$, $\lambda$ being any natural number.
The problem requires that $\frac{1}{2}xy = 2(x+y+z)$. This condition can be written as $\lambda^2(p^2-q^2)(pq) = 2\lambda(p^2-q^2+2pq+p^2+q^2)$ or simply $\lambda(p-q)q = 4$. Since $p-q$ is odd it follows that $p-q = 1$ and the only possibilities for $q$ are $1, 2, 4$.
- If $q = 1$, $p = 2$, $\lambda = 4$, $x = 12$, $y = 16$, $z = 20$.
- If $q = 2$, $p = 3$, $\lambda = 2$, $x = 10$, $y = 24$, $z = 26$.
- If $q = 4$, $p = 5$, $\lambda = 1$, $x = 9$, $y = 40$, $z = 41$. This gives us the final answer as $12+16+20+10+24+26+9+40+41 = \boxed{198}$. | numerical | putnam (modified boxing) | Geometry Number Theory | Prove that there are exactly three right-angled triangles whose sides are integers while the area is numerically equal to twice the perimeter. | All Pythagorean triples can be obtained from $x = \lambda(p^2 - q^2)$, $y = 2\lambda pq$, $z = \lambda(p^2 + q^2)$ where $0 < q < p$, $(p, q) = 1$ and $p \not\equiv q \pmod{2}$, $\lambda$ being any natural number.
The problem requires that $\frac{1}{2}xy = 2(x+y+z)$. This condition can be written as $\lambda^2(p^2-q^2)(pq) = 2\lambda(p^2-q^2+2pq+p^2+q^2)$ or simply $\lambda(p-q)q = 4$. Since $p-q$ is odd it follows that $p-q = 1$ and the only possibilities for $q$ are $1, 2, 4$.
- If $q = 1$, $p = 2$, $\lambda = 4$, $x = 12$, $y = 16$, $z = 20$.
- If $q = 2$, $p = 3$, $\lambda = 2$, $x = 10$, $y = 24$, $z = 26$.
- If $q = 4$, $p = 5$, $\lambda = 1$, $x = 9$, $y = 40$, $z = 41$. | 0 |
1995 | 1995_A2 | Let $S(k)$ be the sum of the coordinates of the first $k$ pairs $(a,b)$ of positive integers for which the improper integral \[ \int_{b}^{\infty} \left( \sqrt{\sqrt{x+a}-\sqrt{x}} - \sqrt{\sqrt{x}-\sqrt{x-b}} \right)\,dx \] converge. Find a closed-form expression for $S(k)$. | The easiest proof uses ``big-O'' notation and the fact that $(1+x)^{1/2} = 1 + x/2 + O(x^{2})$ for $|x|<1$. (Here $O(x^{2})$ means bounded by a constant times $x^{2}$.) So \begin{align*} \sqrt{x+a}-\sqrt{x} &= x^{1/2}(\sqrt{1+a/x} - 1) \\ &= x^{1/2}(1 + a/2x + O(x^{-2})), \end{align*} hence \[ \sqrt{\sqrt{x+a} - \sqrt{x}} = x^{1/4} (a/4x + O(x^{-2})) \] and similarly \[ \sqrt{\sqrt{x} - \sqrt{x-b}} = x^{1/4} (b/4x + O(x^{-2})). \] Hence the integral we're looking at is \[ \int_{b}^{\infty} x^{1/4} ((a-b)/4x + O(x^{-2}))\,dx. \] The term $x^{1/4} O(x^{-2})$ is bounded by a constant times $x^{-7/4}$, whose integral converges. Thus we only have to decide whether $x^{-3/4} (a-b)/4$ converges. But $x^{-3/4}$ has divergent integral, so we get convergence if and only if $a=b$ (in which case the integral telescopes anyway). The integral converges iff $a=b$ making the required expression \boxed{k(k+1)}. | algebraic | putnam (modified boxing) | Analysis Calculus | For what pairs $(a,b)$ of positive real numbers does the improper integral \[ \int_{b}^{\infty} \left( \sqrt{\sqrt{x+a}-\sqrt{x}} - \sqrt{\sqrt{x}-\sqrt{x-b}} \right)\,dx \] converge? | The integral converges \boxed{iff $a=b$}. The easiest proof uses ``big-O'' notation and the fact that $(1+x)^{1/2} = 1 + x/2 + O(x^{2})$ for $|x|<1$. (Here $O(x^{2})$ means bounded by a constant times $x^{2}$.) So \begin{align*} \sqrt{x+a}-\sqrt{x} &= x^{1/2}(\sqrt{1+a/x} - 1) \\ &= x^{1/2}(1 + a/2x + O(x^{-2})), \end{align*} hence \[ \sqrt{\sqrt{x+a} - \sqrt{x}} = x^{1/4} (a/4x + O(x^{-2})) \] and similarly \[ \sqrt{\sqrt{x} - \sqrt{x-b}} = x^{1/4} (b/4x + O(x^{-2})). \] Hence the integral we're looking at is \[ \int_{b}^{\infty} x^{1/4} ((a-b)/4x + O(x^{-2}))\,dx. \] The term $x^{1/4} O(x^{-2})$ is bounded by a constant times $x^{-7/4}$, whose integral converges. Thus we only have to decide whether $x^{-3/4} (a-b)/4$ converges. But $x^{-3/4}$ has divergent integral, so we get convergence if and only if $a=b$ (in which case the integral telescopes anyway). | 0 |
1976 | 1976_B5 | \[ \sum_{k=1}^n (-1)^k \binom{n}{k} (x-k)^n. \] | The sum is \( n! \) since it is an \( n \)-th difference of a monic polynomial, \( x^n \), of degree \( n \) The final answer is \boxed{n!}. | numerical | putnam | Algebra | \[ \sum_{k=1}^n (-1)^k \binom{n}{k} (x-k)^n. \] | The sum is \( \boxed{n!} \) since it is an \( n \)-th difference of a monic polynomial, \( x^n \), of degree \( n \). | 0 |
1973 | 1973_A1 | Let $ABC$ be any triangle. Let $X, Y, Z$ be points on the sides $BC, CA, AB$ respectively. Suppose the distances $BX \leq XC, CY \leq YA, AZ \leq ZB$. Given that the area of the triangle $XYZ$ is $\geq n$ (area of triangle $ABC$) find the lowest non-zero value of $n$. | (a) If $X, Y, Z$ are at the midpoints of the sides, the area of $\triangle XYZ$ is one fourth of the area of $\triangle ABC$. Also, as long as $BX \leq XC, CY \leq YA$ and $AZ \leq ZB$, moving one of $X, Y, Z$ to the midpoint of its side, while leaving the other two fixed, does not increase the area of $\triangle XYZ$ since the altitude to the fixed base of $\triangle XYZ$ decreases or remains constant. This gives us \boxed{\frac{1}{4}}. | numerical | putnam | Geometry | (a) Let $ABC$ be any triangle. Let $X, Y, Z$ be points on the sides $BC, CA, AB$ respectively. Suppose the distances $BX \leq XC, CY \leq YA, AZ \leq ZB$ (see Figure 1). Show that the area of the triangle $XYZ$ is $\geq (1/4)$ (area of triangle $ABC$).
(b) Let $ABC$ be any triangle, and let $X, Y, Z$ be points on the sides $BC, CA, AB$ respectively (but without any assumption about the ratios of the distances $BX/XC$, etc.; see Figures 1 and 2). Using (a) or by any other method, show: One of the three corner triangles $AZY, BXZ, CYX$ has an area $\leq$ area of triangle $XYZ$. | (a) If $X, Y, Z$ are at the midpoints of the sides, the area of $\triangle XYZ$ is one fourth of the area of $\triangle ABC$. Also, as long as $BX \leq XC, CY \leq YA$ and $AZ \leq ZB$, moving one of $X, Y, Z$ to the midpoint of its side, while leaving the other two fixed, does not increase the area of $\triangle XYZ$ since the altitude to the fixed base of $\triangle XYZ$ decreases or remains constant.
(b) Under the hypothesis of (a) the three corner triangles have no more than three fourths of the total area and so one of them must have smaller area than $\triangle XYZ$. All other cases are similar to the one in which $XC < BX$ and $CY < YA$. Then consideration of the altitudes to base $XY$ shows that $\triangle CYX$ has smaller area than $\triangle XYZ$. | 0 |
1969 | 1969_A3 | Let $P$ be a non-self-intersecting closed polygon with $n$ sides. Let its vertices be $P_1, P_2, \ldots, P_n$. Let $m$ other points, $Q_1, Q_2, \ldots, Q_m$ interior to $P$ be given. Let the figure be triangulated. This means that certain pairs of the $(n+m)$ points $P_1, \ldots, Q_m$ are connected by line segments such that (i) the resulting figure consists exclusively of a set $T$ of triangles, (ii) if two different triangles in $T$ have more than a vertex in common then they have exactly a side in common, and (iii) the set of vertices of the triangles in $T$ is precisely the set of $(n+m)$ points $P_1, \ldots, Q_m$. How many triangles in $T$? | Let $t$ be the number of triangles. The sum of all the angles is $\pi t$ (since it is $\pi$ for each triangle) and it is also $\boxed{2\pi m + (n-2)\pi}$. | algebraic | putnam | Geometry | Let $P$ be a non-self-intersecting closed polygon with $n$ sides. Let its vertices be $P_1, P_2, \ldots, P_n$. Let $m$ other points, $Q_1, Q_2, \ldots, Q_m$ interior to $P$ be given. Let the figure be triangulated. This means that certain pairs of the $(n+m)$ points $P_1, \ldots, Q_m$ are connected by line segments such that (i) the resulting figure consists exclusively of a set $T$ of triangles, (ii) if two different triangles in $T$ have more than a vertex in common then they have exactly a side in common, and (iii) the set of vertices of the triangles in $T$ is precisely the set of $(n+m)$ points $P_1, \ldots, Q_m$. How many triangles in $T$? | Let $t$ be the number of triangles. The sum of all the angles is $\pi t$ (since it is $\pi$ for each triangle) and it is also $\boxed{2\pi m + (n-2)\pi}$.
Alternate Solution: Let $t$ be the number of triangles. In Euler’s formula $V - E + F = 2$, $F = t + 1$, and $V = n + m$. Since every edge is on two faces, $2E = 3t + n$. Substitution leads directly to the answer $t = 2m + n - 2$. | 0 |
1992 | 1992_B3 | For any pair $(x, y)$ of real numbers, a sequence $(a_n(x,y))_{n\geq 0}$ is defined as follows: \begin{align*} a_0(x, y) &= x, \\ a_{n+1}(x, y) &= \frac{(a_n(x, y))^2 + y^2}{2}, \qquad \mbox{for $n \geq 0$}. \end{align*} Find the area of the region \[ \{ (x, y) | (a_n(x, y))_{n \geq 0}\ \mbox{converges}\}. \] | The region of convergence is
\[\text{namely, a (closed) square } \{(x, y) | -1 \leq x, y \leq 1\} \text{ of side 2 with (closed) semicircles of radius 1 centered at } (\pm 1, 0) \text{ described on two opposite sides.}\]
If $\lim_{n \to \infty} a_n(x, y) = L$, then $L$ must satisfy $L = \frac{(L^2 + y^2)}{2}$; that is, $L$ must be a root of the equation
\[ r^2 - 2r + y^2 = 0. \quad \text{(1)} \]
In such case, the equation must have real roots, so the discriminant, $4 - 4y^2$, must be nonnegative. Thus, a necessary condition for $(a_n(x, y))$ to converge is that $|y| \leq 1$.
Fix $|y| \leq 1$. The roots of (1) are then $1 - \sqrt{1 - y^2}$ and $1 + \sqrt{1 - y^2}$, which are real and nonnegative. As $a_1(-x, y) = a_1(x, y)$, the interval of convergence is symmetric about $x = 0$. We shall assume then that $x \geq 0$; thus, $a_n(x, y) \geq 0$, for all $n$.
If $r_0 = 1 \pm \sqrt{1 - y^2}$, then $a_{n+1}(x, y)$ is less than, equal to, or greater than $r_0$ according to whether $a_n(x, y)$ is less than, equal to, or greater than $r_0 (= \frac{(r_0^2 + y^2)}{2})$.
If $a_n(x, y)$ lies in the closed interval $[1 - \sqrt{1 - y^2}, 1 + \sqrt{1 - y^2}]$, that is, between the roots of (1), then
\[ (a_n(x, y))^2 - 2a_n(x, y) + y^2 \leq 0, \]
so that
\[ 1 - \sqrt{1 - y^2} \leq a_{n+1}(x, y) \leq a_n(x, y). \]
It follows that $(a_n(x, y))_{n \geq 0}$ converges if $x$ is in the closed interval $[1 - \sqrt{1 - y^2}, 1 + \sqrt{1 - y^2}]$.
If $a_n(x, y)$ does not lie in the interval $[1 - \sqrt{1 - y^2}, 1 + \sqrt{1 - y^2}]$, then
\[ (a_n(x, y))^2 - 2a_n(x, y) + y^2 > 0, \]
so that
\[ a_{n+1}(x, y) > a_n(x, y). \]
Thus, if $x$, and therefore all $a_n(x, y)$, are greater than $1 + \sqrt{1 - y^2}$, then the sequence diverges. On the other hand, if $x$, and therefore all $a_n(x, y)$, lie between 0 and $1 - \sqrt{1 - y^2}$, the sequence converges monotonically to $1 - \sqrt{1 - y^2}$.
To summarize, $(a_n(x, y))_{n \geq 0}$ converges if and only if
\[ -1 \leq y \leq 1 \text{ and } -\left(1 + \sqrt{1 - y^2}\right) \leq x \leq 1 + \sqrt{1 - y^2}. \] The area is $\boxed{4 + \pi}$. | numerical | putnam | Analysis Calculus Geometry | For any pair $(x, y)$ of real numbers, a sequence $(a_n(x,y))_{n\geq 0}$ is defined as follows: \begin{align*} a_0(x, y) &= x, \\ a_{n+1}(x, y) &= \frac{(a_n(x, y))^2 + y^2}{2}, \qquad \mbox{for $n \geq 0$}. \end{align*} Find the area of the region \[ \{ (x, y) | (a_n(x, y))_{n \geq 0}\ \mbox{converges}\}. \] | 0 |
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1990 | 1990_B6 | Let $S$ be a nonempty closed bounded convex set in the plane. Let $K$ be a line and $t$ a positive number. Let $L_1$ and $L_2$ be support lines for $S$ parallel to $K_1$, and let $\overline{L}$ be the line parallel to $K$ and midway between $L_1$ and $L_2$. Let $B_S(K, t)$ be the band of points whose distance from $\overline{L}$ is at most $(t/2)w$, where $w$ is the distance between $L_1$ and $L_2$. What is the smallest $t$ such that \[ S \cap \bigcap_K B_S(K, t) \neq \emptyset \] for all $S$? ($K$ runs over all lines in the plane.) | Consider the dissection of the equilateral triangle (as shown) into 9 similar equilateral triangles. It shows that any $t < 1/3$ produces an empty intersection. We now show that the intersection is nonempty for $t \geq 1/3$, since it always contains the centroid of $S$. Think of $L_1$ as the upper support line (see sketch). Let $P_1$ be a point of contact of $L_1$ with $S$, and $P_2$ a point of contact of $L_2$ with $S$. Extend two lines from $P_1$ to points $Q_1, Q_2$ on $L_2$ so that $A_1$, the region "above" $P_1Q_1$ and in $S$ has area equal to that of $B_1$, the region "below" $P_1Q_1$ above $L_2$ and outside of $S$. If $S$ is perturbed by replacing $A_1$ by $B_1$ (and similarly $A_2$ by $B_2$) the new $S$ (it is a triangle) will have a "lower" centroid. But this new centroid is still $1/3$ of the way above $L_2$ (on the way to $L_1$). Hence if $\beta$ is a band comprising the middle $1/3$ of the strip between $L_1$ and $L_2$, it contains the centroid of $S$. Hence for $t = 1/3$, the intersection is nonempty. Hence $t = \boxed{1/3}$ is the smallest such value. | numerical | putnam | Geometry Analysis | Let $S$ be a nonempty closed bounded convex set in the plane. Let $K$ be a line and $t$ a positive number. Let $L_1$ and $L_2$ be support lines for $S$ parallel to $K_1$, and let $\overline{L}$ be the line parallel to $K$ and midway between $L_1$ and $L_2$. Let $B_S(K, t)$ be the band of points whose distance from $\overline{L}$ is at most $(t/2)w$, where $w$ is the distance between $L_1$ and $L_2$. What is the smallest $t$ such that \[ S \cap \bigcap_K B_S(K, t) \neq \emptyset \] for all $S$? ($K$ runs over all lines in the plane.) | 0 |
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1975 | 1975_A4 | Let $n = 2m$, where $m$ is an odd integer greater than 1. Let $\theta = e^{2\pi i / n}$. Express $(1-\theta)^{-1}$ explicitly as a polynomial in $\theta$,
$a_k \theta^k + a_{k-1} \theta^{k-1} + \cdots + a_1 \theta + a_0$,
with integer coefficients $a_i$.
[Note that $\theta$ is a primitive $n$-th root of unity, and thus it satisfies all of the identities which hold for such roots.] | Let $n = 4k + 2$ with $k > 0$. Then
$$0 = \theta^n - 1 = \theta^{4k+2} - 1 = (\theta^{2k+1} - 1)(\theta + 1)(\theta^{2k} - \theta^{2k-1} + \theta^{2k-2} - \cdots - \theta + 1).$$
Since $\theta$ is a primitive $n$-th root of unity with $n > 2k + 1$ and $n > 2$,
$$(\theta^{2k+1} - 1)(\theta + 1) \neq 0.$$
Hence
(A)
$$\theta^{2k} - \theta^{2k-1} + \theta^{2k-2} - \cdots + \theta^2 - \theta + 1 = 0,$$
$$1 = \theta - \theta^2 + \theta^3 - \cdots - \theta^{2k} = (1 - \theta)(\theta + \theta^2 + \theta^3 + \cdots + \theta^{2k-1}),$$
$$(1 - \theta)^{-1} = \theta + \theta^3 + \cdots + \theta^{2k-1} \quad \text{[where $2k - 1 = (n - 4)/2$]}.$$
Another solution is $(1 - \theta)^{-1} = \boxed{1 + \theta^2 + \theta^4 + \cdots + \theta^{2k}}$ as one sees from (A). | algebraic | putnam | Algebra Complex Numbers | Let $n = 2m$, where $m$ is an odd integer greater than 1. Let $\theta = e^{2\pi i / n}$. Express $(1-\theta)^{-1}$ explicitly as a polynomial in $\theta$,
$a_k \theta^k + a_{k-1} \theta^{k-1} + \cdots + a_1 \theta + a_0$,
with integer coefficients $a_i$.
[Note that $\theta$ is a primitive $n$-th root of unity, and thus it satisfies all of the identities which hold for such roots.] | Let $n = 4k + 2$ with $k > 0$. Then
$$0 = \theta^n - 1 = \theta^{4k+2} - 1 = (\theta^{2k+1} - 1)(\theta + 1)(\theta^{2k} - \theta^{2k-1} + \theta^{2k-2} - \cdots - \theta + 1).$$
Since $\theta$ is a primitive $n$-th root of unity with $n > 2k + 1$ and $n > 2$,
$$(\theta^{2k+1} - 1)(\theta + 1) \neq 0.$$
Hence
(A)
$$\theta^{2k} - \theta^{2k-1} + \theta^{2k-2} - \cdots + \theta^2 - \theta + 1 = 0,$$
$$1 = \theta - \theta^2 + \theta^3 - \cdots - \theta^{2k} = (1 - \theta)(\theta + \theta^2 + \theta^3 + \cdots + \theta^{2k-1}),$$
$$(1 - \theta)^{-1} = \theta + \theta^3 + \cdots + \theta^{2k-1} \quad \text{[where $2k - 1 = (n - 4)/2$]}.$$
Another solution is $(1 - \theta)^{-1} = \boxed{1 + \theta^2 + \theta^4 + \cdots + \theta^{2k}}$ as one sees from (A). | 0 |
1962 | 1962_4 | Assume that $|f(x)| \leq 1$ and $|f''(x)| \leq 1$ for all $x$ on an interval of length at least 2. Find the lower bound of $|f'(x)|$ on the interval. | We may suppose without loss of generality that the interval in question is $[-1, 1]$. Using Taylor's formula to expand about the point $x \in [-1, 1]$, we find:
\[
f(1) = f(x) + (1 - x)f'(x) + \frac{1}{2}(1 - x)^2 f''(\xi),
f(-1) = f(x) + (-1 - x)f'(x) + \frac{1}{2}(-1 - x)^2 f''(\eta),
\]
where $\xi \in (x, 1)$ and $\eta \in (-1, x)$. Hence,
\[ f(1) - f(-1) = 2f'(x) + \frac{1}{2}(1 - x)^2f''(\xi) - \frac{1}{2}(1 + x)^2f''(\eta). \]
Using the given bounds for $f$ and $f''$, we get:
\[
2|f'(x)| \leq |f(1)| + |f(-1)| + \frac{1}{2}(1 - x)^2|f''(\xi)| + \frac{1}{2}(1 + x)^2|f''(\eta)|.\]
Since $|f(1)|, |f(-1)| \leq 1$ and $|f''(\xi)|, |f''(\eta)| \leq 1$, it follows that:
\[
2|f'(x)| \leq 2 + \frac{1}{2}(1 - x)^2 + \frac{1}{2}(1 + x)^2 = 3 + x^2 \leq 4.
\]
Thus:
\[
|f'(x)| \leq \boxed{2}.
\] | numerical | putnam (modified boxing) | Analysis | Assume that $|f(x)| \leq 1$ and $|f''(x)| \leq 1$ for all $x$ on an interval of length at least 2. Show that $|f'(x)| \leq 2$ on the interval. | We may suppose without loss of generality that the interval in question is $[-1, 1]$. Using Taylor's formula to expand about the point $x \in [-1, 1]$, we find:
\[
f(1) = f(x) + (1 - x)f'(x) + \frac{1}{2}(1 - x)^2 f''(\xi),
f(-1) = f(x) + (-1 - x)f'(x) + \frac{1}{2}(-1 - x)^2 f''(\eta),
\]
where $\xi \in (x, 1)$ and $\eta \in (-1, x)$. Hence,
\[ f(1) - f(-1) = 2f'(x) + \frac{1}{2}(1 - x)^2f''(\xi) - \frac{1}{2}(1 + x)^2f''(\eta). \]
Using the given bounds for $f$ and $f''$, we get:
\[
2|f'(x)| \leq |f(1)| + |f(-1)| + \frac{1}{2}(1 - x)^2|f''(\xi)| + \frac{1}{2}(1 + x)^2|f''(\eta)|.\]
Since $|f(1)|, |f(-1)| \leq 1$ and $|f''(\xi)|, |f''(\eta)| \leq 1$, it follows that:
\[
2|f'(x)| \leq 2 + \frac{1}{2}(1 - x)^2 + \frac{1}{2}(1 + x)^2 = 3 + x^2 \leq 4.
\]
Thus:
\[
|f'(x)| \leq 2.
\] | 0 |
2015 | 2015_B5 | Let $P_n$ be the number of permutations $\pi$ of $\{1,2,\dots,n\}$ such that \[ |i-j| = 1 \mbox{ implies } |\pi(i) -\pi(j)| \leq 2 \] for all $i,j$ in $\{1,2,\dots,n\}$. Show that for $n \geq 2$, the quantity \[ P_{n+5} - P_{n+4} - P_{n+3} + P_n \] does not depend on $n$, and find its value. | Assume $n \geq 3$ for the moment. We write the permutations $\pi$ counted by $P_n$ as sequences $\pi(1),\pi(2),\ldots,\pi(n)$. Let $U_n$ be the number of permutations counted by $P_n$ that end with $n-1,n$; let $V_n$ be the number ending in $n,n-1$; let $W_n$ be the number starting with $n-1$ and ending in $n-2,n$; let $T_n$ be the number ending in $n-2,n$ but not starting with $n-1$; and let $S_n$ be the number which has $n-1,n$ consecutively in that order, but not at the beginning or end. It is clear that every permutation $\pi$ counted by $P_n$ either lies in exactly one of the sets counted by $U_n, V_n, W_n, T_n, S_n$, or is the reverse of such a permutation. Therefore \[ P_n = 2 (U_n + V_n + W_n+ T_n+ S_n). \] By examining how each of the elements in the sets counted by $U_{n+1}, V_{n+1}, W_{n+1}, T_{n+1}, S_{n+1}$ can be obtained from a (unique) element in one of the sets counted by $U_n, V_n, W_n, T_n, S_n$ by suitably inserting the element $n+1$, we obtain the recurrence relations \begin{align*} U_{n+1} &= U_n+W_n+T_n, \\ V_{n+1}&=U_n, \\ W_{n+1}&=W_n, \\ T_{n+1}&=V_n, \\ S_{n+1}&=S_n+V_n. \end{align*} Also, it is clear that $W_n=1$ for all $n$. So far we have assumed $n \geq 3$, but it is straightforward to extrapolate the sequences $P_n,U_n,V_n,W_n,T_n,S_n$ back to $n=2$ to preserve the preceding identities. Hence for all $n \geq 2$, \begin{align*} P_{n+5} &= 2(U_{n+5}+V_{n+5}+W_{n+5}+T_{n+5}+S_{n+5}) \\ &= 2((U_{n+4}+W_{n+4}+T_{n+4})+U_{n+4}\\ & \qquad + W_{n+4}+V_{n+4}+(S_{n+4}+V_{n+4})) \\ &= P_{n+4} + 2(U_{n+4}+W_{n+4}+V_{n+4}) \\ &= P_{n+4} + 2((U_{n+3}+W_{n+3}+T_{n+3})+W_{n+3}+U_{n+3}) \\ &= P_{n+4} + P_{n+3} + 2(U_{n+3}-V_{n+3}+W_{n+3}-S_{n+3}) \\ &= P_{n+4} + P_{n+3} + 2((U_{n+2}+W_{n+2}+T_{n+2})-U_{n+2}\\ &\qquad +W_{n+2}-(S_{n+2}-V_{n+2})) \\ &= P_{n+4} + P_{n+3} + 2(2W_{n+2}+T_{n+2}-S_{n+2}-V_{n+2}) \\ &= P_{n+4} + P_{n+3} + 2(2W_{n+1}+V_{n+1}\\ &\qquad -(S_{n+1}+V_{n+1})-U_{n+1}) \\ &= P_{n+4} + P_{n+3} + 2(2W_n+U_n-(S_n+V_n)-U_n\\ &\qquad -(U_n+W_n+T_n)) \\ &= P_{n+4} + P_{n+3} - P_n + 4, \end{align*} as desired. The answer is \boxed{4}. | numerical | putnam | Combinatorics | Let $P_n$ be the number of permutations $\pi$ of $\{1,2,\dots,n\}$ such that \[ |i-j| = 1 \mbox{ implies } |\pi(i) -\pi(j)| \leq 2 \] for all $i,j$ in $\{1,2,\dots,n\}$. Show that for $n \geq 2$, the quantity \[ P_{n+5} - P_{n+4} - P_{n+3} + P_n \] does not depend on $n$, and find its value. | The answer is \boxed{4}. Assume $n \geq 3$ for the moment. We write the permutations $\pi$ counted by $P_n$ as sequences $\pi(1),\pi(2),\ldots,\pi(n)$. Let $U_n$ be the number of permutations counted by $P_n$ that end with $n-1,n$; let $V_n$ be the number ending in $n,n-1$; let $W_n$ be the number starting with $n-1$ and ending in $n-2,n$; let $T_n$ be the number ending in $n-2,n$ but not starting with $n-1$; and let $S_n$ be the number which has $n-1,n$ consecutively in that order, but not at the beginning or end. It is clear that every permutation $\pi$ counted by $P_n$ either lies in exactly one of the sets counted by $U_n, V_n, W_n, T_n, S_n$, or is the reverse of such a permutation. Therefore \[ P_n = 2 (U_n + V_n + W_n+ T_n+ S_n). \] By examining how each of the elements in the sets counted by $U_{n+1}, V_{n+1}, W_{n+1}, T_{n+1}, S_{n+1}$ can be obtained from a (unique) element in one of the sets counted by $U_n, V_n, W_n, T_n, S_n$ by suitably inserting the element $n+1$, we obtain the recurrence relations \begin{align*} U_{n+1} &= U_n+W_n+T_n, \\ V_{n+1}&=U_n, \\ W_{n+1}&=W_n, \\ T_{n+1}&=V_n, \\ S_{n+1}&=S_n+V_n. \end{align*} Also, it is clear that $W_n=1$ for all $n$. So far we have assumed $n \geq 3$, but it is straightforward to extrapolate the sequences $P_n,U_n,V_n,W_n,T_n,S_n$ back to $n=2$ to preserve the preceding identities. Hence for all $n \geq 2$, \begin{align*} P_{n+5} &= 2(U_{n+5}+V_{n+5}+W_{n+5}+T_{n+5}+S_{n+5}) \\ &= 2((U_{n+4}+W_{n+4}+T_{n+4})+U_{n+4}\\ & \qquad + W_{n+4}+V_{n+4}+(S_{n+4}+V_{n+4})) \\ &= P_{n+4} + 2(U_{n+4}+W_{n+4}+V_{n+4}) \\ &= P_{n+4} + 2((U_{n+3}+W_{n+3}+T_{n+3})+W_{n+3}+U_{n+3}) \\ &= P_{n+4} + P_{n+3} + 2(U_{n+3}-V_{n+3}+W_{n+3}-S_{n+3}) \\ &= P_{n+4} + P_{n+3} + 2((U_{n+2}+W_{n+2}+T_{n+2})-U_{n+2}\\ &\qquad +W_{n+2}-(S_{n+2}-V_{n+2})) \\ &= P_{n+4} + P_{n+3} + 2(2W_{n+2}+T_{n+2}-S_{n+2}-V_{n+2}) \\ &= P_{n+4} + P_{n+3} + 2(2W_{n+1}+V_{n+1}\\ &\qquad -(S_{n+1}+V_{n+1})-U_{n+1}) \\ &= P_{n+4} + P_{n+3} + 2(2W_n+U_n-(S_n+V_n)-U_n\\ &\qquad -(U_n+W_n+T_n)) \\ &= P_{n+4} + P_{n+3} - P_n + 4, \end{align*} as desired. | 0 |
1966 | 1966_A3 | Let $0 < x_1 < 1$ and $x_{n+1} = x_n(1 - x_n), \ n = 1, 2, 3, \ldots$. Find the value of
\[
\lim_{n \to \infty} nx_n.
\] | Multiplying the defining relation by $(n+1)$ we get
\[
(n+1)x_{n+1} = nx_n + x_n - (n+1)(x_n)^2 = nx_n + x_n[1 - (n+1)x_n].
\]
To prove that $nx_n$ is increasing, we need to show that $1 - (n+1)x_n \geq 0$. From the graph of $x(1-x)$ we note that $x_2 \leq \frac{1}{4}$ and that $x_n \leq a \leq \frac{1}{2}$ implies $x_{n+1} \leq a(1-a)$. So by induction,
\[
(n+1)x_n \leq (n+1) \frac{1}{n} \left(1 - \frac{1}{n}\right) = 1 - \frac{1}{n^2} \leq 1.
\]
Furthermore, $nx_n < (n+1)x_n \leq 1$ and so $nx_n$ is bounded above by 1. Thus $nx_n$ converges to a limit $\alpha$ with $0 < nx_n < \alpha \leq 1$. Now summing (1) from 2 to $n$ we obtain
\[
1 \geq (n+1)x_{n+1} = 2x_2 + x_2(1 - 3x_2) + x_3(1 - 4x_3) + \cdots + x_n[1 - (n+1)x_n].
\]
If $\alpha \neq 1$ then $[1 - (n+1)x_n] \geq (1-\alpha)/2$ for all large $n$ and thus (2) would show that $\sum x_n$ is convergent. However $nx_n \geq x_1$ and so $\sum x_n \geq x_1 \sum (1/n)$ making the final answer \boxed{1}. | numerical | putnam (modified boxing) | Analysis | Let $0 < x_1 < 1$ and $x_{n+1} = x_n(1 - x_n), \ n = 1, 2, 3, \ldots$. Show that
\[
\lim_{n \to \infty} nx_n = 1.
\] | Multiplying the defining relation by $(n+1)$ we get
\[
(n+1)x_{n+1} = nx_n + x_n - (n+1)(x_n)^2 = nx_n + x_n[1 - (n+1)x_n].
\]
To prove that $nx_n$ is increasing, we need to show that $1 - (n+1)x_n \geq 0$. From the graph of $x(1-x)$ we note that $x_2 \leq \frac{1}{4}$ and that $x_n \leq a \leq \frac{1}{2}$ implies $x_{n+1} \leq a(1-a)$. So by induction,
\[
(n+1)x_n \leq (n+1) \frac{1}{n} \left(1 - \frac{1}{n}\right) = 1 - \frac{1}{n^2} \leq 1.
\]
Furthermore, $nx_n < (n+1)x_n \leq 1$ and so $nx_n$ is bounded above by 1. Thus $nx_n$ converges to a limit $\alpha$ with $0 < nx_n < \alpha \leq 1$. Now summing (1) from 2 to $n$ we obtain
\[
1 \geq (n+1)x_{n+1} = 2x_2 + x_2(1 - 3x_2) + x_3(1 - 4x_3) + \cdots + x_n[1 - (n+1)x_n].
\]
If $\alpha \neq 1$ then $[1 - (n+1)x_n] \geq (1-\alpha)/2$ for all large $n$ and thus (2) would show that $\sum x_n$ is convergent. However $nx_n \geq x_1$ and so $\sum x_n \geq x_1 \sum (1/n)$. | 0 |
1997 | 1997_A1 | A rectangle, $HOMF$, has sides $HO=11$ and $OM=5$. A triangle $ABC$ has $H$ as the intersection of the altitudes, $O$ the center of the circumscribed circle, $M$ the midpoint of $BC$, and $F$ the foot of the altitude from $A$. What is the length of $BC$? | The centroid $G$ of the triangle is collinear with $H$ and $O$ (Euler line), and the centroid lies two-thirds of the way from $A$ to $M$. Therefore $H$ is also two-thirds of the way from $A$ to $F$, so $AF = 15$. Since the triangles $BFH$ and $AFC$ are similar (they're right triangles and \[ \angle HBC = \pi/2 - \angle C = \angle CAF), \] we have \[ BF/FH = AF/FC \] or \[ BF \cdot FC = FH \cdot AF = 75. \] Now \[ BC^2 = (BF + FC)^2 = (BF - FC)^2 + 4 BF \cdot FC, \] but \[ BF - FC = BM+MF-(MC-MF) = 2MF = 22, \] so \[ BC = \sqrt{22^2 + 4 \cdot 75} = \sqrt{784} = \boxed{28}. \] | numerical | putnam | Geometry Trigonometry | A rectangle, $HOMF$, has sides $HO=11$ and $OM=5$. A triangle $ABC$ has $H$ as the intersection of the altitudes, $O$ the center of the circumscribed circle, $M$ the midpoint of $BC$, and $F$ the foot of the altitude from $A$. What is the length of $BC$? | The centroid $G$ of the triangle is collinear with $H$ and $O$ (Euler line), and the centroid lies two-thirds of the way from $A$ to $M$. Therefore $H$ is also two-thirds of the way from $A$ to $F$, so $AF = 15$. Since the triangles $BFH$ and $AFC$ are similar (they're right triangles and \[ \angle HBC = \pi/2 - \angle C = \angle CAF), \] we have \[ BF/FH = AF/FC \] or \[ BF \cdot FC = FH \cdot AF = 75. \] Now \[ BC^2 = (BF + FC)^2 = (BF - FC)^2 + 4 BF \cdot FC, \] but \[ BF - FC = BM+MF-(MC-MF) = 2MF = 22, \] so \[ BC = \sqrt{22^2 + 4 \cdot 75} = \sqrt{784} = \boxed{28}. \] | 0 |
2007 | 2007_A2 | Find the least possible area of a convex set in the plane that intersects both branches of the hyperbola $xy = 1$ and both branches of the hyperbola $xy = -1$. (A set $S$ in the plane is called \emph{convex} if for any two points in $S$ the line segment connecting them is contained in $S$.) | To prove that 4 is a lower bound, let $S$ be a convex set of the desired form. Choose $A,B,C,D \in S$ lying on the branches of the two hyperbolas, with $A$ in the upper right quadrant, $B$ in the upper left, $C$ in the lower left, $D$ in the lower right. Then the area of the quadrilateral $ABCD$ is a lower bound for the area of $S$. Write $A = (a,1/a)$, $B = (-b,1/b)$, $C = (-c,-1/c)$, $D = (d, -1/d)$ with $a,b,c,d > 0$. Then the area of the quadrilateral $ABCD$ is \[ \frac{1}{2}(a/b + b/c + c/d + d/a + b/a + c/b + d/c + a/d), \] which by the arithmetic-geometric mean inequality is at least $4$. The minimum is \boxed{4}, achieved by the square with vertices $(\pm 1, \pm 1)$. | numerical | putnam | Analysis Geometry | Find the least possible area of a convex set in the plane that intersects both branches of the hyperbola $xy = 1$ and both branches of the hyperbola $xy = -1$. (A set $S$ in the plane is called \emph{convex} if for any two points in $S$ the line segment connecting them is contained in $S$.) | The minimum is \boxed{4}, achieved by the square with vertices $(\pm 1, \pm 1)$. To prove that 4 is a lower bound, let $S$ be a convex set of the desired form. Choose $A,B,C,D \in S$ lying on the branches of the two hyperbolas, with $A$ in the upper right quadrant, $B$ in the upper left, $C$ in the lower left, $D$ in the lower right. Then the area of the quadrilateral $ABCD$ is a lower bound for the area of $S$. Write $A = (a,1/a)$, $B = (-b,1/b)$, $C = (-c,-1/c)$, $D = (d, -1/d)$ with $a,b,c,d > 0$. Then the area of the quadrilateral $ABCD$ is \[ \frac{1}{2}(a/b + b/c + c/d + d/a + b/a + c/b + d/c + a/d), \] which by the arithmetic-geometric mean inequality is at least $4$. | 0 |
1971 | 1971_A5 | A game of solitaire is played as follows. After each play, according to the outcome, the player receives either $a$ or $b$ points ($a$ and $b$ are positive integers with $a$ greater than $b$), and his score accumulates from play to play. It has been noticed that there are thirty-five non-attainable scores and that one of these is 58. Find $a+b$. | The attainable scores are those non-negative integers expressible in the form $ax + by$ with $x$ and $y$ non-negative integers. If $a$ and $b$ are not relatively prime there are infinitely many non-attainable scores. Hence $(a, b) = 1$. It will be shown that the number of non-attainable scores is $\frac{1}{2}(a - 1)(b - 1)$.
If $m$ is an attainable score, the line $ax + by = m$ passes through at least one lattice point in the closed first quadrant. Because $a$ and $b$ are relatively prime, the lattice points on a line $ax + by = m$ are at a horizontal distance of $b$. The first-quadrant segment of $ax + by = m$ has a horizontal projection of $m/a$ and thus every score $m \geq ab$ is attainable. Every non-attainable score must satisfy $0 \leq m < ab$.
If $0 \leq m < ab$, the first-quadrant segment of the line $ax + by = m$ has a horizontal projection less than $b$, and so contains at most one lattice point. Thus there is a one-to-one correspondence between lattice points $(x, y)$ with $0 \leq ax + by < ab$ in the first quadrant and attainable scores with $0 \leq m < ab$. The closed rectangle $0 \leq x \leq b$, $0 \leq y \leq a$ contains $(a + 1)(b + 1)$ lattice points, so the number of lattice points in the first quadrant with $0 \leq ax + by < ab$ is $ab - \frac{1}{2}(a + 1)(b + 1) + 1 = \frac{1}{2}(a - 1)(b - 1)$.
In our given example $70 = (a - 1)(b - 1) = 1(70) = 2(35) = 5(14) = 7(10)$. The conditions $a > b$, $(a, b) = 1$ yield two possibilities $a = 71, b = 2$ and $a = 11, b = 8$. Since 58 = 71(0) + 2(29), the first of these alternatives is eliminated. The line $11x + 8y = 58$ passes through $(6, -1)$ and $(-2, 10)$ and thus does not pass through a lattice point in the first quadrant. The unique solution is $a = 11$, $b = 8$ making the final answer \boxed{19}. | numerical | putnam (modified boxing) | Number Theory | A game of solitaire is played as follows. After each play, according to the outcome, the player receives either $a$ or $b$ points ($a$ and $b$ are positive integers with $a$ greater than $b$), and his score accumulates from play to play. It has been noticed that there are thirty-five non-attainable scores and that one of these is 58. Find $a$ and $b$. | The attainable scores are those non-negative integers expressible in the form $ax + by$ with $x$ and $y$ non-negative integers. If $a$ and $b$ are not relatively prime there are infinitely many non-attainable scores. Hence $(a, b) = 1$. It will be shown that the number of non-attainable scores is $\frac{1}{2}(a - 1)(b - 1)$.
If $m$ is an attainable score, the line $ax + by = m$ passes through at least one lattice point in the closed first quadrant. Because $a$ and $b$ are relatively prime, the lattice points on a line $ax + by = m$ are at a horizontal distance of $b$. The first-quadrant segment of $ax + by = m$ has a horizontal projection of $m/a$ and thus every score $m \geq ab$ is attainable. Every non-attainable score must satisfy $0 \leq m < ab$.
If $0 \leq m < ab$, the first-quadrant segment of the line $ax + by = m$ has a horizontal projection less than $b$, and so contains at most one lattice point. Thus there is a one-to-one correspondence between lattice points $(x, y)$ with $0 \leq ax + by < ab$ in the first quadrant and attainable scores with $0 \leq m < ab$. The closed rectangle $0 \leq x \leq b$, $0 \leq y \leq a$ contains $(a + 1)(b + 1)$ lattice points, so the number of lattice points in the first quadrant with $0 \leq ax + by < ab$ is $ab - \frac{1}{2}(a + 1)(b + 1) + 1 = \frac{1}{2}(a - 1)(b - 1)$.
In our given example $70 = (a - 1)(b - 1) = 1(70) = 2(35) = 5(14) = 7(10)$. The conditions $a > b$, $(a, b) = 1$ yield two possibilities $a = 71, b = 2$ and $a = 11, b = 8$. Since 58 = 71(0) + 2(29), the first of these alternatives is eliminated. The line $11x + 8y = 58$ passes through $(6, -1)$ and $(-2, 10)$ and thus does not pass through a lattice point in the first quadrant. The unique solution is $a = 11$, $b = 8$. | 0 |
2009 | 2009_A3 | Let $d_n$ be the determinant of the $n \times n$ matrix whose entries, from left to right and then from top to bottom, are $\cos 1, \cos 2, \dots, \cos n^2$. (For example, \[ d_3 = \left| \begin{matrix} \cos 1 & \cos 2 & \cos 3 \\ \cos 4 & \cos 5 & \cos 6 \\ \cos 7 & \cos 8 & \cos 9 \end{matrix} \right|. \] The argument of $\cos$ is always in radians, not degrees.) Evaluate $\lim_{n\to\infty} d_n$. | We will start by checking that $d_n = 0$ for all $n \geq 3$. Starting from the given matrix, add the third column to the first column; this does not change the determinant. However, thanks to the identity $\cos x + \cos y = 2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}$, the resulting matrix has the form \[ \begin{pmatrix} 2 \cos 2 \cos 1 & \cos 2 & \cdots \\ 2 \cos (n+2) \cos 1 & \cos (n+2) & \cdots \\ 2 \cos (2n+2) \cos 1 & 2 \cos (2n+2) & \cdots \\ \vdots & \vdots & \ddots \end{pmatrix} \] with the first column being a multiple of the second. Hence $d_n = 0$. The limit is $\boxed{0}$. | numerical | putnam | Linear Algebra Trigonometry | Let $d_n$ be the determinant of the $n \times n$ matrix whose entries, from left to right and then from top to bottom, are $\cos 1, \cos 2, \dots, \cos n^2$. (For example, \[ d_3 = \left| \begin{matrix} \cos 1 & \cos 2 & \cos 3 \\ \cos 4 & \cos 5 & \cos 6 \\ \cos 7 & \cos 8 & \cos 9 \end{matrix} \right|. \] The argument of $\cos$ is always in radians, not degrees.) Evaluate $\lim_{n\to\infty} d_n$. | The limit is $\boxed{0}$; we will show this by checking that $d_n = 0$ for all $n \geq 3$. Starting from the given matrix, add the third column to the first column; this does not change the determinant. However, thanks to the identity $\cos x + \cos y = 2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}$, the resulting matrix has the form \[ \begin{pmatrix} 2 \cos 2 \cos 1 & \cos 2 & \cdots \\ 2 \cos (n+2) \cos 1 & \cos (n+2) & \cdots \\ 2 \cos (2n+2) \cos 1 & 2 \cos (2n+2) & \cdots \\ \vdots & \vdots & \ddots \end{pmatrix} \] with the first column being a multiple of the second. Hence $d_n = 0$. | 0 |
1984 | 1984_A2 | Express \( \sum_{k=1}^\infty \frac{6^k}{(3^k - 2^k)(3^{k+1} - 2^{k+1})} \) as a rational number. | Let \( S(n) \) denote the \( n \)-th partial sum of the given series. Then \[ S(n) = \sum_{k=1}^n \left[ \frac{3^k}{3^k - 2^k} - \frac{3^{k+1}}{3^{k+1} - 2^{k+1}} \right] = 3 - \frac{3^{n+1}}{3^{n+1} - 2^{n+1}}, \] and the series converges to \( \lim_{n \to \infty} S(n) = \boxed{2} \). | numerical | putnam | Algebra | Express \( \sum_{k=1}^\infty \frac{6^k}{(3^k - 2^k)(3^{k+1} - 2^{k+1})} \) as a rational number. | Let \( S(n) \) denote the \( n \)-th partial sum of the given series. Then \[ S(n) = \sum_{k=1}^n \left[ \frac{3^k}{3^k - 2^k} - \frac{3^{k+1}}{3^{k+1} - 2^{k+1}} \right] = 3 - \frac{3^{n+1}}{3^{n+1} - 2^{n+1}}, \] and the series converges to \( \lim_{n \to \infty} S(n) = \boxed{2} \). | 0 |
2016 | 2016_A1 | Find the smallest positive integer $j$ such that for every polynomial $p(x)$ with integer coefficients and for every integer $k$, the integer \[ p^{(j)}(k) = \left. \frac{d^j}{dx^j} p(x) \right|_{x=k} \] (the $j$-th derivative of $p(x)$ at $k$) is divisible by 2016. | First suppose that $j$ satisfies the given condition. For $p(x) = x^j$, we have $p^{(j)}(x) = j!$ and thus $j!$ is divisible by $2016$. Since $2016$ is divisible by $2^5$ and $7!$ is not, it follows that $j \geq 8$. Conversely, we claim that $j=8$ works. Indeed, let $p(x) = \sum_{m=0}^n a_m x^m$ be a polynomial with integer coefficients; then if $k$ is any integer, \begin{align*} p^{(8)}(k) &= \sum_{m=8}^n m(m-1)\cdots (m-7) a_m k^{m-8} \\ &= \sum_{m=8}^n {m\choose 8} 8! a_m k^{m-8} \end{align*} is divisible by $8! = 20 \cdot 2016$, and so $p^{(8)}(k)$ is divisible by $2016$. The answer is $j=\boxed{8}$. | numerical | putnam | Calculus Number Theory | Find the smallest positive integer $j$ such that for every polynomial $p(x)$ with integer coefficients and for every integer $k$, the integer \[ p^{(j)}(k) = \left. \frac{d^j}{dx^j} p(x) \right|_{x=k} \] (the $j$-th derivative of $p(x)$ at $k$) is divisible by 2016. | The answer is $j=\boxed{8}$. First suppose that $j$ satisfies the given condition. For $p(x) = x^j$, we have $p^{(j)}(x) = j!$ and thus $j!$ is divisible by $2016$. Since $2016$ is divisible by $2^5$ and $7!$ is not, it follows that $j \geq 8$. Conversely, we claim that $j=8$ works. Indeed, let $p(x) = \sum_{m=0}^n a_m x^m$ be a polynomial with integer coefficients; then if $k$ is any integer, \begin{align*} p^{(8)}(k) &= \sum_{m=8}^n m(m-1)\cdots (m-7) a_m k^{m-8} \\ &= \sum_{m=8}^n {m\choose 8} 8! a_m k^{m-8} \end{align*} is divisible by $8! = 20 \cdot 2016$, and so $p^{(8)}(k)$ is divisible by $2016$. | 0 |
1997 | 1997_B1 | Let $\{x\}$ denote the distance between the real number $x$ and the nearest integer. For each positive integer $n$, evaluate \[F_n=\sum_{m=1}^{6n-1} \min(\{\frac{m}{6n}\},\{\frac{m}{3n}\}).\] (Here $\min(a,b)$ denotes the minimum of $a$ and $b$.) | It is trivial to check that $\frac{m}{6n}=\{\frac{m}{6n}\}\leq \{\frac{m}{3n}\}$ for $1\leq m\leq 2n$, that $1-\frac{m}{3n}=\{\frac{m}{3n}\}\leq \{\frac{m}{6n}\}$ for $2n\leq m\leq 3n$, that $\frac{m}{3n}-1=\{\frac{m}{3n}\}\leq \{\frac{m}{6n}\}$ for $3n\leq m\leq 4n$, and that $1-\frac{m}{6n}=\{\frac{m}{6n}\}\leq \{\frac{m}{3n}\}$ for $4n\leq m\leq 6n$. Therefore the desired sum is \begin{gather*} \sum_{m=1}^{2n-1} \frac{m}{6n} +\sum_{m=2n}^{3n-1} \left(1-\frac{m}{3n} \right) \\ +\sum_{m=3n}^{4n-1} \left(\frac{m}{3n}-1 \right) + \sum_{m=4n}^{6n-1} \left( 1-\frac{m}{6n} \right) =\boxed{n}. \end{gather*} | algebraic | putnam | Algebra Number Theory | Let $\{x\}$ denote the distance between the real number $x$ and the nearest integer. For each positive integer $n$, evaluate \[F_n=\sum_{m=1}^{6n-1} \min(\{\frac{m}{6n}\},\{\frac{m}{3n}\}).\] (Here $\min(a,b)$ denotes the minimum of $a$ and $b$.) | It is trivial to check that $\frac{m}{6n}=\{\frac{m}{6n}\}\leq \{\frac{m}{3n}\}$ for $1\leq m\leq 2n$, that $1-\frac{m}{3n}=\{\frac{m}{3n}\}\leq \{\frac{m}{6n}\}$ for $2n\leq m\leq 3n$, that $\frac{m}{3n}-1=\{\frac{m}{3n}\}\leq \{\frac{m}{6n}\}$ for $3n\leq m\leq 4n$, and that $1-\frac{m}{6n}=\{\frac{m}{6n}\}\leq \{\frac{m}{3n}\}$ for $4n\leq m\leq 6n$. Therefore the desired sum is \begin{gather*} \sum_{m=1}^{2n-1} \frac{m}{6n} +\sum_{m=2n}^{3n-1} \left(1-\frac{m}{3n} \right) \\ +\sum_{m=3n}^{4n-1} \left(\frac{m}{3n}-1 \right) + \sum_{m=4n}^{6n-1} \left( 1-\frac{m}{6n} \right) =\boxed{n}. \end{gather*} | 0 |
1952 | 1952_4 | The flag of the United Nations consists of a polar map of the world, with the North Pole as center, extending approximately to 45° South Latitude. The parallels of latitude are concentric circles with radii proportional to their co-latitudes. Australia is near the periphery of the map and is intersected by the parallel of latitude 30° S. In the very close vicinity of this parallel how much are East and West distances exaggerated as compared to North and South distances? | Let a point on the earth with co-latitude $\alpha$ and longitude $\theta$ be represented by a point on the flag map with polar coordinates $k\alpha$ and $\theta$ where $k$ is a constant of proportionality. \n\nLet $R$ be the radius of the earth. A meridian on the earth has length $\pi R$ and on the (complete) flag map is represented by a segment of length $\pi k$. Hence a short distance $d$ in the North-South direction is represented on the flag map by the distance $d_1 = kd/R$.\n\nOn the earth the small circle of latitude 30° S (= co-latitude $2\pi/3$) is represented on the flag map by a circle of radius $2\pi k/3$ and of circumference $4\pi^2 k/3$. On the earth that circle has length $2\pi R \sin (2\pi/3) = \pi R\sqrt{3}$.\n\nHence a short distance $d$ in the East-West direction is represented on the flag map by a distance \[ \frac{4\pi^2 k/3}{\pi R\sqrt{3}} d = \frac{4\pi kd}{3\sqrt{3}R} = \frac{4\pi}{3\sqrt{3}} d_1. \] \n\nTherefore distances in the East-West direction at latitude 30° S are magnified relative to distances in the North-South direction by the factor $\boxed{\frac{4\pi}{3\sqrt{3}}} \approx 2.42$. | numerical | putnam | Geometry Trigonometry | The flag of the United Nations consists of a polar map of the world, with the North Pole as center, extending approximately to 45° South Latitude. The parallels of latitude are concentric circles with radii proportional to their co-latitudes. Australia is near the periphery of the map and is intersected by the parallel of latitude 30° S. In the very close vicinity of this parallel how much are East and West distances exaggerated as compared to North and South distances? | Let a point on the earth with co-latitude $\alpha$ and longitude $\theta$ be represented by a point on the flag map with polar coordinates $k\alpha$ and $\theta$ where $k$ is a constant of proportionality. \n\nLet $R$ be the radius of the earth. A meridian on the earth has length $\pi R$ and on the (complete) flag map is represented by a segment of length $\pi k$. Hence a short distance $d$ in the North-South direction is represented on the flag map by the distance $d_1 = kd/R$.\n\nOn the earth the small circle of latitude 30° S (= co-latitude $2\pi/3$) is represented on the flag map by a circle of radius $2\pi k/3$ and of circumference $4\pi^2 k/3$. On the earth that circle has length $2\pi R \sin (2\pi/3) = \pi R\sqrt{3}$.\n\nHence a short distance $d$ in the East-West direction is represented on the flag map by a distance \[ \frac{4\pi^2 k/3}{\pi R\sqrt{3}} d = \frac{4\pi kd}{3\sqrt{3}R} = \frac{4\pi}{3\sqrt{3}} d_1. \] \n\nTherefore distances in the East-West direction at latitude 30° S are magnified relative to distances in the North-South direction by the factor $\boxed{\frac{4\pi}{3\sqrt{3}}} \approx 2.42$. | 0 |
2019 | 2019_B4 | Let $\mathcal{F}$ be the set of functions $f(x,y)$ that are twice continuously differentiable for $x \geq 1$, $y \geq 1$ and that satisfy the following two equations (where subscripts denote partial derivatives): \begin{gather*} xf_x + yf_y = xy \ln(xy), \\ x^2 f_{xx} + y^2 f_{yy} = xy. \end{gather*} For each $f \in \mathcal{F}$, let \[ m(f) = \min_{s \geq 1} \left(f(s+1,s+1) - f(s+1,s) - f(s,s+1) + f(s,s) \right). \] Determine $m(f)$, and show that it is independent of the choice of $f$. | Label the given differential equations by (1) and (2). If we write, e.g., $x\frac{\partial}{\partial x}(1)$ for the result of differentiating (1) by $x$ and multiplying the resulting equation by $x$, then the combination $x\frac{\partial}{\partial x}(1)+y\frac{\partial}{\partial y}(1)-(1)-(2)$ gives the equation $2xyf_{xy} = xy\ln(xy)+xy$, whence $f_{xy} = \frac{1}{2} (\ln(x)+\ln(y)+1)$. Now we observe that \begin{align*} \lefteqn{f(s+1,s+1)-f(s+1,s)-f(s,s+1)+f(s,s)} \\ &= \int_s^{s+1} \int_s^{s+1} f_{xy}\,dy\,dx \\ &= \frac{1}{2} \int_s^{s+1} \int_s^{s+1} (\ln(x)+\ln(y)+1)\,dy\,dx \\ &= \frac{1}{2} + \int_s^{s+1} \ln(x)\,dx. \end{align*} Since $\ln(x)$ is increasing, $\int_s^{s+1} \ln(x)\,dx$ is an increasing function of $s$, and so it is minimized over $s \in [1,\infty)$ when $s=1$. We conclude that \[ m(f) = \frac{1}{2} + \int_1^2 \ln(x)\,dx = 2 \ln 2-\frac{1}{2} \] independent of $f$. We compute that $m(f) = \boxed{2 \ln 2 - \frac{1}{2}}$. | numerical | putnam | Analysis Calculus | Let $\mathcal{F}$ be the set of functions $f(x,y)$ that are twice continuously differentiable for $x \geq 1$, $y \geq 1$ and that satisfy the following two equations (where subscripts denote partial derivatives): \begin{gather*} xf_x + yf_y = xy \ln(xy), \\ x^2 f_{xx} + y^2 f_{yy} = xy. \end{gather*} For each $f \in \mathcal{F}$, let \[ m(f) = \min_{s \geq 1} \left(f(s+1,s+1) - f(s+1,s) - f(s,s+1) + f(s,s) \right). \] Determine $m(f)$, and show that it is independent of the choice of $f$. | We compute that $m(f) = \boxed{2 \ln 2 - \frac{1}{2}}$. Label the given differential equations by (1) and (2). If we write, e.g., $x\frac{\partial}{\partial x}(1)$ for the result of differentiating (1) by $x$ and multiplying the resulting equation by $x$, then the combination $x\frac{\partial}{\partial x}(1)+y\frac{\partial}{\partial y}(1)-(1)-(2)$ gives the equation $2xyf_{xy} = xy\ln(xy)+xy$, whence $f_{xy} = \frac{1}{2} (\ln(x)+\ln(y)+1)$. Now we observe that \begin{align*} \lefteqn{f(s+1,s+1)-f(s+1,s)-f(s,s+1)+f(s,s)} \\ &= \int_s^{s+1} \int_s^{s+1} f_{xy}\,dy\,dx \\ &= \frac{1}{2} \int_s^{s+1} \int_s^{s+1} (\ln(x)+\ln(y)+1)\,dy\,dx \\ &= \frac{1}{2} + \int_s^{s+1} \ln(x)\,dx. \end{align*} Since $\ln(x)$ is increasing, $\int_s^{s+1} \ln(x)\,dx$ is an increasing function of $s$, and so it is minimized over $s \in [1,\infty)$ when $s=1$. We conclude that \[ m(f) = \frac{1}{2} + \int_1^2 \ln(x)\,dx = 2 \ln 2-\frac{1}{2} \] independent of $f$. | 0 |
1969 | 1969_B5 | Let $a_1 < a_2 < a_3 < \cdots$ be an increasing sequence of positive integers. Let the series \[ \sum_{n=1}^{\infty} \frac{1}{a_n} \] be convergent. For any number $x$, let $k(x)$ be the number of the $a_n$'s which do not exceed $x$. Find the value of \lim_{x \to \infty} \frac{k(x)}{x}. | The following proof shows it is not necessary to stipulate that the $a_n$ be integers. Suppose for some $\epsilon > 0$ there are $x_j \to \infty$ with $k(x_j)/x_j \geq \epsilon$. Note that if $1 \leq n \leq k(x_j)$, then (because the $a_n$ increase) \[ a_n \leq a_{k(x_j)} \leq x_j \] and \[ \frac{1}{a_n} \geq \frac{1}{x_j}. \] Now for any positive integer $N$, \[ \sum_{n=N}^{\infty} \frac{1}{a_n} \geq \sup_j \sum_{n=N}^{k(x_j)} \frac{1}{a_n} \geq \sup_j \frac{k(x_j) - N}{x_j} \geq \sup_j \epsilon - \frac{N}{x_j} = \epsilon. \] But this contradicts the convergence of $\sum_{n=1}^{\infty} \frac{1}{a_n}$, which implies \[ \lim_{N \to \infty} \sum_{n=N}^{\infty} \frac{1}{a_n} = \boxed{0}. \] | numerical | putnam (modified boxing) | Analysis | Let $a_1 < a_2 < a_3 < \cdots$ be an increasing sequence of positive integers. Let the series \[ \sum_{n=1}^{\infty} \frac{1}{a_n} \] be convergent. For any number $x$, let $k(x)$ be the number of the $a_n$'s which do not exceed $x$. Show that \[ \lim_{x \to \infty} \frac{k(x)}{x} = 0. \] | The following proof shows it is not necessary to stipulate that the $a_n$ be integers. Suppose for some $\epsilon > 0$ there are $x_j \to \infty$ with $k(x_j)/x_j \geq \epsilon$. Note that if $1 \leq n \leq k(x_j)$, then (because the $a_n$ increase) \[ a_n \leq a_{k(x_j)} \leq x_j \] and \[ \frac{1}{a_n} \geq \frac{1}{x_j}. \] Now for any positive integer $N$, \[ \sum_{n=N}^{\infty} \frac{1}{a_n} \geq \sup_j \sum_{n=N}^{k(x_j)} \frac{1}{a_n} \geq \sup_j \frac{k(x_j) - N}{x_j} \geq \sup_j \epsilon - \frac{N}{x_j} = \epsilon. \] But this contradicts the convergence of $\sum_{n=1}^{\infty} \frac{1}{a_n}$, which implies \[ \lim_{N \to \infty} \sum_{n=N}^{\infty} \frac{1}{a_n} = 0. \] | 0 |
2006 | 2006_B5 | For each continuous function $f: [0,1] \to \mathbb{R}$, let $I(f) = \int_0^1 x^2 f(x)\,dx$ and $J(x) = \int_0^1 x \left(f(x)\right)^2\,dx$. Find the maximum value of $I(f) - J(f)$ over all such functions $f$. | We have \begin{align*} &\int_0^1 x^2 f (x)\,dx - \int_0^1 x f(x)^2\,dx \\ &= \int_0^1 (x^3/4 - x ( f(x)-x/2)^2)\,dx \\ &\leq \int_0^1 x^3/4\,dx = 1/16, \end{align*} with equality when $f(x) = x/2$. The answer is $\boxed{1/16}$. | numerical | putnam | Analysis Calculus | For each continuous function $f: [0,1] \to \mathbb{R}$, let $I(f) = \int_0^1 x^2 f(x)\,dx$ and $J(x) = \int_0^1 x \left(f(x)\right)^2\,dx$. Find the maximum value of $I(f) - J(f)$ over all such functions $f$. | The answer is $\boxed{1/16}$. We have \begin{align*} &\int_0^1 x^2 f (x)\,dx - \int_0^1 x f(x)^2\,dx \\ &= \int_0^1 (x^3/4 - x ( f(x)-x/2)^2)\,dx \\ &\leq \int_0^1 x^3/4\,dx = 1/16, \end{align*} with equality when $f(x) = x/2$. | 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 |
1997 | 1997_A6 | For a positive integer $n$ and any real number $c$, define $x_k$ recursively by $x_0=0$, $x_1=1$, and for $k\geq 0$, \[x_{k+2}=\frac{cx_{k+1}-(n-k)x_k}{k+1}.\] Fix $n$ and then take $c$ to be the largest value for which $x_{n+1}=0$. Find $x_k$ in terms of $n$ and $k$, $1\leq k\leq n$. | Clearly $x_{n+1}$ is a polynomial in $c$ of degree $n$, so it suffices to identify $n$ values of $c$ for which $x_{n+1} = 0$. We claim these are $c = n-1-2r$ for $r=0,1,\dots, n-1$; in this case, $x_k$ is the coefficient of $t^{k-1}$ in the polynomial $f(t) = (1-t)^r (1+t)^{n-1-r}$. This can be verified by noticing that $f$ satisfies the differential equation \[ \frac{f'(t)}{f(t)} = \frac{n-1-r}{1+t} - \frac{r}{1-t} \] (by logarithmic differentiation) or equivalently, \begin{align*} (1-t^2) f'(t) &= f(t) [(n-1-r)(1-t) - r(1+t)] \\ &= f(t) [(n-1-2r) - (n-1)t] \end{align*} and then taking the coefficient of $t^{k}$ on both sides: \begin{gather*} (k+1) x_{k+2} - (k-1) x_k = \\ (n-1-2r) x_{k+1} - (n-1) x_{k}. \end{gather*} In particular, the largest such $c$ is $n-1$, and $x_k = \boxed{\binom{n-1}{k-1}}$ for $k= 1, 2, \dots, n$. | algebraic | putnam | Algebra Combinatorics Number Theory | For a positive integer $n$ and any real number $c$, define $x_k$ recursively by $x_0=0$, $x_1=1$, and for $k\geq 0$, \[x_{k+2}=\frac{cx_{k+1}-(n-k)x_k}{k+1}.\] Fix $n$ and then take $c$ to be the largest value for which $x_{n+1}=0$. Find $x_k$ in terms of $n$ and $k$, $1\leq k\leq n$. | Clearly $x_{n+1}$ is a polynomial in $c$ of degree $n$, so it suffices to identify $n$ values of $c$ for which $x_{n+1} = 0$. We claim these are $c = n-1-2r$ for $r=0,1,\dots, n-1$; in this case, $x_k$ is the coefficient of $t^{k-1}$ in the polynomial $f(t) = (1-t)^r (1+t)^{n-1-r}$. This can be verified by noticing that $f$ satisfies the differential equation \[ \frac{f'(t)}{f(t)} = \frac{n-1-r}{1+t} - \frac{r}{1-t} \] (by logarithmic differentiation) or equivalently, \begin{align*} (1-t^2) f'(t) &= f(t) [(n-1-r)(1-t) - r(1+t)] \\ &= f(t) [(n-1-2r) - (n-1)t] \end{align*} and then taking the coefficient of $t^{k}$ on both sides: \begin{gather*} (k+1) x_{k+2} - (k-1) x_k = \\ (n-1-2r) x_{k+1} - (n-1) x_{k}. \end{gather*} In particular, the largest such $c$ is $n-1$, and $x_k = \boxed{\binom{n-1}{k-1}}$ for $k= 1, 2, \dots, n$. | 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 |
1998 | 1998_B3 | let $H$ be the unit hemisphere $\{(x,y,z):x^2+y^2+z^2=1,z\geq 0\}$, $C$ the unit circle $\{(x,y,0):x^2+y^2=1\}$, and $P$ the regular pentagon inscribed in $C$. Determine the surface area of that portion of $H$ lying over the planar region inside $P$, and write your answer in the form $A \sin\alpha + B \cos\beta$, where $A,B,\alpha,\beta$ are real numbers. | We use the well-known result that the surface area of the ``sphere cap'' $\{(x,y,z)\,|\,x^2+y^2+z^2=1,\,z\geq z_0\}$ is simply $2\pi(1-z_0)$. (This result is easily verified using calculus; we omit the derivation here.) Now the desired surface area is just $2\pi$ minus the surface areas of five identical halves of sphere caps; these caps, up to isometry, correspond to $z_0$ being the distance from the center of the pentagon to any of its sides, i.e., $z_0 = \cos \frac{\pi}{5}$. Thus the desired area is $2\pi - \frac{5}{2} \left(2\pi (1-\cos\frac{\pi}{5})\right) = \boxed{5\pi\cos\frac{\pi}{5} - 3\pi}$ (i.e., $B=\pi/2$). | numerical | putnam | Calculus Geometry Trigonometry | let $H$ be the unit hemisphere $\{(x,y,z):x^2+y^2+z^2=1,z\geq 0\}$, $C$ the unit circle $\{(x,y,0):x^2+y^2=1\}$, and $P$ the regular pentagon inscribed in $C$. Determine the surface area of that portion of $H$ lying over the planar region inside $P$, and write your answer in the form $A \sin\alpha + B \cos\beta$, where $A,B,\alpha,\beta$ are real numbers. | We use the well-known result that the surface area of the ``sphere cap'' $\{(x,y,z)\,|\,x^2+y^2+z^2=1,\,z\geq z_0\}$ is simply $2\pi(1-z_0)$. (This result is easily verified using calculus; we omit the derivation here.) Now the desired surface area is just $2\pi$ minus the surface areas of five identical halves of sphere caps; these caps, up to isometry, correspond to $z_0$ being the distance from the center of the pentagon to any of its sides, i.e., $z_0 = \cos \frac{\pi}{5}$. Thus the desired area is $2\pi - \frac{5}{2} \left(2\pi (1-\cos\frac{\pi}{5})\right) = \boxed{5\pi\cos\frac{\pi}{5} - 3\pi}$ (i.e., $B=\pi/2$). | 0 |
2006 | 2006_B4 | Let $Z$ denote the set of points in $\mathbb{R}^n$ whose coordinates are 0 or 1. (Thus $Z$ has $2^n$ elements, which are the vertices of a unit hypercube in $\mathbb{R}^n$.) Given a vector subspace $V$ of $\mathbb{R}^n$, let $Z(V)$ denote the number of members of $Z$ that lie in $V$. Let $k$ be given, $0 \leq k \leq n$. Find the maximum, over all vector subspaces $V \subseteq \mathbb{R}^n$ of dimension $k$, of the number of points in $V \cap Z$. [Editorial note: the proposers probably intended to write $Z(V)$ instead of ``the number of points in $V \cap Z$'', but this changes nothing.] | More generally, we show that any affine $k$-dimensional plane in $\mathbb{R}^n$ can contain at most $2^k$ points in $Z$. The proof is by induction on $k+n$; the case $k=n=0$ is clearly true. Suppose that $V$ is a $k$-plane in $\mathbb{R}^n$. Denote the hyperplanes $\{x_n = 0\}$ and $\{x_n = 1\}$ by $V_0$ and $V_1$, respectively. If $V\cap V_0$ and $V\cap V_1$ are each at most $(k-1)$-dimensional, then $V\cap V_0\cap Z$ and $V\cap V_1 \cap Z$ each have cardinality at most $2^{k-1}$ by the induction assumption, and hence $V\cap Z$ has at most $2^k$ elements. Otherwise, if $V\cap V_0$ or $V\cap V_1$ is $k$-dimensional, then $V \subset V_0$ or $V\subset V_1$; now apply the induction hypothesis on $V$, viewed as a subset of $\mathbb{R}^{n-1}$ by dropping the last coordinate. The maximum is $\boxed{2^k}$, achieved for instance by the subspace \[ \{(x_1, \dots, x_n) \in \mathbb{R}^n: x_1 = \cdots = x_{n-k} = 0\}. \] | algebraic | putnam | Combinatorics Linear Algebra | Let $Z$ denote the set of points in $\mathbb{R}^n$ whose coordinates are 0 or 1. (Thus $Z$ has $2^n$ elements, which are the vertices of a unit hypercube in $\mathbb{R}^n$.) Given a vector subspace $V$ of $\mathbb{R}^n$, let $Z(V)$ denote the number of members of $Z$ that lie in $V$. Let $k$ be given, $0 \leq k \leq n$. Find the maximum, over all vector subspaces $V \subseteq \mathbb{R}^n$ of dimension $k$, of the number of points in $V \cap Z$. [Editorial note: the proposers probably intended to write $Z(V)$ instead of ``the number of points in $V \cap Z$'', but this changes nothing.] | The maximum is $\boxed{2^k}$, achieved for instance by the subspace \[ \{(x_1, \dots, x_n) \in \mathbb{R}^n: x_1 = \cdots = x_{n-k} = 0\}. \] More generally, we show that any affine $k$-dimensional plane in $\mathbb{R}^n$ can contain at most $2^k$ points in $Z$. The proof is by induction on $k+n$; the case $k=n=0$ is clearly true. Suppose that $V$ is a $k$-plane in $\mathbb{R}^n$. Denote the hyperplanes $\{x_n = 0\}$ and $\{x_n = 1\}$ by $V_0$ and $V_1$, respectively. If $V\cap V_0$ and $V\cap V_1$ are each at most $(k-1)$-dimensional, then $V\cap V_0\cap Z$ and $V\cap V_1 \cap Z$ each have cardinality at most $2^{k-1}$ by the induction assumption, and hence $V\cap Z$ has at most $2^k$ elements. Otherwise, if $V\cap V_0$ or $V\cap V_1$ is $k$-dimensional, then $V \subset V_0$ or $V\subset V_1$; now apply the induction hypothesis on $V$, viewed as a subset of $\mathbb{R}^{n-1}$ by dropping the last coordinate. | 0 |
2014 | 2014_B3 | Let $A$ be an $m \times n$ matrix with rational entries. Suppose that there are at least $m+n$ distinct prime numbers among the absolute values of the entries of $A$. Find the least possible rank of $A$. | Assume by way of contradiction that $A$ has rank at most 1; in this case, we can find rational numbers $a_1,\dots,a_m$, $b_1,\dots,b_n$ such that $A_{ij} = a_i b_j$ for all $i,j$. By deleting rows or columns, we may reduce to the case where the $a_i$'s and $b_j$'s are all nonzero. Recall that any nonzero rational number $q$ has a unique prime factorization \[ q = \pm 2^{c_1} 3^{c_2} 5^{c_3} \cdots \] with exponents in $\ZZ$. Set \[ c(q) = (c_1, c_2, c_3, \dots). \] Note that $|a_i b_j|$ is prime if and only if $c(a_i) + c(b_j)$ has one entry equal to 1 and all others equal to 0. The condition that $m+n$ distinct primes appear in the matrix implies that the vector space \[ \left\{ \sum_i x_i c(a_i) + \sum_j y_i c(b_j) : x_i, y_j \in \RR, \sum_i x_i = \sum_j y_j \right\} \] contains a linearly independent set of size $m+n$. But that space evidently has dimension at most $m+n-1$, contradiction. This gives is the least possible rank of $A$ as \boxed{2}. | numerical | putnam (modified boxing) | Linear Algebra Number Theory | Let $A$ be an $m \times n$ matrix with rational entries. Suppose that there are at least $m+n$ distinct prime numbers among the absolute values of the entries of $A$. Show that the rank of $A$ is at least 2. | Assume by way of contradiction that $A$ has rank at most 1; in this case, we can find rational numbers $a_1,\dots,a_m$, $b_1,\dots,b_n$ such that $A_{ij} = a_i b_j$ for all $i,j$. By deleting rows or columns, we may reduce to the case where the $a_i$'s and $b_j$'s are all nonzero. Recall that any nonzero rational number $q$ has a unique prime factorization \[ q = \pm 2^{c_1} 3^{c_2} 5^{c_3} \cdots \] with exponents in $\ZZ$. Set \[ c(q) = (c_1, c_2, c_3, \dots). \] Note that $|a_i b_j|$ is prime if and only if $c(a_i) + c(b_j)$ has one entry equal to 1 and all others equal to 0. The condition that $m+n$ distinct primes appear in the matrix implies that the vector space \[ \left\{ \sum_i x_i c(a_i) + \sum_j y_i c(b_j) : x_i, y_j \in \RR, \sum_i x_i = \sum_j y_j \right\} \] contains a linearly independent set of size $m+n$. But that space evidently has dimension at most $m+n-1$, contradiction. | 0 |
1980 | 1980_A1 | Let $b$ and $c$ be fixed real numbers and let the ten points $(j, y_j), j = 1, 2, \dots, 10,$ lie on the parabola $y = x^2 + bx + c$. For $j = 1, 2, \dots, 9$, let $I_j$ be the point of intersection of the tangents to the given parabola at $(j, y_j)$ and $(j+1, y_{j+1})$. Determine the polynomial function $y = g(x)$ of least degree whose graph passes through all nine points $I_j$. | The equation of the tangent to the given parabola at $P_j = (j, y_j)$ is easily seen to be $y = L_j$, where $L_j = (2j + b)x - j^2 + c$. Solving $y = L_j$ and $y = L_{j+1}$ simultaneously, one finds that $x = (2j + 1)/2$ and so $j = (2x - 1)/2$ at $I_j$. Substituting this expression for $j$ into $L_j$ gives the $g(x)$ above. Thus we show that $g(x) = \boxed{x^2 + bx + c - (1/4)}$. | algebraic | putnam | Algebra Geometry | Let $b$ and $c$ be fixed real numbers and let the ten points $(j, y_j), j = 1, 2, \dots, 10,$ lie on the parabola $y = x^2 + bx + c$. For $j = 1, 2, \dots, 9$, let $I_j$ be the point of intersection of the tangents to the given parabola at $(j, y_j)$ and $(j+1, y_{j+1})$. Determine the polynomial function $y = g(x)$ of least degree whose graph passes through all nine points $I_j$. | We show that $g(x) = \boxed{x^2 + bx + c - (1/4)}$. The equation of the tangent to the given parabola at $P_j = (j, y_j)$ is easily seen to be $y = L_j$, where $L_j = (2j + b)x - j^2 + c$. Solving $y = L_j$ and $y = L_{j+1}$ simultaneously, one finds that $x = (2j + 1)/2$ and so $j = (2x - 1)/2$ at $I_j$. Substituting this expression for $j$ into $L_j$ gives the $g(x)$ above. | 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 |
2015 | 2015_B2 | Given a list of the positive integers $1,2,3,4,\dots$, take the first three numbers $1,2,3$ and their sum $6$ and cross all four numbers off the list. Repeat with the three smallest remaining numbers $4,5,7$ and their sum $16$. Continue in this way, crossing off the three smallest remaining numbers and their sum, and consider the sequence of sums produced: $6, 16, 27, 36, \dots$. Find a number in the sequence whose base 10 representation ends with $2015$. \, | We will prove that 42015 is such a number in the sequence. Label the sequence of sums $s_0, s_1, \dots$, and let $a_n, b_n, c_n$ be the summands of $s_n$ in ascending order. We prove the following two statements for each nonnegative integer $n$: \begin{enumerate} \item[(a)$_n$] The sequence \[ a_{3n}, b_{3n}, c_{3n}, a_{3n+1}, b_{3n+1}, c_{3n+1}, a_{3n+2}, b_{3n+2}, c_{3n+2} \] is obtained from the sequence $10n+1, \dots, 10n+10$ by removing one of $10n+5, 10n+6, 10n+7$. \item[(b)$_n$] We have \begin{align*} s_{3n} &= 30n+6,\ s_{3n+1} &\in \{30n+15, 30n+16, 30n+17\}, \ s_{3n+2} &= 30n+27. \end{align*} \end{enumerate} These statements follow by induction from the following simple observations: \begin{itemize} \item by computing the table of values \[ \begin{array}{c|cccc} n & a_n & b_n & c_n & s_n \ \hline 0 & 1 & 2 & 3 & 6 \ 1 & 4 & 5 & 7 & 16 \ 2 & 8 & 9 & 10 & 27 \end{array} \] we see that (a)$_0$ holds; \item (a)$_n$ implies (b)$_n$; \item (a)$_n$ and (b)$_1$, \dots, (b)$_n$ together imply (a)$_{n+1}$. \end{itemize} To produce a value of $n$ for which $s_n \equiv 2015 \pmod{10000}$, we take $n = 3m+1$ for some nonnegative integer $m$ for which $s_{3m+1} = 30m+15$. We must also have $30m \equiv 2000 \pmod{10000}$, or equivalently $m \equiv 400 \pmod{1000}$. By taking $m = 1400$, we ensure that $m \equiv 2 \pmod{3}$, so $s_m = 10m + 7$; this ensures that $s_{n}$ does indeed equal $30m+15 = \boxed{42015}$, as desired. | numerical | putnam | Combinatorics Number Theory | Given a list of the positive integers $1,2,3,4,\dots$, take the first three numbers $1,2,3$ and their sum $6$ and cross all four numbers off the list. Repeat with the three smallest remaining numbers $4,5,7$ and their sum $16$. Continue in this way, crossing off the three smallest remaining numbers and their sum, and consider the sequence of sums produced: $6, 16, 27, 36, \dots$. Prove or disprove that there is some number in the sequence whose base 10 representation ends with $2015$. \, | We will prove that 42015 is such a number in the sequence. Label the sequence of sums $s_0, s_1, \dots$, and let $a_n, b_n, c_n$ be the summands of $s_n$ in ascending order. We prove the following two statements for each nonnegative integer $n$: \begin{enumerate} \item[(a)$_n$] The sequence \[ a_{3n}, b_{3n}, c_{3n}, a_{3n+1}, b_{3n+1}, c_{3n+1}, a_{3n+2}, b_{3n+2}, c_{3n+2} \] is obtained from the sequence $10n+1, \dots, 10n+10$ by removing one of $10n+5, 10n+6, 10n+7$. \item[(b)$_n$] We have \begin{align*} s_{3n} &= 30n+6,\ s_{3n+1} &\in \{30n+15, 30n+16, 30n+17\}, \ s_{3n+2} &= 30n+27. \end{align*} \end{enumerate} These statements follow by induction from the following simple observations: \begin{itemize} \item by computing the table of values \[ \begin{array}{c|cccc} n & a_n & b_n & c_n & s_n \ \hline 0 & 1 & 2 & 3 & 6 \ 1 & 4 & 5 & 7 & 16 \ 2 & 8 & 9 & 10 & 27 \end{array} \] we see that (a)$_0$ holds; \item (a)$_n$ implies (b)$_n$; \item (a)$_n$ and (b)$_1$, \dots, (b)$_n$ together imply (a)$_{n+1}$. \end{itemize} To produce a value of $n$ for which $s_n \equiv 2015 \pmod{10000}$, we take $n = 3m+1$ for some nonnegative integer $m$ for which $s_{3m+1} = 30m+15$. We must also have $30m \equiv 2000 \pmod{10000}$, or equivalently $m \equiv 400 \pmod{1000}$. By taking $m = 1400$, we ensure that $m \equiv 2 \pmod{3}$, so $s_m = 10m + 7$; this ensures that $s_{n}$ does indeed equal $30m+15 = 42015$, as desired. | 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 |
1990 | 1990_A3 | Given any convex pentagon whose vertices (no three of which are collinear) have integer coordinates find the minimum possible areas it can cover? | By Pick’s formula, the area is $I + B/2 - 1$, where $I$ is the number of internal lattice points and $B$ is the number on the boundary. Clearly, $I \geq 0$ and $B \geq 5$. If $I \geq 1$ we are done.
If $I = 0$, then separate the vertices $v_1, v_2, v_3, v_4, v_5$ into four classes according to the parity of their coordinates. At least one class must have at least two elements, say $v_1$ and $v_2$. Hence the mid-point $\frac{1}{2}(v_1 + v_2)$, is also a lattice point; call it $v_0$. Since $I = 0$, $v_0$ is on the boundary of the pentagon. Now consider the five points $\{v_0, v_2, v_3, v_4, v_5\}$. The same reasoning produces a second lattice point $v_0^*$ which is not $v_1$ (since $v_0^*$ is a mid-point) and not in the interior (since $I = 0$). Thus we have a second new lattice point on the boundary. Therefore, $B \geq 7$, so again the area is $\geq \boxed{5/2}$. | numerical | putnam (modified boxing) | Geometry Number Theory | Prove that any convex pentagon whose vertices (no three of which are collinear) have integer coordinates must have area greater than or equal to 5/2. | 0 |
|
1981 | 1981_B1 | Find
\[\lim_{n \to \infty}\left[ \frac{1}{n^5} \sum_{h=1}^n \sum_{k=1}^n \left(5h^4 - 18h^2k^2 + 5k^4\right) \right].\] | Let $S_k(n) = 1^k + 2^k + \cdots + n^k$. Using standard methods of calculus texts one finds that \[ S_2(n) = \frac{n^3}{3} + \frac{n^2}{2} + an \] and \[ S_4(n) = \frac{n^5}{5} + \frac{n^4}{2} + bn^3 + cn^2 + dn, \] with $a$, $b$, $c$, $d$ constants. Then the double sum is \[ 10nS_4(n) - 18\left[S_2(n)\right]^2 = \left(2n^6 + 5n^5 + \cdots\right) - \left(2n^6 + 6n^5 + \cdots\right) = -n^5 + \cdots \] and the desired limit is \(\boxed{-1}\). | numerical | putnam | Calculus Algebra | Find
\[\lim_{n \to \infty}\left[ \frac{1}{n^5} \sum_{h=1}^n \sum_{k=1}^n \left(5h^4 - 18h^2k^2 + 5k^4\right) \right].\] | Let $S_k(n) = 1^k + 2^k + \cdots + n^k$. Using standard methods of calculus texts one finds that \[ S_2(n) = \frac{n^3}{3} + \frac{n^2}{2} + an \] and \[ S_4(n) = \frac{n^5}{5} + \frac{n^4}{2} + bn^3 + cn^2 + dn, \] with $a$, $b$, $c$, $d$ constants. Then the double sum is \[ 10nS_4(n) - 18\left[S_2(n)\right]^2 = \left(2n^6 + 5n^5 + \cdots\right) - \left(2n^6 + 6n^5 + \cdots\right) = -n^5 + \cdots \] and the desired limit is \(\boxed{-1}\). | 0 |
1965 | 1965_B2 | In a round-robin tournament with $n$ players $P_1, P_2, \ldots, P_n$ (where $n > 1$), each player plays one game with each of the other players and the rules are such that no ties can occur. Let $\omega_r$ and $l_r$ be the number of games won and lost, respectively, by $P_r$. Find the difference between
\[
\sum_{r=1}^n \omega_r^2] & [\sum_{r=1}^n l_r^2.
\]. | Clearly $\omega_r + l_r = n-1$ for $r=1, 2, \ldots, n$ and $\sum_{i=1}^n \omega_r = \sum_{i=1}^n l_r$. Hence,
\[
\sum_{r=1}^n \omega_r^2 - \sum_{r=1}^n l_r^2 = \sum_{r=1}^n (\omega_r - l_r)(\omega_r + l_r) = (n-1)\sum_{r=1}^n (\omega_r - l_r) = (n-1) \cdot 0 = \boxed{0}.
\] | numerical | putnam (modified boxing) | Algebra | In a round-robin tournament with $n$ players $P_1, P_2, \ldots, P_n$ (where $n > 1$), each player plays one game with each of the other players and the rules are such that no ties can occur. Let $\omega_r$ and $l_r$ be the number of games won and lost, respectively, by $P_r$. Show that
\[
\sum_{r=1}^n \omega_r^2 = \sum_{r=1}^n l_r^2.
\] | Clearly $\omega_r + l_r = n-1$ for $r=1, 2, \ldots, n$ and $\sum_{i=1}^n \omega_r = \sum_{i=1}^n l_r$. Hence,
\[
\sum_{r=1}^n \omega_r^2 - \sum_{r=1}^n l_r^2 = \sum_{r=1}^n (\omega_r - l_r)(\omega_r + l_r) = (n-1)\sum_{r=1}^n (\omega_r - l_r) = (n-1) \cdot 0 = 0.
\] | 0 |
1985 | 1985_A4 | Define a sequence $\{a_i\}$ by $a_1=3$ and $a_{i+1}=3^{a_i}$ for $i\geq 1$. Which integers between 00 and 99 inclusive occur as the last two digits in the decimal expansion of infinitely many $a_i$? | We wish to consider $a_i \equiv 3^{a_{i-1}} \pmod{100}$. Recall that $a^{\phi(n)} \equiv 1 \pmod{n}$ whenever $a$ is relatively prime to $n$ ($\phi$ is the Euler $\phi$-function). Therefore, since $\phi(100) = 40$, we can find $a_i \pmod{100}$ by knowing $a_{i-1} \pmod{40}$. Similarly, we can know $a_{i-1} \pmod{40}$ by finding $a_{i-2} \pmod{16}$, because $a_{i-1} = 3^{a_{i-2}}$ and $\phi(40) = 16$. Again, $a_{i-2} = 3^{a_{i-3}}$ and $\phi(16) = 8$, and therefore \[ a_{i-3} \equiv 3^{a_{i-4}} \equiv 3^{\text{odd integer}} \equiv 3 \pmod{8}. \] It follows that $a_{i-2} \equiv 3^3 \equiv 11 \pmod{16}$, and from this, \[ a_{i-1} \equiv 3^{11} \equiv 27 \pmod{40}, \] and finally, \[ a_i \equiv 3^{27} \equiv 87 \pmod{100}. \] All of this is valid for all $i \geq 4$. Thus, $\boxed{87}$ is the only integer that occurs as the last two digits in the decimal expansions of infinitely many $a_i$. | numerical | putnam | Number Theory | Define a sequence $\{a_i\}$ by $a_1=3$ and $a_{i+1}=3^{a_i}$ for $i\geq 1$. Which integers between 00 and 99 inclusive occur as the last two digits in the decimal expansion of infinitely many $a_i$? | 0 |
|
1958 | 1958_3_Sp | Under the assumption that the following set of relations has a unique solution for $u(t)$, determine it.
\[ \frac{du(t)}{dt} = u(t) + \int_0^1 u(s) \, ds, \quad u(0) = 1. \] | Put
\[ b = \int_0^1 u(s) \, ds. \]
Then $b$ is a constant and the given equation becomes the familiar linear differential equation
\[ u'(t) = u(t) + b \]
with the general solution
\[ u(t) = -b + ce^t. \]
There remains the problem of determining $b$ and $c$. We have
\[ b = \int_0^1 (-b + ce^s) \, ds = -b + c(e - 1) \]
and
\[ u(0) = -b + c = 1. \]
These two equations imply that
\[ b = \frac{e - 1}{3 - e} \quad \text{and} \quad c = \frac{2}{3 - e} \]
so
\[ u(t) = \boxed{\frac{1}{3 - e} (2e^t - e + 1)}. \]
It is readily checked that this function is indeed a solution.
The above derivation proves the uniqueness of the solution function $u$, so this assumption in the statement of the problem is unnecessary. | algebraic | putnam | Differential Equations Analysis | Under the assumption that the following set of relations has a unique solution for $u(t)$, determine it.
\[ \frac{du(t)}{dt} = u(t) + \int_0^1 u(s) \, ds, \quad u(0) = 1. \] | Put
\[ b = \int_0^1 u(s) \, ds. \]
Then $b$ is a constant and the given equation becomes the familiar linear differential equation
\[ u'(t) = u(t) + b \]
with the general solution
\[ u(t) = -b + ce^t. \]
There remains the problem of determining $b$ and $c$. We have
\[ b = \int_0^1 (-b + ce^s) \, ds = -b + c(e - 1) \]
and
\[ u(0) = -b + c = 1. \]
These two equations imply that
\[ b = \frac{e - 1}{3 - e} \quad \text{and} \quad c = \frac{2}{3 - e} \]
so
\[ u(t) = \boxed{\frac{1}{3 - e} (2e^t - e + 1)}. \]
It is readily checked that this function is indeed a solution.
The above derivation proves the uniqueness of the solution function $u$, so this assumption in the statement of the problem is unnecessary. | 0 |
2016 | 2016_A2 | Given a positive integer $n$, let $M(n)$ be the largest integer $m$ such that \[ \binom{m}{n-1} > \binom{m-1}{n}. \] Evaluate \[ \lim_{n \to \infty} \frac{M(n)}{n}. \] | Note that for $m > n+1$, both binomial coefficients are nonzero and their ratio is \begin{align*} {m\choose n-1}/{m-1\choose n} &= \frac{m!n!(m-n-1)!}{(m-1)!(n-1)!(m-n+1)!} \\ &= \frac{mn}{(m-n+1)(m-n)}. \end{align*} Thus the condition ${m\choose{n-1}} > {{m-1}\choose n}$ is equivalent to $(m-n+1)(m-n)-mn < 0$. The left hand side of this last inequality is a quadratic function of $m$ with roots \begin{align*} \alpha(n) &= \frac{3n-1+\sqrt{5n^2-2n+1}}{2}, \\ \beta(n) &= \frac{3n-1-\sqrt{5n^2-2n+1}}{2}, \end{align*} both of which are real since $5n^2-2n+1 = 4n^2+(n-1)^2 > 0$; it follows that $m$ satisfies the given inequality if and only if $\beta(n) < m < \alpha(n)$. (Note in particular that since $\alpha(n)-\beta(n) = \sqrt{5n^2-2n+1} > 1$, there is always some integer $m$ between $\beta(n)$ and $\alpha(n)$.) We conclude that $M(n)$ is the greatest integer strictly less than $\alpha(n)$, and thus that $\alpha(n)-1 \leq M(n) < \alpha(n)$. Now \[ \lim_{n\to\infty} \frac{\alpha(n)}{n} = \lim_{n\to\infty} \frac{3-\frac{1}{n}+\sqrt{5-\frac{2}{n}+\frac{1}{n^2}}}{2} = \frac{3+\sqrt{5}}{2} \] and similarly $\lim_{n\to\infty} \frac{\alpha(n)-1}{n} = \frac{3+\sqrt{5}}{2}$, and so by the sandwich theorem, $\lim_{n\to\infty} \frac{M(n)}{n} = \frac{3+\sqrt{5}}{2}$. The answer is $\boxed{\frac{3+\sqrt{5}}{2}}$. | numerical | putnam | Analysis Calculus Combinatorics | Given a positive integer $n$, let $M(n)$ be the largest integer $m$ such that \[ \binom{m}{n-1} > \binom{m-1}{n}. \] Evaluate \[ \lim_{n \to \infty} \frac{M(n)}{n}. \] | The answer is $\boxed{\frac{3+\sqrt{5}}{2}}$. Note that for $m > n+1$, both binomial coefficients are nonzero and their ratio is \begin{align*} {m\choose n-1}/{m-1\choose n} &= \frac{m!n!(m-n-1)!}{(m-1)!(n-1)!(m-n+1)!} \\ &= \frac{mn}{(m-n+1)(m-n)}. \end{align*} Thus the condition ${m\choose{n-1}} > {{m-1}\choose n}$ is equivalent to $(m-n+1)(m-n)-mn < 0$. The left hand side of this last inequality is a quadratic function of $m$ with roots \begin{align*} \alpha(n) &= \frac{3n-1+\sqrt{5n^2-2n+1}}{2}, \\ \beta(n) &= \frac{3n-1-\sqrt{5n^2-2n+1}}{2}, \end{align*} both of which are real since $5n^2-2n+1 = 4n^2+(n-1)^2 > 0$; it follows that $m$ satisfies the given inequality if and only if $\beta(n) < m < \alpha(n)$. (Note in particular that since $\alpha(n)-\beta(n) = \sqrt{5n^2-2n+1} > 1$, there is always some integer $m$ between $\beta(n)$ and $\alpha(n)$.) We conclude that $M(n)$ is the greatest integer strictly less than $\alpha(n)$, and thus that $\alpha(n)-1 \leq M(n) < \alpha(n)$. Now \[ \lim_{n\to\infty} \frac{\alpha(n)}{n} = \lim_{n\to\infty} \frac{3-\frac{1}{n}+\sqrt{5-\frac{2}{n}+\frac{1}{n^2}}}{2} = \frac{3+\sqrt{5}}{2} \] and similarly $\lim_{n\to\infty} \frac{\alpha(n)-1}{n} = \frac{3+\sqrt{5}}{2}$, and so by the sandwich theorem, $\lim_{n\to\infty} \frac{M(n)}{n} = \frac{3+\sqrt{5}}{2}$. | 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 |
1957 | 1957_8 | Consider the determinant $|a_{ij}|$ of order 100 with $a_{ij} = i \times j$. Given that the absolute value of each of the 100! terms in the expansion of this determinant is divided by 101, find the maximum remainder across all cases. | Solution. Each term in the expansion of the given determinant is, except for sign, the product of all possible row indices and all possible column indices, that is, $(100!)^2$, and this is the absolute value of every term.
Now 101 is a prime, so by Wilson's theorem $100! \equiv -1 \pmod{101}$. Hence $(100!)^2 \equiv (-1)^2 \equiv 1 \pmod{101}$, as required. Thus the remainder will be \boxed{1}. | numerical | putnam (modified boxing) | Number Theory Linear Algebra | Consider the determinant $|a_{ij}|$ of order 100 with $a_{ij} = i \times j$. Prove that if the absolute value of each of the 100! terms in the expansion of this determinant is divided by 101 then the remainder in each case is 1. | Solution. Each term in the expansion of the given determinant is, except for sign, the product of all possible row indices and all possible column indices, that is, $(100!)^2$, and this is the absolute value of every term.
Now 101 is a prime, so by Wilson's theorem $100! \equiv -1 \pmod{101}$. Hence $(100!)^2 \equiv (-1)^2 \equiv 1 \pmod{101}$, as required. | 0 |
1976 | 1976_B1 | \text{Evaluate } \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \left( \left\lfloor \frac{2n}{k} \right\rfloor - 2 \left\lfloor \frac{n}{k} \right\rfloor \right) \text{ and express your answer in the form } \log a - b, \text{ with } a \text{ and } b \text{ positive integers.} \\ \text{Here } \lfloor x \rfloor \text{ is defined to be the integer such that } \lfloor x \rfloor \leq x < \lfloor x \rfloor + 1 \text{ and } \log x \text{ is the logarithm of } x \text{ to base } e. | It is shown below that $a = 4$ and $b = 1$. Let $f(x) = \lfloor 2/x \rfloor - 2\lfloor 1/x \rfloor$. Then the desired limit $L$ equals $\int_0^1 f(x)dx$. For $n = 1, 2, \dots$, $f(x) = 0$ on $2/(2n + 1) < x \leq 1/n$ and $f(x) = 1$ on $1/(n + 1) < x \leq 2/(2n + 1)$. Hence \\ \[ L = \left( \frac{2}{3} - \frac{2}{4} \right) + \left( \frac{2}{5} - \frac{2}{6} \right) + \cdots = -1 + 2 \left( 1 - \frac{1}{2} + \frac{1}{3} - \cdots \right) \] \\ \[ = -1 + 2 \int_0^1 \frac{dx}{1+x} = -1 + 2 \ln 2 = \boxed{\ln 4 - 1}. \] | algebraic | putnam | Analysis | \text{Evaluate } \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \left( \left\lfloor \frac{2n}{k} \right\rfloor - 2 \left\lfloor \frac{n}{k} \right\rfloor \right) \text{ and express your answer in the form } \log a - b, \text{ with } a \text{ and } b \text{ positive integers.} \\ \text{Here } \lfloor x \rfloor \text{ is defined to be the integer such that } \lfloor x \rfloor \leq x < \lfloor x \rfloor + 1 \text{ and } \log x \text{ is the logarithm of } x \text{ to base } e. | It is shown below that $a = 4$ and $b = 1$. Let $f(x) = \lfloor 2/x \rfloor - 2\lfloor 1/x \rfloor$. Then the desired limit $L$ equals $\int_0^1 f(x)dx$. For $n = 1, 2, \dots$, $f(x) = 0$ on $2/(2n + 1) < x \leq 1/n$ and $f(x) = 1$ on $1/(n + 1) < x \leq 2/(2n + 1)$. Hence \\ \[ L = \left( \frac{2}{3} - \frac{2}{4} \right) + \left( \frac{2}{5} - \frac{2}{6} \right) + \cdots = -1 + 2 \left( 1 - \frac{1}{2} + \frac{1}{3} - \cdots \right) \] \\ \[ = -1 + 2 \int_0^1 \frac{dx}{1+x} = -1 + 2 \ln 2 = \boxed{\ln 4 - 1}. \] | 0 |
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