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1955 | 1955_1 | Find the number of sets of integers $m$, $n$, $p$ except $0$, $0$, $0$ for which
\[
m + n\sqrt{2} + p\sqrt{3} = 0.
\] | We recall that $(1)$ if $a$ is a positive integer and $\sqrt{a}$ is not an integer, then $\sqrt{a}$ is irrational.
Suppose that $m$, $n$, $p$ are integers such that
\[
m + n\sqrt{2} + p\sqrt{3} = 0.
\]
If both $n$ and $p$ are zero, so is $m$. If just one of $n$ and $p$ is zero, we have either $\sqrt{2} = -m/n$ or $\sqrt{3} = -m/p$, both contrary to $(1)$. If neither $n$ nor $p$ is zero, then
\[
m^2 = (n\sqrt{2} + p\sqrt{3})^2 = 2n^2 + 3p^2 + 2np\sqrt{6},
\]
and
\[
\sqrt{6} = (m^2 - 2n^2 - 3p^2)/2np,
\]
again contrary to $(1)$. So the only integer triplet for which $(2)$ is true is $m = 0$, $n = 0$, $p = 0$.
For completeness we include a proof of $(1)$. Suppose $a$ is a positive integer and $\sqrt{a} = b/c$ where $b$ and $c$ are positive integers. We may assume $b$ and $c$ are relatively prime. Then $ac^2 = b^2$. Consider a prime $q$ that divides $c$. Then $q$ divides $b^2$ and hence $b$; so if there is such a prime $q$, then $b$ and $c$ have a common factor, which is a contradiction. Hence $c$ has no prime factors, so $c = 1$, and $\sqrt{a} = b$, an integer. Thus there exists \boxed{0} solutions except (0,0,0). | numerical | putnam (modified boxing) | Number Theory | Prove that there is no set of integers $m$, $n$, $p$ except $0$, $0$, $0$ for which
\[
m + n\sqrt{2} + p\sqrt{3} = 0.
\] | We recall that $(1)$ if $a$ is a positive integer and $\sqrt{a}$ is not an integer, then $\sqrt{a}$ is irrational.
Suppose that $m$, $n$, $p$ are integers such that
\[
m + n\sqrt{2} + p\sqrt{3} = 0.
\]
If both $n$ and $p$ are zero, so is $m$. If just one of $n$ and $p$ is zero, we have either $\sqrt{2} = -m/n$ or $\sqrt{3} = -m/p$, both contrary to $(1)$. If neither $n$ nor $p$ is zero, then
\[
m^2 = (n\sqrt{2} + p\sqrt{3})^2 = 2n^2 + 3p^2 + 2np\sqrt{6},
\]
and
\[
\sqrt{6} = (m^2 - 2n^2 - 3p^2)/2np,
\]
again contrary to $(1)$. So the only integer triplet for which $(2)$ is true is $m = 0$, $n = 0$, $p = 0$.
For completeness we include a proof of $(1)$. Suppose $a$ is a positive integer and $\sqrt{a} = b/c$ where $b$ and $c$ are positive integers. We may assume $b$ and $c$ are relatively prime. Then $ac^2 = b^2$. Consider a prime $q$ that divides $c$. Then $q$ divides $b^2$ and hence $b$; so if there is such a prime $q$, then $b$ and $c$ have a common factor, which is a contradiction. Hence $c$ has no prime factors, so $c = 1$, and $\sqrt{a} = b$, an integer. | 0 |
1955 | 1955_4 | On a circle, $n$ points are selected and the chords joining them in pairs are drawn. Assuming that no three of these chords are concurrent (except at the endpoints), how many points of intersection are there? | Any four points on a circle determine just one pair of chords that intersect at an interior point. Since the hypothesis implies that all such points of intersection are distinct, there are \(
\boxed{\binom{n}{4}}\) points of intersection in the interior of the circle. | algebraic | putnam | Combinatorics | On a circle, $n$ points are selected and the chords joining them in pairs are drawn. Assuming that no three of these chords are concurrent (except at the endpoints), how many points of intersection are there? | Any four points on a circle determine just one pair of chords that intersect at an interior point. Since the hypothesis implies that all such points of intersection are distinct, there are \(
\boxed{\binom{n}{4}}\) points of intersection in the interior of the circle. | 0 |
1955 | 1955_6 | Determine a necessary and sufficient condition on the positive integer \(n\) such that the equation \(x^n + (2+x)^n + (2-x)^n = 0\) has a rational root. | There can be no real root if \(n\) is even, since for real \(x\), each term is non-negative and they cannot vanish simultaneously. If \(n = 1\), there is obviously a unique root \(x = -4\). Suppose \(n\) is odd and at least 3. When the terms of the equation are expanded and collected, the result is monic with all coefficients non-negative integers and constant term \(2^{n+1}\). The only possible roots, therefore, are of the form \(-2^p\). For \(x = -1\), all three terms of the given expression are odd, so \(-1\) is not a root. Putting \(x = -2\), we find \((-2)^n + 0 + 4^n \neq 0\), so \(-2\) is not a root. If we put \(x = -2^{p+1}\) where \(p \geq 1\), the left member of \(1\) becomes:
\[
2^n \left[-2^{pn} + 2\left(1 + \binom{n}{2} 2^{2p} + \binom{n}{4} 2^{4p} + \ldots \right)\right]
\]
The expression in the brackets is \(\equiv 2 \pmod{4}\) (recall \(n \geq 3\)), so \(-2^{p+1}\) is not a root for \(p \geq 1\). Therefore, there are no roots if \(n > 1\). Summarizing, relation \(1\) has a rational root if and only if \(\boxed{n = 1}\). | algebraic | putnam | Number Theory | Find a necessary and sufficient condition on the positive integer $n$ that the equation \(x^n + (2+x)^n + (2-x)^n = 0\) has a rational root. | There can be no real root if \(n\) is even, since for real \(x\), each term is non-negative and they cannot vanish simultaneously. If \(n = 1\), there is obviously a unique root \(x = -4\). Suppose \(n\) is odd and at least 3. When the terms of the equation are expanded and collected, the result is monic with all coefficients non-negative integers and constant term \(2^{n+1}\). The only possible roots, therefore, are of the form \(-2^p\). For \(x = -1\), all three terms of the given expression are odd, so \(-1\) is not a root. Putting \(x = -2\), we find \((-2)^n + 0 + 4^n \neq 0\), so \(-2\) is not a root. If we put \(x = -2^{p+1}\) where \(p \geq 1\), the left member of \(1\) becomes:
\[
2^n \left[-2^{pn} + 2\left(1 + \binom{n}{2} 2^{2p} + \binom{n}{4} 2^{4p} + \ldots \right)\right]
\]
The expression in the brackets is \(\equiv 2 \pmod{4}\) (recall \(n \geq 3\)), so \(-2^{p+1}\) is not a root for \(p \geq 1\). Therefore, there are no roots if \(n > 1\). Summarizing, relation \(1\) has a rational root if and only if \(\boxed{n = 1}\). | 0 |
1955 | 1955_8 | A sphere rolls along two intersecting straight lines. Find the locus of its center. | We choose axes so that the given lines lie in the $x, y$-plane and the $x$-axis bisects the angle between them. Then the normal forms of the equations of the lines are
\[x \sin \theta - y \cos \theta = 0\]
and
\[x \sin \theta + y \cos \theta = 0,\]
where $0 < \theta < \pi/2$.
The squared distance from the point $(x, y, z)$ to these lines is $z^2 + (x \sin \theta - y \cos \theta)^2$ and $z^2 + (x \sin \theta + y \cos \theta)^2$. The center of the sphere must be at distance $r$ from both lines; hence the desired locus is given by the two equations
\[z^2 + (x \sin \theta - y \cos \theta)^2 = r^2\]
and
\[z^2 + (x \sin \theta + y \cos \theta)^2 = r^2.\]
Subtracting these equations, we obtain $4xy \sin \theta \cos \theta = 0$. Therefore the locus lies on the union of the planes $x = 0$ and $y = 0$, i.e., the $y,z$-plane and the $x,z$-plane. The part on the $y,z$-plane has the equation
\[z^2 + y^2 \cos^2 \theta = r^2,\]
so it is an ellipse having minor axis on the $z$-axis of length $2r$ and having major axis on the $y$-axis of length $2r \sec \theta$. The part on the $x,z$-plane has the equation
\[z^2 + x^2 \sin^2 \theta = r^2.\]
It is therefore an ellipse having minor axis of length $2r$ lying on the $z$-axis and major axis of length $2r \csc \theta$ lying on the $x$-axis. The locus is the union of these two ellipses and is given by \boxed{z^2 + (x \sin \theta \pm y \cos \theta)^2 = r^2}. | algebraic | putnam (modified boxing) | Geometry Algebra | A sphere rolls along two intersecting straight lines. Find the locus of its center. | We choose axes so that the given lines lie in the $x, y$-plane and the $x$-axis bisects the angle between them. Then the normal forms of the equations of the lines are
\[x \sin \theta - y \cos \theta = 0\]
and
\[x \sin \theta + y \cos \theta = 0,\]
where $0 < \theta < \pi/2$.
The squared distance from the point $(x, y, z)$ to these lines is $z^2 + (x \sin \theta - y \cos \theta)^2$ and $z^2 + (x \sin \theta + y \cos \theta)^2$. The center of the sphere must be at distance $r$ from both lines; hence the desired locus is given by the two equations
\\boxed{[z^2 + (x \sin \theta - y \cos \theta)^2 = r^2\]
and
\[z^2 + (x \sin \theta + y \cos \theta)^2 = r^2.]}
Subtracting these equations, we obtain $4xy \sin \theta \cos \theta = 0$. Therefore the locus lies on the union of the planes $x = 0$ and $y = 0$, i.e., the $y,z$-plane and the $x,z$-plane. The part on the $y,z$-plane has the equation
\[z^2 + y^2 \cos^2 \theta = r^2,\]
so it is an ellipse having minor axis on the $z$-axis of length $2r$ and having major axis on the $y$-axis of length $2r \sec \theta$. The part on the $x,z$-plane has the equation
\[z^2 + x^2 \sin^2 \theta = r^2.\]
It is therefore an ellipse having minor axis of length $2r$ lying on the $z$-axis and major axis of length $2r \csc \theta$ lying on the $x$-axis. The locus is the union of these two ellipses. | 0 |
1956 | 1956_1 | Evaluate
\[ \lim_{x \to +\infty} \left[ \frac{1}{x} \frac{a^x - 1}{a - 1} \right]^{1/x} \]
where $a > 0, a \neq 1$. | Let
\[ f(x) = \left[ \frac{1}{x} \frac{a^x - 1}{a - 1} \right]^{1/x}. \]
Then for $x > 0$ and $a > 1$, we have
\[ \log f(x) = - \frac{\log x}{x} - \frac{\log (a - 1)}{x} + \frac{\log (a^x - 1)}{x}. \]
As $x \to +\infty$,
\[ \frac{\log x}{x} \to 0, \quad \frac{\log (a - 1)}{x} \to 0, \]
and
\[ \frac{\log (a^x - 1)}{x} = \frac{\log (1 - a^{-x})}{x} + \log a \to \log a. \]
Hence $\log f(x) \to \log a$.
On the other hand, if $0 < a < 1$ and $x > 0$, then
\[ \log f(x) = - \frac{\log x}{x} - \frac{\log (1 - a)}{x} + \frac{\log (1 - a^x)}{x}, \]
and it is clear that all three terms approach zero as $x \to +\infty$.
Since $\exp$ is a continuous function, we have
\[ \lim_{x \to +\infty} f(x) = \exp \lim_{x \to +\infty} \log f(x) = \begin{cases} \exp \log a = a & \text{if } a > 1, \\ \exp 0 = 1 & \text{if } 0 < a < 1. \end{cases} \]
More concisely, $\lim_{x \to +\infty} f(x) = \boxed{\max (a, 1)}$. | algebraic | putnam | Analysis Calculus | Evaluate
\[ \lim_{x \to +\infty} \left[ \frac{1}{x} \frac{a^x - 1}{a - 1} \right]^{1/x} \]
where $a > 0, a \neq 1$. | Let
\[ f(x) = \left[ \frac{1}{x} \frac{a^x - 1}{a - 1} \right]^{1/x}. \]
Then for $x > 0$ and $a > 1$, we have
\[ \log f(x) = - \frac{\log x}{x} - \frac{\log (a - 1)}{x} + \frac{\log (a^x - 1)}{x}. \]
As $x \to +\infty$,
\[ \frac{\log x}{x} \to 0, \quad \frac{\log (a - 1)}{x} \to 0, \]
and
\[ \frac{\log (a^x - 1)}{x} = \frac{\log (1 - a^{-x})}{x} + \log a \to \log a. \]
Hence $\log f(x) \to \log a$.
On the other hand, if $0 < a < 1$ and $x > 0$, then
\[ \log f(x) = - \frac{\log x}{x} - \frac{\log (1 - a)}{x} + \frac{\log (1 - a^x)}{x}, \]
and it is clear that all three terms approach zero as $x \to +\infty$.
Since $\exp$ is a continuous function, we have
\[ \lim_{x \to +\infty} f(x) = \exp \lim_{x \to +\infty} \log f(x) = \begin{cases} \exp \log a = a & \text{if } a > 1, \\ \exp 0 = 1 & \text{if } 0 < a < 1. \end{cases} \]
More concisely, $\lim_{x \to +\infty} f(x) = \boxed{\max (a, 1)}$. | 0 |
1956 | 1956_3 | A particle falls in a vertical plane from rest under the influence of gravity and a force perpendicular to and proportional to its velocity. Obtain the equations of the trajectory and sum them together. | Suppose the position of the particle at time $t$ is $\langle x(t), y(t) \rangle$. Then the velocity vector is given by $\langle x'(t), y'(t) \rangle$ and the acceleration vector by $\langle x''(t), y''(t) \rangle$. We choose the coordinate system so that the particle starts at the origin with $x$ running horizontally and $y$ vertically as usual.
Since the vector $\langle y'(t), -x'(t) \rangle$ is perpendicular to and of the same length as the velocity vector the differential equation governing the motion is
\[ \langle x''(t), y''(t) \rangle = c \langle y'(t), -x'(t) \rangle + \langle 0, -g \rangle \]
where $c$ is a constant incorporating the mass of the given particle. The initial conditions are $x(0) = y(0) = x'(0) = y'(0) = 0$. Separating $(1)$ into two equations we have
\[ x''(t) = c y'(t) \]
\[ y''(t) = -c x'(t) - g \]
Differentiating the first of these equations, we get
\[ x'''(t) = c y''(t) = -c^2 x'(t) - cg, \]
which in standard form is
\[ x'''(t) + c^2 x'(t) = -cg. \]
The corresponding homogeneous differential equation, $x'''(t) + c^2 x'(t) = 0$, has the three linearly independent solutions $1, \sin ct, \cos ct$. Since the right member of $(2)$ is a solution of the homogeneous differential equation, we look for a particular solution of the form $kt$, and find that $-gt/c$ is such a solution. Hence the general solution of $(2)$ is
\[ x(t) = -\frac{gt}{c} + \alpha + \beta \cos ct + \gamma \sin ct. \]
The initial conditions are $x(0) = x'(0) = x''(0) = 0$ (the latter from $y'(0) = 0$). Hence
\[ x(t) = -\frac{gt}{c} + \frac{g}{c^2} \sin ct. \]
Therefore $y' = x''/c = -g/c \sin ct$ and
\[ y(t) = \frac{g}{c^2} \left(-1 + \cos ct \right), \]
using the initial condition $y(0) = 0$.
Thus equations $(3)$ and $(4)$ describe the motion. They are the parametric equations of a cycloid traced by a point on the rim of a circle of radius $g/c^2$ rolling along the underside of the $x$-axis with velocity $-g/c$. Thus the final answer is \boxed{-\frac{gt}{c} + \frac{g}{c^2} \sin ct - \frac{g}{c^2} + \frac{g}{c^2} \cos ct}}. | algebraic | putnam (modified boxing) | Differential Equations Calculus | A particle falls in a vertical plane from rest under the influence of gravity and a force perpendicular to and proportional to its velocity. Obtain the equations of the trajectory and identify the curve. | Suppose the position of the particle at time $t$ is $\langle x(t), y(t) \rangle$. Then the velocity vector is given by $\langle x'(t), y'(t) \rangle$ and the acceleration vector by $\langle x''(t), y''(t) \rangle$. We choose the coordinate system so that the particle starts at the origin with $x$ running horizontally and $y$ vertically as usual.
Since the vector $\langle y'(t), -x'(t) \rangle$ is perpendicular to and of the same length as the velocity vector the differential equation governing the motion is
\[ \langle x''(t), y''(t) \rangle = c \langle y'(t), -x'(t) \rangle + \langle 0, -g \rangle \]
where $c$ is a constant incorporating the mass of the given particle. The initial conditions are $x(0) = y(0) = x'(0) = y'(0) = 0$. Separating $(1)$ into two equations we have
\[ x''(t) = c y'(t) \]
\[ y''(t) = -c x'(t) - g \]
Differentiating the first of these equations, we get
\[ x'''(t) = c y''(t) = -c^2 x'(t) - cg, \]
which in standard form is
\[ x'''(t) + c^2 x'(t) = -cg. \]
The corresponding homogeneous differential equation, $x'''(t) + c^2 x'(t) = 0$, has the three linearly independent solutions $1, \sin ct, \cos ct$. Since the right member of $(2)$ is a solution of the homogeneous differential equation, we look for a particular solution of the form $kt$, and find that $-gt/c$ is such a solution. Hence the general solution of $(2)$ is
\[ x(t) = -\frac{gt}{c} + \alpha + \beta \cos ct + \gamma \sin ct. \]
The initial conditions are $x(0) = x'(0) = x''(0) = 0$ (the latter from $y'(0) = 0$). Hence
\[ \boxed{x(t) = -\frac{gt}{c} + \frac{g}{c^2} \sin ct. \]
Therefore $y' = x''/c = -g/c \sin ct$ and
\[ y(t) = \frac{g}{c^2} \left(-1 + \cos ct \right)}, \]
using the initial condition $y(0) = 0$.
Thus equations $(3)$ and $(4)$ describe the motion. They are the parametric equations of a cycloid traced by a point on the rim of a circle of radius $g/c^2$ rolling along the underside of the $x$-axis with velocity $-g/c$. | 0 |
1956 | 1956_4 | Suppose the $n$ times differentiable real function $f(x)$ has at least $n+1$ distinct zeros in the closed interval $[a,b]$ and that the polynomial $P(z) \equiv z^n + C_{n-1} z^{n-1} + \cdots + C_0$ has only real zeros. Find the number of zero's that $(D^n + C_{n-1} D^{n-1} + \cdots + C_0)f(x)$ has in the interval $[a,b]$ where $D^n$ denotes, as usual, $d^n/dx^n$. | We first prove a lemma.
**Lemma.** Suppose $f$ is differentiable on $[a,b]$ and has $m+1$ distinct zeros there. Then for any real number $\lambda$, $(D - \lambda)f$ has at least $m$ distinct zeros on $[a,b]$.
**Proof.** Consider the identity
\[ (D - \lambda)f(x) = e^{\lambda x}D(e^{-\lambda x}f(x)). \]
Applying Rolle's theorem to the right member of this identity we see that there is a zero of $(D - \lambda)f$ between any two consecutive zeros of $f$. Therefore there are at least $m$ distinct zeros on $[a,b]$. \(\blacksquare\)
We can now prove the result stated in the problem by induction on $n$.
If $n = 1$, this is just the lemma with $m = 1$.
Assume the result is true for $n = k$. Suppose $P$ has degree $k+1$ and that $f$ is $k+1$ times differentiable and has $k+2$ distinct zeros on $[a,b]$. Since $P$ has all real roots, we can write $P(z) = Q(z)(z - \lambda)$, where $\lambda$ is real and $Q$ is a polynomial of degree $k$ with all real roots. Then $g = (D - \lambda)f$ is $k$ times differentiable on $[a,b]$ and has at least $k+1$ distinct zeros on $[a,b]$, by the lemma with $m = k+1$. Hence by the inductive hypothesis $Q(D)g$ has at least one zero on $[a,b]$. But
\[ Q(D)g \equiv Q(D)(D - \lambda)f = P(D)f. \]
Thus the result is true for $n = k+1$. This completes the induction giving us the final answer of \boxed{0}. | numerical | putnam (modified boxing) | Analysis Differential Equations | Suppose the $n$ times differentiable real function $f(x)$ has at least $n+1$ distinct zeros in the closed interval $[a,b]$ and that the polynomial $P(z) \equiv z^n + C_{n-1} z^{n-1} + \cdots + C_0$ has only real zeros. Show that $(D^n + C_{n-1} D^{n-1} + \cdots + C_0)f(x)$ has at least one zero in the interval $[a,b]$ where $D^n$ denotes, as usual, $d^n/dx^n$. | We first prove a lemma.
**Lemma.** Suppose $f$ is differentiable on $[a,b]$ and has $m+1$ distinct zeros there. Then for any real number $\lambda$, $(D - \lambda)f$ has at least $m$ distinct zeros on $[a,b]$.
**Proof.** Consider the identity
\[ (D - \lambda)f(x) = e^{\lambda x}D(e^{-\lambda x}f(x)). \]
Applying Rolle's theorem to the right member of this identity we see that there is a zero of $(D - \lambda)f$ between any two consecutive zeros of $f$. Therefore there are at least $m$ distinct zeros on $[a,b]$. \(\blacksquare\)
We can now prove the result stated in the problem by induction on $n$.
If $n = 1$, this is just the lemma with $m = 1$.
Assume the result is true for $n = k$. Suppose $P$ has degree $k+1$ and that $f$ is $k+1$ times differentiable and has $k+2$ distinct zeros on $[a,b]$. Since $P$ has all real roots, we can write $P(z) = Q(z)(z - \lambda)$, where $\lambda$ is real and $Q$ is a polynomial of degree $k$ with all real roots. Then $g = (D - \lambda)f$ is $k$ times differentiable on $[a,b]$ and has at least $k+1$ distinct zeros on $[a,b]$, by the lemma with $m = k+1$. Hence by the inductive hypothesis $Q(D)g$ has at least one zero on $[a,b]$. But
\[ Q(D)g \equiv Q(D)(D - \lambda)f = P(D)f. \]
Thus the result is true for $n = k+1$. This completes the induction. | 0 |
1956 | 1956_5 | Given $n$ objects arranged in a row. A subset of these objects is called unfriendly if no two of its elements are consecutive. Find a binomial expression for the number of unfriendly subsets each having $k$ elements. | For each subset $S$ of an ordered set of $n$ objects we form a linear arrangement of $A$'s and $B$'s by writing an $A$ in positions corresponding to members of $S$ and $B$'s elsewhere. We describe such an arrangement as unfriendly if no two $A$'s are consecutive. Then an unfriendly subset corresponds to an unfriendly arrangement, and we must show that the number of unfriendly arrangements of $k$ $A$'s and $n-k$ $B$'s is \( \binom{n-k+1}{k}. \)
We shall establish a bijective correspondence between all unfriendly arrangements of $k$ $A$'s and $n-k$ $B$'s and all arrangements of $k$ $A$'s and $n-2k+1$ $B$'s.
Given an unfriendly arrangement of $k$ $A$'s and $n-k$ $B$'s, remove the $B$ standing immediately to the right of each of the first $k-1$ $A$'s (i.e., all but the last $A$). We obtain in this way an arrangement of $k$ $A$'s and $n-2k+1$ $B$'s.
Conversely, given an arrangement of $k$ $A$'s and $n-2k+1$ $B$'s, insert a $B$ after each of the $A$'s but the last. We obtain in this way an unfriendly arrangement of $k$ $A$'s and $n-k$ $B$'s.
Obviously these two transformations are inverses of one another. Therefore the number of unfriendly arrangements of $k$ $A$'s and $n-k$ $B$'s is the same as the number of arrangements of $k$ $A$'s and $n-2k+1$ $B$'s, namely \( \boxed{\binom{n-k+1}{k}}. \). | algebraic | putnam (modified boxing) | Combinatorics | Given $n$ objects arranged in a row. A subset of these objects is called unfriendly if no two of its elements are consecutive. Show that the number of unfriendly subsets each having $k$ elements is \[ \binom{n-k+1}{k}. \] | For each subset $S$ of an ordered set of $n$ objects we form a linear arrangement of $A$'s and $B$'s by writing an $A$ in positions corresponding to members of $S$ and $B$'s elsewhere. We describe such an arrangement as unfriendly if no two $A$'s are consecutive. Then an unfriendly subset corresponds to an unfriendly arrangement, and we must show that the number of unfriendly arrangements of $k$ $A$'s and $n-k$ $B$'s is \( \binom{n-k+1}{k}. \)
We shall establish a bijective correspondence between all unfriendly arrangements of $k$ $A$'s and $n-k$ $B$'s and all arrangements of $k$ $A$'s and $n-2k+1$ $B$'s.
Given an unfriendly arrangement of $k$ $A$'s and $n-k$ $B$'s, remove the $B$ standing immediately to the right of each of the first $k-1$ $A$'s (i.e., all but the last $A$). We obtain in this way an arrangement of $k$ $A$'s and $n-2k+1$ $B$'s.
Conversely, given an arrangement of $k$ $A$'s and $n-2k+1$ $B$'s, insert a $B$ after each of the $A$'s but the last. We obtain in this way an unfriendly arrangement of $k$ $A$'s and $n-k$ $B$'s.
Obviously these two transformations are inverses of one another. Therefore the number of unfriendly arrangements of $k$ $A$'s and $n-k$ $B$'s is the same as the number of arrangements of $k$ $A$'s and $n-2k+1$ $B$'s, namely \( \binom{n-k+1}{k}. \). | 0 |
1956 | 1956_7 | Given that the number of odd binomial coefficients in any finite binomial expansion is a power of $n$, find $n$. | First we note that $(1+x)^2 \equiv 1+x^2 \pmod{2}$ and hence, by induction $(1+x)^\beta \equiv 1+x^\beta \pmod{2}$ if $\beta$ is any power of 2.
Let the exponent of the binomial be $n$, a positive integer, and represent $n$ in the form
\[ n = 2^{\alpha_1} + 2^{\alpha_2} + \cdots + 2^{\alpha_s}, \]
where the $\alpha$'s are integers and $0 \leq \alpha_1 < \alpha_2 < \cdots < \alpha_s$. [Note that in effect this is the dyadic expansion of $n$.] For convenience let $\beta_i = 2^{\alpha_i}$. Then we write the finite binomial expansion in the form
\[ (1+x)^n = (1+x)^{\beta_1}(1+x)^{\beta_2}\cdots(1+x)^{\beta_s} \equiv (1+x^{\beta_1})(1+x^{\beta_2})\cdots(1+x^{\beta_s}) \pmod{2}. \]
When the latter expression is multiplied out, there are clearly $2^s$ terms, each involving a different power of $x$ because each non-negative integer has exactly one representation as a sum (possibly empty) of distinct powers of 2. Hence exactly $2^s$ terms of $(1+x)^n$ have odd integers as coefficients, that is, exactly $2^s$ of the binomial coefficients \( \binom{n}{i} \), $i=0,1,\ldots,n$, are odd. The proof shows that $s$ is the number of unit digits which appear in the dyadic expansion of $n$. The result is also valid for $n=0$.Thus makes the value of $n = \boxed{2}$. | numerical | putnam (modified boxing) | Combinatorics Number Theory | Prove that the number of odd binomial coefficients in any finite binomial expansion is a power of 2. | First we note that $(1+x)^2 \equiv 1+x^2 \pmod{2}$ and hence, by induction $(1+x)^\beta \equiv 1+x^\beta \pmod{2}$ if $\beta$ is any power of 2.
Let the exponent of the binomial be $n$, a positive integer, and represent $n$ in the form
\[ n = 2^{\alpha_1} + 2^{\alpha_2} + \cdots + 2^{\alpha_s}, \]
where the $\alpha$'s are integers and $0 \leq \alpha_1 < \alpha_2 < \cdots < \alpha_s$. [Note that in effect this is the dyadic expansion of $n$.] For convenience let $\beta_i = 2^{\alpha_i}$. Then we write the finite binomial expansion in the form
\[ (1+x)^n = (1+x)^{\beta_1}(1+x)^{\beta_2}\cdots(1+x)^{\beta_s} \equiv (1+x^{\beta_1})(1+x^{\beta_2})\cdots(1+x^{\beta_s}) \pmod{2}. \]
When the latter expression is multiplied out, there are clearly $2^s$ terms, each involving a different power of $x$ because each non-negative integer has exactly one representation as a sum (possibly empty) of distinct powers of 2. Hence exactly $2^s$ terms of $(1+x)^n$ have odd integers as coefficients, that is, exactly $2^s$ of the binomial coefficients \( \binom{n}{i} \), $i=0,1,\ldots,n$, are odd. The proof shows that $s$ is the number of unit digits which appear in the dyadic expansion of $n$. The result is also valid for $n=0$. | 0 |
1956 | 1956_13 | Given $T_1 = 2$, $T_{n+1} = T_n^2 - T_n + 1$, $n > 0$, find the value of $\sum_{i=1}^\infty \frac{1}{T_i}$. | The first few members of the sequences are $T_1 = 2$, $T_2 = 3$, $T_3 = 7$. We shall prove by induction that
\[ T_{n+1} = 1 + \prod_{i=1}^n T_i \text{ for } n \geq 1. \]
This is true for $n=1$. Suppose it is true for $n=k$. Then
\[ T_{k+2} = 1 + T_{k+1}(T_{k+1} - 1) = 1 + T_{k+1} \left[ \prod_{i=1}^k T_i \right] = 1 + \prod_{i=1}^{k+1} T_i. \]
(The first step by the given recursion, the second by the inductive hypothesis.) This completes the inductive proof of (1).
Now suppose $m \neq n$, say $m < n$. Then (1) shows that $T_m$ divides $T_n - 1$, so $T_m$ and $T_n$ are relatively prime. This is (i).
Next we prove by induction that
\[ \sum_{i=1}^n \frac{1}{T_i} = 1 - \frac{1}{T_{n+1} - 1} \text{ for all } n. \]
This is true for $n=1$ and, if it is true for $n=k$, we have
\[ \sum_{i=1}^{k+1} \frac{1}{T_i} = 1 - \frac{1}{T_{k+1} - 1} + \frac{1}{T_{k+1}} = 1 - \frac{1}{T_{k+1}(T_{k+1} - 1)} = 1 - \frac{1}{T_{k+2} - 1}. \]
Thus (2) is established.
Since $T_n \to \infty$ as $n \to \infty$, it follows from (2) that
\[ \sum_{i=1}^\infty \frac{1}{T_i} = \boxed{1}, \]
as required. | numerical | putnam (modified boxing) | Number Theory | Given $T_1 = 2$, $T_{n+1} = T_n^2 - T_n + 1$, $n > 0$, Prove:
\begin{enumerate}
\item[(i)] If $m \neq n$, $T_m$ and $T_n$ have no common factor greater than 1.
\item[(ii)] $\sum_{i=1}^\infty \frac{1}{T_i} = 1$.
\end{enumerate} | The first few members of the sequences are $T_1 = 2$, $T_2 = 3$, $T_3 = 7$. We shall prove by induction that
\[ T_{n+1} = 1 + \prod_{i=1}^n T_i \text{ for } n \geq 1. \]
This is true for $n=1$. Suppose it is true for $n=k$. Then
\[ T_{k+2} = 1 + T_{k+1}(T_{k+1} - 1) = 1 + T_{k+1} \left[ \prod_{i=1}^k T_i \right] = 1 + \prod_{i=1}^{k+1} T_i. \]
(The first step by the given recursion, the second by the inductive hypothesis.) This completes the inductive proof of (1).
Now suppose $m \neq n$, say $m < n$. Then (1) shows that $T_m$ divides $T_n - 1$, so $T_m$ and $T_n$ are relatively prime. This is (i).
Next we prove by induction that
\[ \sum_{i=1}^n \frac{1}{T_i} = 1 - \frac{1}{T_{n+1} - 1} \text{ for all } n. \]
This is true for $n=1$ and, if it is true for $n=k$, we have
\[ \sum_{i=1}^{k+1} \frac{1}{T_i} = 1 - \frac{1}{T_{k+1} - 1} + \frac{1}{T_{k+1}} = 1 - \frac{1}{T_{k+1}(T_{k+1} - 1)} = 1 - \frac{1}{T_{k+2} - 1}. \]
Thus (2) is established.
Since $T_n \to \infty$ as $n \to \infty$, it follows from (2) that
\[ \sum_{i=1}^\infty \frac{1}{T_i} = 1, \]
as required. | 0 |
1957 | 1957_3 | A and B are real numbers and k a positive integer. Find the upper bound of
\[ \left| \frac{\cos kB \cos A - \cos kA \cos B}{\cos B - \cos A} \right| \]
in terms of $k$ whenever the left side is defined. | Let $x = (A - B)/2$, $y = (A + B)/2$. The numerator of the expression on the left is
\[ \frac{1}{2}[\cos (kB + A) + \cos (kB - A) - \cos (kA + B) - \cos (kA - B)] \]
\[ = \frac{1}{2}[\cos (kB + A) - \cos (kA + B)] + \frac{1}{2}[\cos (kB - A) - \cos (kA - B)] \]
\[ = \sin (k - 1)x \sin (k + 1)y + \sin (k + 1)x \sin (k - 1)y. \]
The denominator is $2 \sin x \sin y$, and we may assume this is not zero. Hence we have
\[ \left| \frac{\cos kB \cos A - \cos kA \cos B}{\cos B - \cos A} \right| \leq \frac{1}{2} \left| \frac{\sin (k - 1)x}{\sin x} \cdot \frac{\sin (k + 1)y}{\sin y} \right|
+ \frac{1}{2} \left| \frac{\sin (k + 1)x}{\sin x} \cdot \frac{\sin (k - 1)y}{\sin y} \right|. \]
Now since $|\sin nz| \leq n |\sin z|$ for all $z$ and all positive integers $n$ with strict inequality unless $n = 1$ or $\sin z = 0$ (this is proved below), both terms are less than $(k^2 - 1)/2$ provided $k > 1$. Therefore
\[ \left| \frac{\cos kB \cos A - \cos kA \cos B}{\cos B - \cos A} \right| < k^2 - 1 \]
provided $k > 1$ and $\cos B \neq \cos A$. Obviously, we have equality if $k = 1$, so the problem is not accurately phrased.
We now prove that $|\sin nz| \leq n |\sin z|$ for all positive integers $n$. If $\sin z = 0$, equality holds for every integer $n$, so we assume from now on that $\sin z \neq 0$; then $|\cos z| < 1$. We continue by induction on $n$. Clearly there is equality if $n = 1$. For $n = 2$, we have $|\sin 2z| = |2 \cos z \cdot \sin z| < 2 |\sin z|$. Suppose we have strict inequality for $n = k$. Then
\[ |\sin (k + 1)z| = |\sin kz \cos z + \cos kz \sin z| \leq |\sin kz| + |\sin z| < (k + 1) |\sin z|. \]
Thus we have strict inequality for $n = k + 1$. Hence we have proved
\[ |\sin nz| \leq n |\sin z| \]
for all real $z$ and all positive integers $n$ with strict inequality unless $n = 1$ or $\sin z = 0$. Thus the lower bound is \boxed{k^2 - 1}. | algebraic | putnam (modified boxing) | Trigonometry Algebra | A and B are real numbers and k a positive integer. Show that
\[ \left| \frac{\cos kB \cos A - \cos kA \cos B}{\cos B - \cos A} \right| < k^2 - 1 \]
whenever the left side is defined. | Let $x = (A - B)/2$, $y = (A + B)/2$. The numerator of the expression on the left is
\[ \frac{1}{2}[\cos (kB + A) + \cos (kB - A) - \cos (kA + B) - \cos (kA - B)] \]
\[ = \frac{1}{2}[\cos (kB + A) - \cos (kA + B)] + \frac{1}{2}[\cos (kB - A) - \cos (kA - B)] \]
\[ = \sin (k - 1)x \sin (k + 1)y + \sin (k + 1)x \sin (k - 1)y. \]
The denominator is $2 \sin x \sin y$, and we may assume this is not zero. Hence we have
\[ \left| \frac{\cos kB \cos A - \cos kA \cos B}{\cos B - \cos A} \right| \leq \frac{1}{2} \left| \frac{\sin (k - 1)x}{\sin x} \cdot \frac{\sin (k + 1)y}{\sin y} \right|
+ \frac{1}{2} \left| \frac{\sin (k + 1)x}{\sin x} \cdot \frac{\sin (k - 1)y}{\sin y} \right|. \]
Now since $|\sin nz| \leq n |\sin z|$ for all $z$ and all positive integers $n$ with strict inequality unless $n = 1$ or $\sin z = 0$ (this is proved below), both terms are less than $(k^2 - 1)/2$ provided $k > 1$. Therefore
\[ \left| \frac{\cos kB \cos A - \cos kA \cos B}{\cos B - \cos A} \right| < k^2 - 1 \]
provided $k > 1$ and $\cos B \neq \cos A$. Obviously, we have equality if $k = 1$, so the problem is not accurately phrased.
We now prove that $|\sin nz| \leq n |\sin z|$ for all positive integers $n$. If $\sin z = 0$, equality holds for every integer $n$, so we assume from now on that $\sin z \neq 0$; then $|\cos z| < 1$. We continue by induction on $n$. Clearly there is equality if $n = 1$. For $n = 2$, we have $|\sin 2z| = |2 \cos z \cdot \sin z| < 2 |\sin z|$. Suppose we have strict inequality for $n = k$. Then
\[ |\sin (k + 1)z| = |\sin kz \cos z + \cos kz \sin z| \leq |\sin kz| + |\sin z| < (k + 1) |\sin z|. \]
Thus we have strict inequality for $n = k + 1$. Hence we have proved
\[ |\sin nz| \leq n |\sin z| \]
for all real $z$ and all positive integers $n$ with strict inequality unless $n = 1$ or $\sin z = 0$. | 0 |
1957 | 1957_5 | Let $S$ be a set of $n$ points in the plane such that the greatest distance between two points of $S$ is $1$. Find the maximum number of times this distance can occur between pairs of points of $S$. | Induction on $n$. Obviously true for $n \leq 3$. Suppose it is true for $n$. Take $n+1$ points. If no point is a distance 1 from more than 2 points, then we are done. So assume that $A, B,$ and $C$ are all a distance 1 from $P$. Without loss of generality, the largest of the three angles $\angle APB, \angle APC, \angle BPC$ is $\angle APB$. It must be at most $60^\circ$, since $AB \leq 1$. So the ray $PC$ lies between the rays $PA$ and $PB$.
Now suppose there is another point $D$ (apart from $P$) such that $CD = 1$. Then $CD$ must intersect $PA$, because otherwise one of $CP, CA, DP,$ and $DA$ would exceed 1. Similarly, it must intersect $PB$. But that is impossible. So there is no such point $D$. Hence, if we remove $C$, we lose only one realization of the distance 1. But the remaining $n$ points have at most $n$ realizations, so the result is established. There can be at most $\boxed{n}$ realizations of the distance 1. | algebraic | putnam (modified boxing) | Geometry Combinatorics | Given $n$ points in the plane, show that the largest distance determined by these points cannot occur more than $n$ times. | Suppose $\mathcal{S}$ is a set in the plane. By a diameter of $\mathcal{S}$ we mean a segment connecting two points of $\mathcal{S}$ that are as far apart as any two points of $\mathcal{S}$. We are asked to prove (1) a set of $n$ points in a plane can have at most $n$ diameters. We shall use induction on $n$. Clearly, (1) is true for $n = 2$ or $3$.
Assume that (1) is true for $n = k$. Let $\mathcal{S}$ be a set in a plane with $k + 1$ points.
Suppose some point of $\mathcal{S}$, say $X$, is the endpoint of at most one diameter of $\mathcal{S}$. Then $\mathcal{S} \setminus \{X\}$ is a set of $k$ points in the plane and has at most $k$ diameters by the inductive hypothesis. Then $\mathcal{S}$ has at most $k + 1$ diameters.
Suppose some point of $\mathcal{S}$, say $P$, is the endpoint of at least three diameters, $PQ$, $PR$, and $PS$. Let $r = |PQ|$. Then $Q$, $R$, and $S$ lie on a circle of radius $r$ about $P$ and, in fact, on a minor arc of that circle. We choose the notation so that $R$ is between $Q$ and $S$ on that arc. Since every two points of $\mathcal{S}$ are at most $r$ apart, $\mathcal{S}$ lies in the intersection $\mathcal{G}$ of three closed circular disks of radius $r$ and centers $P$, $Q$, and $S$. Now, except for the point $P$, $\mathcal{G}$ is inside the circle of radius $r$ about $R$, and therefore $R$ is the endpoint of just one diameter of $\mathcal{S}$, namely, $RP$. Therefore, the previous paragraph applies, and $\mathcal{S}$ has at most $k + 1$ diameters.
Finally, suppose each point of $\mathcal{S}$ is the endpoint of exactly two diameters. Then the diameters have altogether $2(k + 1)$ endpoints, so there are exactly $k + 1$ of them.
Thus, in any case, $\mathcal{S}$ has at most $k + 1$ diameters. Hence (1) is true for $n = k + 1$. It follows by induction that (1) is true for all $n$. | 0 |
1957 | 1957_6 | $S_1 = \ln a$ and $S_n = \sum_{i=1}^{n-1} \ln (a - S_i), n > 1$. Find the value of
\[
\lim_{n \to \infty} S_n.
\] | The given recursion can be written
\[ S_{n+1} = S_n + \ln (a - S_n). \]
The polygonal representation of this recursion is shown in the figure [see p. 223]. It is clear that, with any choice of $S_1 < a$, we have
\[
S_2 \leq S_3 \leq S_4 \leq \cdots \leq a - 1,
\]
and that the sequence converges to $a - 1$.
To prove this analytically, we let
\[ f(x) = x + \ln (a - x) \]
for $x < a$. Then $f'(x) = 1 - 1/(a - x)$, which is positive for $x < a - 1$ and negative for $x > a - 1$. Since $f(a - 1) = a - 1$, it follows that $f(x) \leq a - 1$ for all $x < a$. Also, if $x \leq a - 1$, then $\ln (a - x) \geq 0$, so $f(x) \geq x$. Then (1) follows immediately, so the sequence $\{S_n\}$ has a limit, say $T$. Clearly, $T \leq a - 1$, so $T$ is a point of continuity for $f$. Hence
\[ f(T) = f(\lim S_n) = \lim f(S_n) = \lim S_{n+1} = T. \]
This gives $\ln (a - T) = 0$, and therefore $T = a - 1$.
We have proved $\lim S_n = \boxed{a - 1}$, as required. | algebraic | putnam (modified boxing) | Analysis Sequences | $S_1 = \ln a$ and $S_n = \sum_{i=1}^{n-1} \ln (a - S_i), n > 1$. Show that
\[
\lim_{n \to \infty} S_n = a - 1.
\] | The given recursion can be written
\[ S_{n+1} = S_n + \ln (a - S_n). \]
The polygonal representation of this recursion is shown in the figure [see p. 223]. It is clear that, with any choice of $S_1 < a$, we have
\[
S_2 \leq S_3 \leq S_4 \leq \cdots \leq a - 1,
\]
and that the sequence converges to $a - 1$.
To prove this analytically, we let
\[ f(x) = x + \ln (a - x) \]
for $x < a$. Then $f'(x) = 1 - 1/(a - x)$, which is positive for $x < a - 1$ and negative for $x > a - 1$. Since $f(a - 1) = a - 1$, it follows that $f(x) \leq a - 1$ for all $x < a$. Also, if $x \leq a - 1$, then $\ln (a - x) \geq 0$, so $f(x) \geq x$. Then (1) follows immediately, so the sequence $\{S_n\}$ has a limit, say $T$. Clearly, $T \leq a - 1$, so $T$ is a point of continuity for $f$. Hence
\[ f(T) = f(\lim S_n) = \lim f(S_n) = \lim S_{n+1} = T. \]
This gives $\ln (a - T) = 0$, and therefore $T = a - 1$.
We have proved $\lim S_n = a - 1$, as required. | 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 |
1957 | 1957_9 | If facilities for division are not available, it is sometimes convenient in determining the decimal expansion of $1/A$, $A > 0$ to use the iteration $X_{k+1} = X_k(2 - AX_k)$, $k = 0, 1, 2, \dots$ where $X_0$ is a selected "starting" value. Find the sum of the upper and lower bounds, if any, on the starting value $X_0$ in order that the above iteration converges to the desired value $1/A$. | The polygonal representation of this recursion is shown in the figure. (See p. 223 for an explanation.) It is clear that if $X_0$ lies in $(0, 2/A)$ then
\[
0 < X_1 \leq X_2 \leq X_3 \leq \cdots \leq \frac{1}{A},
\]
and the sequence converges to $1/A$. If $X_0 = 0$ or $2/A$, then obviously
\[
X_1 = X_2 = X_3 = \cdots = 0,
\]
and if $X_0$ lies outside $(0, 2/A)$, then
\[
0 > X_1 > X_2 > X_3 > \cdots
\]
and the sequence diverges to $-\infty$.
To make this rigorous, we define $f(x) = x(2 - Ax)$. We find that $f$ achieves its maximum value $1/A$ for $x = 1/A$. Moreover, $f(x) > x$ for $0 < x < 1/A$ and $f(x) < x$ for $x < 0$. Then $(1)$, $(2)$, and $(3)$ follow immediately. In case $(1)$ the sequence must be convergent and the limit must be a positive root of $f(x) = x$, and there is only one, namely, $x = 1/A$. In case $(2)$ the sequence obviously converges to $0$. In case $(3)$ it either diverges or converges to a negative root of $f(x) = x$, but there is no such root.
Therefore, the sequence converges to $1/A$ if and only if $0 < X_0 < 2/A$. Thus the final answer is \boxed{2/A}. | algebraic | putnam (modified boxing) | Analysis Calculus | If facilities for division are not available, it is sometimes convenient in determining the decimal expansion of $1/A$, $A > 0$ to use the iteration $X_{k+1} = X_k(2 - AX_k)$, $k = 0, 1, 2, \dots$ where $X_0$ is a selected "starting" value. Find the limitations, if any, on the starting value $X_0$ in order that the above iteration converges to the desired value $1/A$. | The polygonal representation of this recursion is shown in the figure. (See p. 223 for an explanation.) It is clear that if $X_0$ lies in $(0, 2/A)$ then
\[
0 < X_1 \leq X_2 \leq X_3 \leq \cdots \leq \frac{1}{A},
\]
and the sequence converges to $1/A$. If $X_0 = 0$ or $2/A$, then obviously
\[
X_1 = X_2 = X_3 = \cdots = 0,
\]
and if $X_0$ lies outside $(0, 2/A)$, then
\[
0 > X_1 > X_2 > X_3 > \cdots
\]
and the sequence diverges to $-\infty$.
To make this rigorous, we define $f(x) = x(2 - Ax)$. We find that $f$ achieves its maximum value $1/A$ for $x = 1/A$. Moreover, $f(x) > x$ for $0 < x < 1/A$ and $f(x) < x$ for $x < 0$. Then $(1)$, $(2)$, and $(3)$ follow immediately. In case $(1)$ the sequence must be convergent and the limit must be a positive root of $f(x) = x$, and there is only one, namely, $x = 1/A$. In case $(2)$ the sequence obviously converges to $0$. In case $(3)$ it either diverges or converges to a negative root of $f(x) = x$, but there is no such root.
Therefore, the sequence converges to $1/A$ if and only if $0 < X_0 < 2/A$. | 0 |
1957 | 1957_11 | Let $a(n)$ be the number of representations of the positive integer $n$ as the sums of $1$'s and $2$'s taking order into account. For example, since
\[
4 = 1 + 1 + 2 = 1 + 2 + 1 = 2 + 1 + 1 \\
= 2 + 2 = 1 + 1 + 1 + 1,
\]
then $a(4) = 5$. Let $b(n)$ be the number of representations of $n$ as the sum of integers greater than $1$, again taking order into account and counting the summand $n$. For example, since $6 = 4 + 2 = 2 + 4 = 3 + 3 = 2 + 2 + 2$, we have $b(6) = 5$. Find an expression to relate functions $a$ and $b$ for each $n$. | By direct counting we can find the first few values of the two sequences:
\[
\begin{array}{c|c|c}
n & a(n) & b(n) \\\ \hline
1 & 1 & 0 \\
2 & 2 & 1 \\
3 & 3 & 1 \\
4 & 5 & 2 \\
5 & 8 & 3 \\
6 & 13 & 5 \\
\end{array}
\]
Thus it appears that the sequences are the well-known Fibonacci sequence satisfying the recursion
\[
a(n + 1) = a(n) + a(n - 1), \quad n = 2, 3, \dots.
\]
To prove that this is indeed the case, note that any representation of $n + 1$ as an ordered sum of $1$'s and $2$'s becomes, on removal of the last summand, either a representation of $n$ (if the removed term is a $1$) or a representation of $n - 1$ (if the removed term is a $2$). Conversely, any representation of either $n$ or $n - 1$ becomes a representation of $n + 1$ on adjoining either a $1$ or $2$, as appropriate. It follows that the $a$-sequence does indeed satisfy the recursion $(1)$.
Next we show that the $b$'s satisfy the same recursion. Delete the last term from a representation of $n + 1$ as a sum of integers greater than $1$. We obtain then either a representation of $n - 1$, $n - 2$, $\dots$, $2$, or a vacuous sum. Conversely, any such representation extends to a representation of $n + 1$. Hence for $n \geq 1$
\[
b(n + 1) = b(n - 1) + b(n - 2) + \cdots + b(2) + 1.
\]
If $n$ is at least $2$, we have also
\[
b(n) = b(n - 2) + \cdots + b(2) + 1.
\]
Comparing $(2)$ and $(3)$, we find
\[
b(n + 1) = b(n - 1) + b(n), \quad n \geq 2.
\]
So the $b$-sequence satisfies the same recursion as the $a$-sequence. Then since $b(3) = a(1)$, $b(4) = a(2)$, it follows by induction that $\boxed{b(n + 2) = a(n)}$ for all $n \geq 1$. | algebraic | putnam (modified boxing) | Combinatorics | Let $a(n)$ be the number of representations of the positive integer $n$ as the sums of $1$'s and $2$'s taking order into account. For example, since
\[
4 = 1 + 1 + 2 = 1 + 2 + 1 = 2 + 1 + 1 \\
= 2 + 2 = 1 + 1 + 1 + 1,
\]
then $a(4) = 5$. Let $b(n)$ be the number of representations of $n$ as the sum of integers greater than $1$, again taking order into account and counting the summand $n$. For example, since $6 = 4 + 2 = 2 + 4 = 3 + 3 = 2 + 2 + 2$, we have $b(6) = 5$. Show that for each $n$, $a(n) = b(n+2)$. | By direct counting we can find the first few values of the two sequences:
\[
\begin{array}{c|c|c}
n & a(n) & b(n) \\\ \hline
1 & 1 & 0 \\
2 & 2 & 1 \\
3 & 3 & 1 \\
4 & 5 & 2 \\
5 & 8 & 3 \\
6 & 13 & 5 \\
\end{array}
\]
Thus it appears that the sequences are the well-known Fibonacci sequence satisfying the recursion
\[
a(n + 1) = a(n) + a(n - 1), \quad n = 2, 3, \dots.
\]
To prove that this is indeed the case, note that any representation of $n + 1$ as an ordered sum of $1$'s and $2$'s becomes, on removal of the last summand, either a representation of $n$ (if the removed term is a $1$) or a representation of $n - 1$ (if the removed term is a $2$). Conversely, any representation of either $n$ or $n - 1$ becomes a representation of $n + 1$ on adjoining either a $1$ or $2$, as appropriate. It follows that the $a$-sequence does indeed satisfy the recursion $(1)$.
Next we show that the $b$'s satisfy the same recursion. Delete the last term from a representation of $n + 1$ as a sum of integers greater than $1$. We obtain then either a representation of $n - 1$, $n - 2$, $\dots$, $2$, or a vacuous sum. Conversely, any such representation extends to a representation of $n + 1$. Hence for $n \geq 1$
\[
b(n + 1) = b(n - 1) + b(n - 2) + \cdots + b(2) + 1.
\]
If $n$ is at least $2$, we have also
\[
b(n) = b(n - 2) + \cdots + b(2) + 1.
\]
Comparing $(2)$ and $(3)$, we find
\[
b(n + 1) = b(n - 1) + b(n), \quad n \geq 2.
\]
So the $b$-sequence satisfies the same recursion as the $a$-sequence. Then since $b(3) = a(1)$, $b(4) = a(2)$, it follows by induction that $b(n + 2) = a(n)$ for all $n \geq 1$. | 0 |
1958 | 1958_1 | If $a_0, a_1, \dots, a_n$ are real numbers satisfying
\[
\frac{a_0}{1} + \frac{a_1}{2} + \dots + \frac{a_n}{n+1} = 0,
\]
find the least number of real roots $a_0 + a_1x + a_2x^2 + \dots + a_nx^n = 0$ has. | If $f(x) = a_0 + a_1x + \dots + a_nx^n$, then
\[
\int_0^1 f(x) \, dx = \frac{a_0}{1} + \frac{a_1}{2} + \dots + \frac{a_n}{n+1} = 0.
\]
Hence, by the mean value theorem for integrals, there exists a number $\xi$ between $0$ and $1$ such that
\[
f(\xi) = \int_0^1 f(x) \, dx = 0.
\] Thus it must have atleast \boxed{1} real root. | numerical | putnam (modified boxing) | Analysis Algebra | If $a_0, a_1, \dots, a_n$ are real numbers satisfying
\[
\frac{a_0}{1} + \frac{a_1}{2} + \dots + \frac{a_n}{n+1} = 0,
\]
show that the equation $a_0 + a_1x + a_2x^2 + \dots + a_nx^n = 0$ has at least one real root. | If $f(x) = a_0 + a_1x + \dots + a_nx^n$, then
\[
\int_0^1 f(x) \, dx = \frac{a_0}{1} + \frac{a_1}{2} + \dots + \frac{a_n}{n+1} = 0.
\]
Hence, by the mean value theorem for integrals, there exists a number $\xi$ between $0$ and $1$ such that
\[
f(\xi) = \int_0^1 f(x) \, dx = 0.
\] | 0 |
1958 | 1958_2 | Two uniform solid spheres of equal radii are so placed that one is directly above the other. The bottom sphere is fixed, and the top sphere, initially at rest, rolls off. At what point will contact between the two spheres be "lost"? Assume the coefficient of friction is such that no slipping occurs. | Let $S$ be the rolling sphere; let $a$ be its radius and $M$ its mass. Recall that the moment of inertia of a uniform solid sphere is $I = \frac{2}{5}Ma^2$.
We assume that the motion is essentially two-dimensional. Then the state of the system is defined by the angle $\theta$ between the line of centers and the vertical as shown in the figure. We denote the time derivative of $\theta$ by $\dot{\theta}$. Let $v$ be the translational speed of the center of $S$ and let $\omega$ be the speed of rotation of $S$.
As long as contact is maintained between the two spheres, the center of $S$ moves along a circle of radius $2a$, so $v = 2a\dot{\theta}$ and $\omega = 2\dot{\theta}$. The kinetic energy of $S$ is given by
\[
\frac{1}{2} Mv^2 + \frac{1}{2} I \omega^2 = \frac{14}{5} M a^2 \dot{\theta}^2
\]
and the potential energy by
\[
2M a g \cos \theta
\]
relative to the level of the center of the lower sphere. The total energy is constant, so
\[
2M a g \cos \theta + \frac{14}{5} M a^2 \dot{\theta}^2 = 2M a g.
\]
(The right member is the left evaluated for $\theta = 0$.) Therefore we have
\[
\frac{7}{5} a \dot{\theta}^2 = g(1 - \cos \theta).
\]
To keep $S$ in a circular orbit of radius $2a$, a force toward the center of magnitude $2aM\dot{\theta}^2$ is required. This force is supplied by the component of the gravitational force along the line of centers, which is $Mg \cos \theta$. As long as this component exceeds the required centripetal force, the excess is opposed by the force of contact between the two spheres. When the component of gravity becomes inadequate to supply the necessary force, contact is lost and $S$ goes into a parabolic path while rotating at constant speed. The loss of contact occurs when
\[
2aM\dot{\theta}^2 = Mg \cos \theta.
\]
Combining this with $(1)$, we obtain
\[
\frac{7}{10} \cos \theta = 1 - \cos \theta;
\]
Hence contact is lost when $\cos \theta = 10/17$, i.e., when $\theta = \boxed{\arccos (10/17)}$. | numerical | putnam | Calculus | Two uniform solid spheres of equal radii are so placed that one is directly above the other. The bottom sphere is fixed, and the top sphere, initially at rest, rolls off. At what point will contact between the two spheres be "lost"? Assume the coefficient of friction is such that no slipping occurs. | Let $S$ be the rolling sphere; let $a$ be its radius and $M$ its mass. Recall that the moment of inertia of a uniform solid sphere is $I = \frac{2}{5}Ma^2$.
We assume that the motion is essentially two-dimensional. Then the state of the system is defined by the angle $\theta$ between the line of centers and the vertical as shown in the figure. We denote the time derivative of $\theta$ by $\dot{\theta}$. Let $v$ be the translational speed of the center of $S$ and let $\omega$ be the speed of rotation of $S$.
As long as contact is maintained between the two spheres, the center of $S$ moves along a circle of radius $2a$, so $v = 2a\dot{\theta}$ and $\omega = 2\dot{\theta}$. The kinetic energy of $S$ is given by
\[
\frac{1}{2} Mv^2 + \frac{1}{2} I \omega^2 = \frac{14}{5} M a^2 \dot{\theta}^2
\]
and the potential energy by
\[
2M a g \cos \theta
\]
relative to the level of the center of the lower sphere. The total energy is constant, so
\[
2M a g \cos \theta + \frac{14}{5} M a^2 \dot{\theta}^2 = 2M a g.
\]
(The right member is the left evaluated for $\theta = 0$.) Therefore we have
\[
\frac{7}{5} a \dot{\theta}^2 = g(1 - \cos \theta).
\]
To keep $S$ in a circular orbit of radius $2a$, a force toward the center of magnitude $2aM\dot{\theta}^2$ is required. This force is supplied by the component of the gravitational force along the line of centers, which is $Mg \cos \theta$. As long as this component exceeds the required centripetal force, the excess is opposed by the force of contact between the two spheres. When the component of gravity becomes inadequate to supply the necessary force, contact is lost and $S$ goes into a parabolic path while rotating at constant speed. The loss of contact occurs when
\[
2aM\dot{\theta}^2 = Mg \cos \theta.
\]
Combining this with $(1)$, we obtain
\[
\frac{7}{10} \cos \theta = 1 - \cos \theta;
\]
Hence contact is lost when $\cos \theta = 10/17$, i.e., when $\theta = \boxed{\arccos (10/17)}$. | 0 |
1958 | 1958_3 | Real numbers are chosen at random from the interval $[0,1]$. If after choosing the nth number the sum of the numbers exceeds 1, find the expected value of $n$. | We assume that the phrase "real numbers are chosen at random" means that the x's are independent and each has the uniform distribution on [0, 1]. Then the probability that \((x_1, x_2, \dots, x_n)\) falls in a region $S$ of the cube [0, 1]^n is the $n$-dimensional content of $S$.
Let $p_n$ be the probability that $x_1 + x_2 + \cdots + x_n \leq 1$. The probability that $x_1 + x_2 + \cdots + x_n > 1$ but $x_1 + x_2 + \cdots + x_{n-1} \leq 1$ is then
\[
q_n = p_{n-1} - p_n.
\]
It is proved below that $p_n = 1/n!$ Hence the expected number of choices required to make the sum exceed one is
\[
E = \sum_{n=1}^\infty n q_n = \sum_{n=1}^\infty n \left( \frac{1}{(n-1)!} - \frac{1}{n!} \right) = \sum_{n=1}^\infty \frac{1}{(n-1)!} - \sum_{n=1}^\infty \frac{1}{n!}.
\]
Shifting indices in the second term, we have
\[
E = \sum_{n=1}^\infty \frac{1}{(n-1)!} - \sum_{n=2}^\infty \frac{1}{(n-1)!} = \frac{1}{0!} + \sum_{n=2}^\infty \frac{1}{(n-1)!} - \sum_{n=2}^\infty \frac{1}{(n-1)!} = 1 = e.
\]
**Lemma**. The $n$-dimensional content $V_n(a)$ of the region in $\mathbb{R}^n$ determined by the inequalities $x_i \geq 0$, $i = 1, 2, \dots, n$ and $x_1 + x_2 + \cdots + x_n \leq a$ is $a^n / n!$.
**Proof**. This is evidently true for $n = 1$. Suppose it is true for $n = k$. Then $V_{k+1}(a)$ can be formed by "slicing" perpendicular to the $x_{k+1}$-axis. The slice for $x_{k+1} = b$ is the $k$-dimensional region determined by the inequalities $x_i \geq 0$, $i = 1, 2, \dots, k$ and $x_1 + x_2 + \cdots + x_k \leq a - b$ for $0 \leq b \leq a$. By the inductive hypothesis its $k$-dimensional content is $V_k(a-b) = \frac{(a-b)^k}{k!}$. Hence
\[
V_{k+1}(a) = \int_0^a V_k(a - x_{k+1})dx_{k+1} = \int_0^a \frac{(a - x_{k+1})^k}{k!} dx_{k+1} = \frac{a^{k+1}}{(k+1)!}.
\]
Thus the formula $V_n(a) = \frac{a^n}{n!}$ is established for all positive integers $n$. Evidently $p_n = V_n(1) = \frac{1}{n!}$. Thus the final answer is \boxed{e}. | numerical | putnam | Probability | Real numbers are chosen at random from the interval (0 ≤ x ≤ 1). If after choosing the nth number the sum of the numbers so chosen first exceeds 1, show that the expected or average value for n is e. | We assume that the phrase "real numbers are chosen at random" means that the x's are independent and each has the uniform distribution on [0, 1]. Then the probability that \((x_1, x_2, \dots, x_n)\) falls in a region $S$ of the cube [0, 1]^n is the $n$-dimensional content of $S$.
Let $p_n$ be the probability that $x_1 + x_2 + \cdots + x_n \leq 1$. The probability that $x_1 + x_2 + \cdots + x_n > 1$ but $x_1 + x_2 + \cdots + x_{n-1} \leq 1$ is then
\[
q_n = p_{n-1} - p_n.
\]
It is proved below that $p_n = 1/n!$ Hence the expected number of choices required to make the sum exceed one is
\[
E = \sum_{n=1}^\infty n q_n = \sum_{n=1}^\infty n \left( \frac{1}{(n-1)!} - \frac{1}{n!} \right) = \sum_{n=1}^\infty \frac{1}{(n-1)!} - \sum_{n=1}^\infty \frac{1}{n!}.
\]
Shifting indices in the second term, we have
\[
E = \sum_{n=1}^\infty \frac{1}{(n-1)!} - \sum_{n=2}^\infty \frac{1}{(n-1)!} = \frac{1}{0!} + \sum_{n=2}^\infty \frac{1}{(n-1)!} - \sum_{n=2}^\infty \frac{1}{(n-1)!} = 1 = \boxed{e}.
\]
**Lemma**. The $n$-dimensional content $V_n(a)$ of the region in $\mathbb{R}^n$ determined by the inequalities $x_i \geq 0$, $i = 1, 2, \dots, n$ and $x_1 + x_2 + \cdots + x_n \leq a$ is $a^n / n!$.
**Proof**. This is evidently true for $n = 1$. Suppose it is true for $n = k$. Then $V_{k+1}(a)$ can be formed by "slicing" perpendicular to the $x_{k+1}$-axis. The slice for $x_{k+1} = b$ is the $k$-dimensional region determined by the inequalities $x_i \geq 0$, $i = 1, 2, \dots, k$ and $x_1 + x_2 + \cdots + x_k \leq a - b$ for $0 \leq b \leq a$. By the inductive hypothesis its $k$-dimensional content is $V_k(a-b) = \frac{(a-b)^k}{k!}$. Hence
\[
V_{k+1}(a) = \int_0^a V_k(a - x_{k+1})dx_{k+1} = \int_0^a \frac{(a - x_{k+1})^k}{k!} dx_{k+1} = \frac{a^{k+1}}{(k+1)!}.
\]
Thus the formula $V_n(a) = \frac{a^n}{n!}$ is established for all positive integers $n$. Evidently $p_n = V_n(1) = \frac{1}{n!}.$ | 0 |
1958 | 1958_4 | If $a_1, a_2, \dots, a_n$ are complex numbers such that
\[
|a_1| = |a_2| = \cdots = |a_n| = r \neq 0,
\]
and if $T_s$ denotes the sum of all products of these $n$ numbers taken $s$ at a time, find an expression for
\[
\left| \frac{n T_s}{n T_{n-s}} \right|,
\]
in terms of $r$ whenever the denominator of the left-hand side is different from zero. | For any non-zero complex number $z$, we have $z \cdot \bar{z} = |z|^2$ where the bar denotes complex conjugation. If $J$ is a set of $s$ indices selected from $\{1, 2, \dots, n\}$, then
\[
\prod_{i \in J} a_i = a_1 a_2 \cdots a_n \prod_{i \in J} a_i^{-1} = a_1 a_2 \cdots a_n \prod_{i \in J} \frac{\bar{a}_i}{r^2} = \frac{a_1 a_2 \cdots a_n \prod_{i \in J} \bar{a}_i}{r^{2s}}.
\]
Therefore,
\[
n T_{n-s} = \sum_J \prod_{i \in J} a_i = \frac{a_1 a_2 \cdots a_n}{r^{2s}} \sum_J \prod_{i \in J} \bar{a}_i = \frac{a_1 a_2 \cdots a_n}{r^{2s}} \cdot n T_s.
\]
Hence,
\[
|n T_{n-s}| = \frac{r^n}{r^{2s}} |n T_s| = r^{n - 2s} |n T_s|.
\]
Thus,
\[
\left| \frac{n T_s}{n T_{n-s}} \right| = \boxed{r^{2s - n}},
\]
which proves the required formula. | algebraic | putnam (modified boxing) | Complex Numbers Algebra | If $a_1, a_2, \dots, a_n$ are complex numbers such that
\[
|a_1| = |a_2| = \cdots = |a_n| = r \neq 0,
\]
and if $T_s$ denotes the sum of all products of these $n$ numbers taken $s$ at a time, prove that
\[
\left| \frac{n T_s}{n T_{n-s}} \right| = r^{2s - n},
\]
whenever the denominator of the left-hand side is different from zero. | For any non-zero complex number $z$, we have $z \cdot \bar{z} = |z|^2$ where the bar denotes complex conjugation. If $J$ is a set of $s$ indices selected from $\{1, 2, \dots, n\}$, then
\[
\prod_{i \in J} a_i = a_1 a_2 \cdots a_n \prod_{i \in J} a_i^{-1} = a_1 a_2 \cdots a_n \prod_{i \in J} \frac{\bar{a}_i}{r^2} = \frac{a_1 a_2 \cdots a_n \prod_{i \in J} \bar{a}_i}{r^{2s}}.
\]
Therefore,
\[
n T_{n-s} = \sum_J \prod_{i \in J} a_i = \frac{a_1 a_2 \cdots a_n}{r^{2s}} \sum_J \prod_{i \in J} \bar{a}_i = \frac{a_1 a_2 \cdots a_n}{r^{2s}} \cdot n T_s.
\]
Hence,
\[
|n T_{n-s}| = \frac{r^n}{r^{2s}} |n T_s| = r^{n - 2s} |n T_s|.
\]
Thus,
\[
\left| \frac{n T_s}{n T_{n-s}} \right| = r^{2s - n},
\]
which proves the required formula. | 0 |
1958 | 1958_5 | Given the integral equation \[ f(x, y) = 1 + \int_0^x \int_0^y f(u, v)\, du\, dv \] find the maximum number of solutions continuous for $0 \leq x \leq 1$, $0 \leq y \leq 1$. | Suppose there are two continuous solutions and let $g$ be their difference. Then $g$ is continuous and \[ g(x, y) = \int_0^x \int_0^y g(u, v)\, du\, dv. \] Since $g$ is continuous it is bounded on the given square. Let $M$ be a bound. Then \[ |g(x, y)| \leq \int_0^x \int_0^y |g(u, v)|\, du\, dv \leq \int_0^x \int_0^y M\, du\, dv = Mxy \] for $0 \leq x \leq 1$, $0 \leq y \leq 1$. We now prove that \[ (1) \quad |g(x, y)| \leq M \frac{x^n}{n!} \frac{y^n}{n!} \] for any positive integer $n$. This has been proved for $n = 1$. Assume that it is true for $n = k$; then \[ |g(x, y)| \leq \int_0^x \int_0^y |g(u, v)|\, du\, dv \leq \int_0^x \int_0^y M \frac{u^k}{k!} \frac{v^k}{k!}\, du\, dv = M \frac{x^{k+1}}{(k+1)!} \frac{y^{k+1}}{(k+1)!}. \] Thus $(1)$ is established by mathematical induction. But for any fixed $x$ and $y$, \[ \lim_{n \to \infty} M \frac{x^n}{n!} \frac{y^n}{n!} = 0. \] Hence $|g(x, y)| \leq 0$, that is $g(x, y) = 0$. Thus there cannot be two different continuous solutions making the answer \boxed{1}. | algebraic | putnam (modified boxing) | Analysis | Show that the integral equation \[ f(x, y) = 1 + \int_0^x \int_0^y f(u, v)\, du\, dv \] has at most one solution continuous for $0 \leq x \leq 1$, $0 \leq y \leq 1$. | Suppose there are two continuous solutions and let $g$ be their difference. Then $g$ is continuous and \[ g(x, y) = \int_0^x \int_0^y g(u, v)\, du\, dv. \] Since $g$ is continuous it is bounded on the given square. Let $M$ be a bound. Then \[ |g(x, y)| \leq \int_0^x \int_0^y |g(u, v)|\, du\, dv \leq \int_0^x \int_0^y M\, du\, dv = Mxy \] for $0 \leq x \leq 1$, $0 \leq y \leq 1$. We now prove that \[ (1) \quad |g(x, y)| \leq M \frac{x^n}{n!} \frac{y^n}{n!} \] for any positive integer $n$. This has been proved for $n = 1$. Assume that it is true for $n = k$; then \[ |g(x, y)| \leq \int_0^x \int_0^y |g(u, v)|\, du\, dv \leq \int_0^x \int_0^y M \frac{u^k}{k!} \frac{v^k}{k!}\, du\, dv = M \frac{x^{k+1}}{(k+1)!} \frac{y^{k+1}}{(k+1)!}. \] Thus $(1)$ is established by mathematical induction. But for any fixed $x$ and $y$, \[ \lim_{n \to \infty} M \frac{x^n}{n!} \frac{y^n}{n!} = 0. \] Hence $|g(x, y)| \leq 0$, that is $g(x, y) = 0$. Thus there cannot be two different continuous solutions. | 0 |
1958 | 1958_6 | What is the smallest amount that may be invested at interest rate $i$, compounded annually, in order that one may withdraw 1 dollar at the end of the first year, 4 dollars at the end of the second year, $\dots$, $n^2$ dollars at the end of the $n$th year, in perpetuity? | The present value of one dollar to be paid after $n$ years is $(1 + i)^{-n}$ dollars. Hence the value in dollars of the given annuity is \[ \sum_{n=1}^\infty n^2 (1 + i)^{-n}. \] Since \[ \frac{1}{1 - x} = \sum_{n=0}^\infty x^n, \] we have \[ \frac{x}{(1-x)^2} = x \frac{d}{dx} \left( \frac{1}{1-x} \right) = \sum_{n=1}^\infty n x^n \] and \[ \frac{x + x^2}{(1-x)^3} = x \frac{d}{dx} \left( \frac{x}{(1-x)^2} \right) = \sum_{n=1}^\infty n^2 x^n \] for $|x| < 1$. Putting $x = 1/(1+i)$, we obtain \[ \sum_{n=1}^\infty n^2 (1 + i)^{-n} = \boxed{\frac{(1+i)(2+i)}{i^3}}. \] (At 6 percent interest, the cost of the annuity would be $10,109.26$.) | numerical | putnam | Analysis Algebra | What is the smallest amount that may be invested at interest rate $i$, compounded annually, in order that one may withdraw 1 dollar at the end of the first year, 4 dollars at the end of the second year, $\dots$, $n^2$ dollars at the end of the $n$th year, in perpetuity? | The present value of one dollar to be paid after $n$ years is $(1 + i)^{-n}$ dollars. Hence the value in dollars of the given annuity is \[ \sum_{n=1}^\infty n^2 (1 + i)^{-n}. \] Since \[ \frac{1}{1 - x} = \sum_{n=0}^\infty x^n, \] we have \[ \frac{x}{(1-x)^2} = x \frac{d}{dx} \left( \frac{1}{1-x} \right) = \sum_{n=1}^\infty n x^n \] and \[ \frac{x + x^2}{(1-x)^3} = x \frac{d}{dx} \left( \frac{x}{(1-x)^2} \right) = \sum_{n=1}^\infty n^2 x^n \] for $|x| < 1$. Putting $x = 1/(1+i)$, we obtain \[ \sum_{n=1}^\infty n^2 (1 + i)^{-n} = \boxed{\frac{(1+i)(2+i)}{i^3}}. \] (At 6 percent interest, the cost of the annuity would be $10,109.26$.) | 0 |
1958 | 1958_8_ii | Given any acute-angled triangle $ABC$ and one altitude $AH$, select any point $D$ on $AH$, then draw $BD$ and extend until it intersects $AC$ in $E$, and draw $CD$ and extend until it intersects $AB$ in $F$. Find the difference in value between \angle $AHE and \angle AH$. | Draw $l$ through $A$ parallel to $BC$. Since angles $ABC$ and $ACB$ are acute, the foot $H$ of the altitude falls between $B$ and $C$. Assuming $D \neq A, H$, which are trivial cases, complete the diagram as shown.
Considering several pairs of similar triangles, we see that \[ \frac{|AX|}{|BH|} = \frac{|AF|}{|BF|} = \frac{|AW|}{|BC|}, \quad \frac{|AY|}{|CH|} = \frac{|AE|}{|CE|} = \frac{|AZ|}{|CB|}, \] and \[ \frac{|AW|}{|HC|} = \frac{|AD|}{|HD|} = \frac{|AZ|}{|HB|}. \]
Therefore, \[ |AX| \cdot |BC| = |AW| \cdot |BH| = |AZ| \cdot |HC| = |AY| \cdot |BC|, \] whence $|AX| = |AY|$. So right triangles $AHX$ and $AHY$ are congruent and $\angle AHX = \angle AHY$, as required making the difference \boxed{0}. | numerical | putnam (modified boxing) | Geometry Trigonometry | Given any acute-angled triangle $ABC$ and one altitude $AH$, select any point $D$ on $AH$, then draw $BD$ and extend until it intersects $AC$ in $E$, and draw $CD$ and extend until it intersects $AB$ in $F$. Prove angle $AHE = \angle AH$. | Draw $l$ through $A$ parallel to $BC$. Since angles $ABC$ and $ACB$ are acute, the foot $H$ of the altitude falls between $B$ and $C$. Assuming $D \neq A, H$, which are trivial cases, complete the diagram as shown.
Considering several pairs of similar triangles, we see that \[ \frac{|AX|}{|BH|} = \frac{|AF|}{|BF|} = \frac{|AW|}{|BC|}, \quad \frac{|AY|}{|CH|} = \frac{|AE|}{|CE|} = \frac{|AZ|}{|CB|}, \] and \[ \frac{|AW|}{|HC|} = \frac{|AD|}{|HD|} = \frac{|AZ|}{|HB|}. \]
Therefore, \[ |AX| \cdot |BC| = |AW| \cdot |BH| = |AZ| \cdot |HC| = |AY| \cdot |BC|, \] whence $|AX| = |AY|$. So right triangles $AHX$ and $AHY$ are congruent and $\angle AHX = \angle AHY$, as required. | 0 |
1958 | 1958_11 | What is the average straight line distance between two points on a sphere of radius $1$? | We take the radius of the sphere to be $a$ to help keep our dimensions in order. By symmetry we need only compute the average length of all chords emanating from some fixed point which we take to be the north pole $N$ of a spherical coordinate system. Slice the sphere into zones by parallels of co-latitude. The zone between co-latitude $\theta$ and co-latitude $\theta + \Delta \theta$ has area approximately \[ (2\pi a \sin \theta)(a \Delta \theta), \] and all the chords from $N$ to points in this zone have length approximately \[ 2a \sin (\theta/2). \]
Therefore, the average length is \[ L = \frac{4\pi a^3}{S} \int_0^{\pi} \sin \frac{\theta}{2} \sin \theta \, d\theta, \] where $S = 4\pi a^2$ is the surface area of the sphere. Since $a = 1$, we have \[ L = \int_0^{\pi} \sin \frac{\theta}{2} \left( 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \right) \, d\theta = \frac{4}{3} \sin^3 \frac{\theta}{2} \bigg|_0^{\pi} = \boxed{\frac{4}{3}}. \] | numerical | putnam | Geometry Calculus | What is the average straight line distance between two points on a sphere of radius $1$? | We take the radius of the sphere to be $a$ to help keep our dimensions in order. By symmetry we need only compute the average length of all chords emanating from some fixed point which we take to be the north pole $N$ of a spherical coordinate system. Slice the sphere into zones by parallels of co-latitude. The zone between co-latitude $\theta$ and co-latitude $\theta + \Delta \theta$ has area approximately \[ (2\pi a \sin \theta)(a \Delta \theta), \] and all the chords from $N$ to points in this zone have length approximately \[ 2a \sin (\theta/2). \]
Therefore, the average length is \[ L = \frac{4\pi a^3}{S} \int_0^{\pi} \sin \frac{\theta}{2} \sin \theta \, d\theta, \] where $S = 4\pi a^2$ is the surface area of the sphere. Since $a = 1$, we have \[ L = \int_0^{\pi} \sin \frac{\theta}{2} \left( 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \right) \, d\theta = \frac{4}{3} \sin^3 \frac{\theta}{2} \bigg|_0^{\pi} = \boxed{\frac{4}{3}}. \] | 0 |
1958 | 1958_1_Sp | Let $f(m, 1) = f(1, n) = 1$ for $m \geq 1$, $n \geq 1$, and let \[ f(m, n) = f(m-1, n) + f(m-1, n-1) + f(m, n-1) \] for $m > 1$ and $n > 1$. Also let \[ S(n) = \sum_{a+b=n} f(a, b), \quad a \geq 1 \text{ and } b \geq 1. \] Find an expression for S(n+2) in terms of S \text{for } n \geq 2. \] | If we write the value of $f(m, n)$ at the point $\langle m, n \rangle$ in the plane and border the resulting array with zeros as in the diagram, \[
\begin{array}{ccccccc}
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & f(m-1, n) & f(m, n) & \cdots & \cdots & 0 \\
0 & 1 & 3 & f(m-1, n-1) & \cdots & \cdots & 0 \\
0 & 1 & 5 & 13 & 25 & \cdots & 0 \\
0 & 1 & 7 & \cdots & \cdots & \cdots & 0 \\
\end{array} \]
we see that the recursion relation together with the given values for $f(1, n)$ and $f(m, 1)$ amount to the assertion that every non-zero entry in this array (except $f(1, 1)$) is the sum of the entry immediately to its left, the entry just below it, and the entry diagonally below it to the left.
Now $S(n+2)$ is the sum of the terms on the $(n+2)$nd diagonal, $x+y = n+2$, and it is clear from the diagram that each non-zero term on the $(n+1)$st diagonal enters this sum twice while each term on the $n$th diagonal enters once; hence, $S(n+2) = 2S(n+1) + S(n)$.
This argument can be carried out formally as follows: \[ S(n+2) = \sum_{j=1}^{n+1} f(n+2-j, j). \] Expanding using the recursive definition, \[ S(n+2) = f(n+1, 1) + \sum_{j=2}^n \left( f(n+1-j, j) + f(n+2-j, j-1) \right). \] Rearranging terms, \[ S(n+2) = \boxed{2S(n+1) + S(n)}. \] | algebraic | putnam (modified boxing) | Algebra Combinatorics | Let $f(m, 1) = f(1, n) = 1$ for $m \geq 1$, $n \geq 1$, and let \[ f(m, n) = f(m-1, n) + f(m-1, n-1) + f(m, n-1) \] for $m > 1$ and $n > 1$. Also let \[ S(n) = \sum_{a+b=n} f(a, b), \quad a \geq 1 \text{ and } b \geq 1. \] Prove that \[ S(n+2) = S(n) + 2S(n+1) \quad \text{for } n \geq 2. \] | If we write the value of $f(m, n)$ at the point $\langle m, n \rangle$ in the plane and border the resulting array with zeros as in the diagram, \[
\begin{array}{ccccccc}
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & f(m-1, n) & f(m, n) & \cdots & \cdots & 0 \\
0 & 1 & 3 & f(m-1, n-1) & \cdots & \cdots & 0 \\
0 & 1 & 5 & 13 & 25 & \cdots & 0 \\
0 & 1 & 7 & \cdots & \cdots & \cdots & 0 \\
\end{array} \]
we see that the recursion relation together with the given values for $f(1, n)$ and $f(m, 1)$ amount to the assertion that every non-zero entry in this array (except $f(1, 1)$) is the sum of the entry immediately to its left, the entry just below it, and the entry diagonally below it to the left.
Now $S(n+2)$ is the sum of the terms on the $(n+2)$nd diagonal, $x+y = n+2$, and it is clear from the diagram that each non-zero term on the $(n+1)$st diagonal enters this sum twice while each term on the $n$th diagonal enters once; hence, $S(n+2) = 2S(n+1) + S(n)$.
This argument can be carried out formally as follows: \[ S(n+2) = \sum_{j=1}^{n+1} f(n+2-j, j). \] Expanding using the recursive definition, \[ S(n+2) = f(n+1, 1) + \sum_{j=2}^n \left( f(n+1-j, j) + f(n+2-j, j-1) \right). \] Rearranging terms, \[ S(n+2) = S(n+1) + S(n+1) + S(n). \] | 0 |
1958 | 1958_2_Sp | Let \[ R_1 = 1, \quad R_{n+1} = 1 + \frac{n}{R_n}, \quad n \geq 1. \] Given $n \geq 1$, find the sum of the lower and upper bound of R_n. \] | We shall use induction on $n$. Evidently $\sqrt{1} = 1 = R_1 \leq \sqrt{1} + 1$, so the given formula is true for $n = 1$. Suppose we know that it is true for $n = k$, i.e., for $k$ a positive integer, \[ \sqrt{k} \leq R_k \leq \sqrt{k} + 1. \] Then \[ \sqrt{k+1} - 1 = \frac{k}{\sqrt{k+1} + 1} < \frac{k}{\sqrt{k+1}} \leq \frac{k}{R_k} \leq \frac{k}{\sqrt{k}} = \sqrt{k} < \sqrt{k+1}. \] Hence \[ \sqrt{k+1} < 1 + \frac{k}{R_k} = R_{k+1} < \sqrt{k+1} + 1, \] and the induction is complete.
Note that the proof shows that the inequalities are strict for all integers $n > 1$. Thus the final answer is \boxed{2\sqrt{n}+1}. | algebraic | putnam (modified boxing) | Analysis Inequalities | Let \[ R_1 = 1, \quad R_{n+1} = 1 + \frac{n}{R_n}, \quad n \geq 1. \] Show that for $n \geq 1$, \[ \sqrt{n} \leq R_n \leq \sqrt{n} + 1. \] | We shall use induction on $n$. Evidently $\sqrt{1} = 1 = R_1 \leq \sqrt{1} + 1$, so the given formula is true for $n = 1$. Suppose we know that it is true for $n = k$, i.e., for $k$ a positive integer, \[ \sqrt{k} \leq R_k \leq \sqrt{k} + 1. \] Then \[ \sqrt{k+1} - 1 = \frac{k}{\sqrt{k+1} + 1} < \frac{k}{\sqrt{k+1}} \leq \frac{k}{R_k} \leq \frac{k}{\sqrt{k}} = \sqrt{k} < \sqrt{k+1}. \] Hence \[ \sqrt{k+1} < 1 + \frac{k}{R_k} = R_{k+1} < \sqrt{k+1} + 1, \] and the induction is complete.
Note that the proof shows that the inequalities are strict for all integers $n > 1$. | 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 |
1958 | 1958_4_Sp | In assigning dormitory rooms, a college gives preference to pairs of students in this order:
\[ \text{AA, AB, AC, BB, BC, AD, CC, BD, CD, DD,} \]
in which $\text{AA}$ means two seniors, $\text{AB}$ means a senior and a junior, etc. Determine numerical values to assign to $A, B, C, D$ so that the set of numbers $A + A, A + B, A + C, B + B, A + D$, etc., corresponding to the order above will be in descending magnitude. Find the solution in least positive integers and give the sum as the final answer. | The inequalities to be solved are
\[ \begin{aligned}
& 2A > A + B > A + C > 2B > B + C > A + D > \
& 2C > B + D > C + D > 2D.
\end{aligned} \]
Evidently we must have $A > B > C > D$, so put
\[ A = \alpha + \beta, \; B = \beta + \gamma, \; C = \gamma + \delta, \; D = \delta. \]
where $\alpha, \beta, \gamma$ are positive. The first, second, fourth, eighth, and ninth inequalities in (1) are now satisfied, while the third, fifth, sixth, and seventh inequalities become, respectively
\[ \alpha > \beta, \; \gamma > \alpha, \; \alpha + \beta > \gamma, \; \gamma > \beta. \]
The inequality $\gamma > \beta$ is a consequence of $\gamma > \alpha$ and $\alpha > \beta$, so it can be dropped from this system. Now put $\alpha = \beta + \delta, \; \gamma = \alpha + \epsilon$ where $\delta$ and $\epsilon$ are positive. Then $\alpha + \beta > \gamma$ becomes $\beta > \epsilon$, so we can put $\beta = \epsilon + \zeta$ with $\zeta$ positive. Then
\[ \alpha = \delta + \epsilon + \zeta, \; \beta = \epsilon + \zeta, \; \gamma = \delta + 2\epsilon + \zeta. \]
and finally,
\[ \begin{aligned}
A & = 2\delta + 4\epsilon + 3\zeta + D, \\
B & = \delta + 3\epsilon + 2\zeta + D, \\
C & = \delta + 2\epsilon + \zeta + D, \\
D & = D.
\end{aligned} \]
Here $D$ can be chosen arbitrarily, while $\delta, \epsilon, \zeta$ must be positive. It follows either from the derivation or from a direct check that if $A, B, C, D$ satisfy (2), where $\delta, \epsilon, \zeta$ are positive, then they satisfy the inequalities (1).
To find integral solutions, we note that if we start with integral $A, B, C, D$ satisfying (1), then all the numbers $\alpha, \beta, \gamma, \delta, \epsilon, \zeta$ will be positive integers. Hence the least solution in positive integers is obtained by choosing $\delta = \epsilon = \zeta = D = 1$. Hence $A = 10, B = 7, C = 5, D = 1$ is the least solution in positive integers. This makes the final answer \boxed{23}. | numerical | putnam (modified boxing) | Algebra Linear Algebra | In assigning dormitory rooms, a college gives preference to pairs of students in this order:
\[ \text{AA, AB, AC, BB, BC, AD, CC, BD, CD, DD,} \]
in which $\text{AA}$ means two seniors, $\text{AB}$ means a senior and a junior, etc. Determine numerical values to assign to $A, B, C, D$ so that the set of numbers $A + A, A + B, A + C, B + B, A + D$, etc., corresponding to the order above will be in descending magnitude. Find the general solution and the solution in least positive integers. | The inequalities to be solved are
\[ \begin{aligned}
& 2A > A + B > A + C > 2B > B + C > A + D > \
& 2C > B + D > C + D > 2D.
\end{aligned} \]
Evidently we must have $A > B > C > D$, so put
\[ A = \alpha + \beta, \; B = \beta + \gamma, \; C = \gamma + \delta, \; D = \delta. \]
where $\alpha, \beta, \gamma$ are positive. The first, second, fourth, eighth, and ninth inequalities in (1) are now satisfied, while the third, fifth, sixth, and seventh inequalities become, respectively
\[ \alpha > \beta, \; \gamma > \alpha, \; \alpha + \beta > \gamma, \; \gamma > \beta. \]
The inequality $\gamma > \beta$ is a consequence of $\gamma > \alpha$ and $\alpha > \beta$, so it can be dropped from this system. Now put $\alpha = \beta + \delta, \; \gamma = \alpha + \epsilon$ where $\delta$ and $\epsilon$ are positive. Then $\alpha + \beta > \gamma$ becomes $\beta > \epsilon$, so we can put $\beta = \epsilon + \zeta$ with $\zeta$ positive. Then
\[ \alpha = \delta + \epsilon + \zeta, \; \beta = \epsilon + \zeta, \; \gamma = \delta + 2\epsilon + \zeta. \]
and finally,
\[ \begin{aligned}
A & = 2\delta + 4\epsilon + 3\zeta + D, \\
B & = \delta + 3\epsilon + 2\zeta + D, \\
C & = \delta + 2\epsilon + \zeta + D, \\
D & = D.
\end{aligned} \]
Here $D$ can be chosen arbitrarily, while $\delta, \epsilon, \zeta$ must be positive. It follows either from the derivation or from a direct check that if $A, B, C, D$ satisfy (2), where $\delta, \epsilon, \zeta$ are positive, then they satisfy the inequalities (1).
To find integral solutions, we note that if we start with integral $A, B, C, D$ satisfying (1), then all the numbers $\alpha, \beta, \gamma, \delta, \epsilon, \zeta$ will be positive integers. Hence the least solution in positive integers is obtained by choosing $\delta = \epsilon = \zeta = D = 1$. Hence $\boxed{A = 10, B = 7, C = 5, D = 1}$ is the least solution in positive integers. | 0 |
1958 | 1958_7_Sp | Let $a$ and $b$ be relatively prime positive integers, $b$ even. For each positive integer $q$, let $p = p(q)$ be chosen so that
\[ \left| \frac{p}{q} - \frac{a}{b} \right| \]
is a minimum. Find the value of
\[ \lim_{n \to \infty} \frac{1}{n} \sum_{q=1}^n q \left| \frac{p}{q} - \frac{a}{b} \right|. \] | Rewrite $q \left| \frac{p}{q} - \frac{a}{b} \right|$ in the form
\[ \frac{1}{b} |pb - qa|. \]
For each $q$, we are to choose $p$ to minimize this; then $pb - qa$ will be the absolutely least residue of $qa$ modulo $b$. Since $a$ is relatively prime to $b$, as $q$ varies through a complete set of residues modulo $b$, so will $qa$, and therefore $pb - qa$ will take the values
\[ -C + 1, -C + 2, \ldots, -1, 0, 1, \ldots, C - 1, C \]
where $b = 2C$ (recall that $b$ is even) and the contribution to the sum will be
\[ \frac{1}{b} (0 + 1 + 2 + \cdots + C - 1 + C + C - 1 + \cdots + 1) = \frac{C^2}{2C} = \frac{b}{4}. \]
Thus if $n = b \cdot r + s$ where $0 \leq s < b$, we have
\[ \sum_{q=1}^n q \left| \frac{p}{q} - \frac{a}{b} \right| = r \frac{b}{4} + \sum_{q=br+1}^{br+s} q \left| \frac{p}{q} - \frac{a}{b} \right|, \]
since $q$ runs through $r$ complete residue systems (mod $b$) and then the integers $br + 1, br + 2, \ldots, br + s$. Since $p(q)$ is the integer nearest to $qa / b$,
\[ q \left| \frac{p}{q} - \frac{a}{b} \right| = q \left| p - \frac{qa}{b} \right| \leq \frac{1}{2}. \]
We have therefore
\[ \left| \frac{1}{n} \sum_{q=1}^n q \left| \frac{p}{q} - \frac{a}{b} \right| - \frac{1}{4} \right| = \frac{s}{4} - \sum_{q=br+1}^{br+s} q \left| \frac{p}{q} - \frac{a}{b} \right|. \]
\[ \leq \frac{s}{4} + s \sup_q q \left| \frac{p}{q} - \frac{a}{b} \right| \leq \frac{3}{4}s < \frac{3}{4}b. \]
Dividing by $n$, we get
\[ \left| \frac{1}{4} - \frac{1}{n} \sum_{q=1}^n q \left| \frac{p}{q} - \frac{a}{b} \right| \right| < \frac{3b}{4n}. \]
Letting $n \to \infty$, we see that
\[ \lim_{n \to \infty} \frac{1}{n} \sum_{q=1}^n q \left| \frac{p}{q} - \frac{a}{b} \right| = \boxed{\frac{1}{4}}. \] | numerical | putnam (modified boxing) | Number Theory Analysis | Let $a$ and $b$ be relatively prime positive integers, $b$ even. For each positive integer $q$, let $p = p(q)$ be chosen so that
\[ \left| \frac{p}{q} - \frac{a}{b} \right| \]
is a minimum. Prove that
\[ \lim_{n \to \infty} \frac{1}{n} \sum_{q=1}^n q \left| \frac{p}{q} - \frac{a}{b} \right| = \frac{1}{4}. \] | Rewrite $q \left| \frac{p}{q} - \frac{a}{b} \right|$ in the form
\[ \frac{1}{b} |pb - qa|. \]
For each $q$, we are to choose $p$ to minimize this; then $pb - qa$ will be the absolutely least residue of $qa$ modulo $b$. Since $a$ is relatively prime to $b$, as $q$ varies through a complete set of residues modulo $b$, so will $qa$, and therefore $pb - qa$ will take the values
\[ -C + 1, -C + 2, \ldots, -1, 0, 1, \ldots, C - 1, C \]
where $b = 2C$ (recall that $b$ is even) and the contribution to the sum will be
\[ \frac{1}{b} (0 + 1 + 2 + \cdots + C - 1 + C + C - 1 + \cdots + 1) = \frac{C^2}{2C} = \frac{b}{4}. \]
Thus if $n = b \cdot r + s$ where $0 \leq s < b$, we have
\[ \sum_{q=1}^n q \left| \frac{p}{q} - \frac{a}{b} \right| = r \frac{b}{4} + \sum_{q=br+1}^{br+s} q \left| \frac{p}{q} - \frac{a}{b} \right|, \]
since $q$ runs through $r$ complete residue systems (mod $b$) and then the integers $br + 1, br + 2, \ldots, br + s$. Since $p(q)$ is the integer nearest to $qa / b$,
\[ q \left| \frac{p}{q} - \frac{a}{b} \right| = q \left| p - \frac{qa}{b} \right| \leq \frac{1}{2}. \]
We have therefore
\[ \left| \frac{1}{n} \sum_{q=1}^n q \left| \frac{p}{q} - \frac{a}{b} \right| - \frac{1}{4} \right| = \frac{s}{4} - \sum_{q=br+1}^{br+s} q \left| \frac{p}{q} - \frac{a}{b} \right|. \]
\[ \leq \frac{s}{4} + s \sup_q q \left| \frac{p}{q} - \frac{a}{b} \right| \leq \frac{3}{4}s < \frac{3}{4}b. \]
Dividing by $n$, we get
\[ \left| \frac{1}{4} - \frac{1}{n} \sum_{q=1}^n q \left| \frac{p}{q} - \frac{a}{b} \right| \right| < \frac{3b}{4n}. \]
Letting $n \to \infty$, we see that
\[ \lim_{n \to \infty} \frac{1}{n} \sum_{q=1}^n q \left| \frac{p}{q} - \frac{a}{b} \right| = \frac{1}{4}. \] | 0 |
1958 | 1958_8_Sp | Given
\[ b_n = \sum_{k=0}^n \binom{n}{k}^{-1}, \quad n \geq 1, \] find the value of \lim_{n \to \infty}. | We have
\[ n! b_n = \sum_{k=0}^n k! (n-k)! = \sum_{k=1}^{n+1} (k-1)! (n-k+1)! \]
so
\[ 2n! b_n = 0!n! + \sum_{k=1}^n [k! (n-k)! + (k-1)! (n-k+1)!] + n!0! \]
\[ = 2n! + (n+1) \sum_{k=1}^n (k-1)! (n-k)! \]
\[ = 2n! + (n+1)(n-1)!b_{n-1}. \]
Dividing by $2n!$, we get
\[ b_n = 1 + \frac{n+1}{2n} b_{n-1} \]
as required.
Let $b_n = 2 + C_n$. Then
\[ nC_n = 1 + \frac{n+1}{2(n-1)}(n-1)C_{n-1}. \]
We shall prove by induction that $nC_n \leq 6$ for all $n$. This is clearly true for $n = 1, 2, 3$ ($C_1 = 0$, $C_2 = \frac{1}{2}$, $C_3 = \frac{2}{3}$). Assume it is true for $n = k-1$ where $k \geq 4$. Then
\[ kC_k \leq 1 + \frac{k+1}{2(k-1)} 6 \leq 1 + \frac{5}{2 \cdot 3}6 = 6, \]
where we used the fact that $(x+1)/(x-1) = 1 + 2/(x-1)$ increases as $x$ decreases for $x > 1$. This completes the induction.
From the inequality $C_n \leq 6/n$ and the obvious fact that $C_n \geq 0$, it follows that $C_n \to 0$, and therefore $b_n \to \boxed{2}$. | numerical | putnam (modified boxing) | Combinatorics Analysis | Given
\[ b_n = \sum_{k=0}^n \binom{n}{k}^{-1}, \quad n \geq 1, \]
prove that
\[ b_n = \frac{n+1}{2n} b_{n-1} + 1, \quad n \geq 2, \]
and hence, as a corollary,
\[ \lim_{n \to \infty} b_n = 2. \] | We have
\[ n! b_n = \sum_{k=0}^n k! (n-k)! = \sum_{k=1}^{n+1} (k-1)! (n-k+1)! \]
so
\[ 2n! b_n = 0!n! + \sum_{k=1}^n [k! (n-k)! + (k-1)! (n-k+1)!] + n!0! \]
\[ = 2n! + (n+1) \sum_{k=1}^n (k-1)! (n-k)! \]
\[ = 2n! + (n+1)(n-1)!b_{n-1}. \]
Dividing by $2n!$, we get
\[ b_n = 1 + \frac{n+1}{2n} b_{n-1} \]
as required.
Let $b_n = 2 + C_n$. Then
\[ nC_n = 1 + \frac{n+1}{2(n-1)}(n-1)C_{n-1}. \]
We shall prove by induction that $nC_n \leq 6$ for all $n$. This is clearly true for $n = 1, 2, 3$ ($C_1 = 0$, $C_2 = \frac{1}{2}$, $C_3 = \frac{2}{3}$). Assume it is true for $n = k-1$ where $k \geq 4$. Then
\[ kC_k \leq 1 + \frac{k+1}{2(k-1)} 6 \leq 1 + \frac{5}{2 \cdot 3}6 = 6, \]
where we used the fact that $(x+1)/(x-1) = 1 + 2/(x-1)$ increases as $x$ decreases for $x > 1$. This completes the induction.
From the inequality $C_n \leq 6/n$ and the obvious fact that $C_n \geq 0$, it follows that $C_n \to 0$, and therefore $b_n \to 2$. | 0 |
1958 | 1958_9_Sp | Given a set of $n+1$ positive integers, none of which exceeds $2n$, find the least number of elements of the set such that one divide another member of the set. | Every positive integer can be written uniquely in the form $2^p q$ where $p$ is a non-negative integer and $q$ is a positive odd integer, the odd part of $n$. The odd part of an integer in the set $S = \{1, 2, 3, \dots, 2n\}$ must be one of the $n$ integers $1, 3, 5, \dots, (2n-1)$. Given $(n+1)$ integers in $S$, at least two must have the same odd part, that is they must be of the form
\[ 2^{p_1}q \text{ and } 2^{p_2}q, \]
where $p_1 \neq p_2$. We can choose the notation so that $p_1 < p_2$; then $2^{p_1}q$ divides $2^{p_2}q$. Thus the least number is \boxed{1}. | numerical | putnam (modified boxing) | Combinatorics Number Theory | Given a set of $n+1$ positive integers, none of which exceeds $2n$, show that at least one member of the set must divide another member of the set. | Every positive integer can be written uniquely in the form $2^p q$ where $p$ is a non-negative integer and $q$ is a positive odd integer, the odd part of $n$. The odd part of an integer in the set $S = \{1, 2, 3, \dots, 2n\}$ must be one of the $n$ integers $1, 3, 5, \dots, (2n-1)$. Given $(n+1)$ integers in $S$, at least two must have the same odd part, that is they must be of the form
\[ 2^{p_1}q \text{ and } 2^{p_2}q, \]
where $p_1 \neq p_2$. We can choose the notation so that $p_1 < p_2$; then $2^{p_1}q$ divides $2^{p_2}q$. | 0 |
1958 | 1958_10_Sp | If a square of unit side be partitioned into two sets, what is the minimum possible value of the larger diameter among the two sets? (The diameter of a set is defined as the least upper bound of distances between pairs of points in that set.) | Suppose the square is $ABCD$ (with unit side) and the midpoints of the sides $AB$ and $BC$ are $E$ and $F$, respectively. Then $|AF| = |DF| = |DE| = |CE| = \sqrt{5}/2$.
Suppose that the square is partitioned into two sets $S$ and $T$ of diameter less than $\sqrt{5}/2$, and choose the notation so that $A \in S$. Then $F \in T, D \in S, E \in T, C \in S$, since $A$ and $F$, for example, are too far apart to be both members of $S$. Thus $A$ and $C$ are in the same subset $S$, but $|AC| = \sqrt{2} > \sqrt{5}/2$, contradicting the fact that the diameter of $S$ is $< \sqrt{5}/2$.
On the other hand, one can clearly partition the square into two rectangular sets of diameter $\sqrt{5}/2$, as indicated.
The points of the dividing segment $EG$ can be apportioned to either subset. (It is obvious that the diameter of a rectangle is the length of its diagonal, but an analytic proof can easily be given.) Thus the final answer is \boxed{\sqrt{5}/2}. | numerical | putnam (modified boxing) | Geometry | If a square of unit side be partitioned into two sets, then the diameter (least upper bound of the distances between pairs of points) of one of the sets is not less than $\sqrt{5}/2$. Show also that no larger number will do. | Suppose the square is $ABCD$ (with unit side) and the midpoints of the sides $AB$ and $BC$ are $E$ and $F$, respectively. Then $|AF| = |DF| = |DE| = |CE| = \sqrt{5}/2$.
Suppose that the square is partitioned into two sets $S$ and $T$ of diameter less than $\sqrt{5}/2$, and choose the notation so that $A \in S$. Then $F \in T, D \in S, E \in T, C \in S$, since $A$ and $F$, for example, are too far apart to be both members of $S$. Thus $A$ and $C$ are in the same subset $S$, but $|AC| = \sqrt{2} > \sqrt{5}/2$, contradicting the fact that the diameter of $S$ is $< \sqrt{5}/2$.
On the other hand, one can clearly partition the square into two rectangular sets of diameter $\sqrt{5}/2$, as indicated.
The points of the dividing segment $EG$ can be apportioned to either subset. (It is obvious that the diameter of a rectangle is the length of its diagonal, but an analytic proof can easily be given.) | 0 |
1958 | 1958_11_Sp | The lengths of successive segments of a broken line are represented by the successive terms of the harmonic progression $1, \frac{1}{2}, \frac{1}{3}, \dots, \frac{1}{n}, \dots$. Each segment makes with the preceding segment a given angle $\theta$. What is the distance of the limiting point from the initial point of the first segment? | We may identify the plane with the complex number plane in such a way that the first segment extends from $0$ to $1$. Then the next segment extends from $1$ to $1 + \frac{1}{2} e^{i\theta}$, since this segment represents the complex number $\frac{1}{2} e^{i\theta}$.
The $n$th segment of the path represents the complex number $\frac{1}{n} e^{i(n-1)\theta}$, and when added to the previous $(n-1)$ segments, the sum ends at the point $\sum_{p=1}^\infty \frac{1}{p} e^{i(p-1)\theta}$. Thus the question really concerns the convergence and evaluation of
\[ \sum_{p=1}^\infty \frac{1}{p} e^{i(p-1)\theta}. \]
This becomes the harmonic series if $\theta = 0$ (or a multiple of $2\pi$) and does not converge in this case. We shall show that the series $(1)$ converges in all other cases.
A theorem of Abel asserts: If $b_1, b_2, b_3, \dots$ is a real sequence that decreases monotonically to zero, and if the partial sums of the series $\sum_{p=1}^\infty a_p b_p$ are bounded, then $\sum_{p=1}^\infty a_p b_p$ converges. A proof of this theorem was given in the discussion of Problem A.M. 7 of the Eleventh Competition (p. 325). It is also proved in a number of advanced calculus texts, for example, Louis Brand, *Advanced Calculus: An Introduction to Classical Analysis,* Wiley, New York, 1955, pages 418–420; or T. M. Apostol, *Mathematical Analysis,* Addison-Wesley, Reading, Mass., 1957, page 365.
For the present case, take $b_p = \frac{1}{p}$ and $a_p = e^{i(p-1)\theta}$. Since the $a$'s form a geometric progression, and since we are assuming $e^{i\theta} \neq 1$, we have
\[ \left| \sum_{p=1}^N a_p \right| = \left| \frac{1 - e^{iN\theta}}{1 - e^{i\theta}} \right| \leq \frac{2}{|1 - e^{i\theta}|}. \]
Thus the partial sums are bounded, Abel's theorem applies, and the series $(1)$ converges when $e^{i\theta} \neq 1$.
To evaluate $(1)$ we can use another theorem of Abel. If $\sum_{p=1}^\infty c_p$ converges, then
\[ \lim_{r \to 1^-} \sum_{p=1}^\infty c_p r^p = \sum_{p=1}^\infty c_p, \]
where the limit is taken for $r$ increasing to $1$ through real values. See Brand, pages 423–424, or Apostol, page 421.
In the present case,
\[ \sum_{p=1}^\infty \frac{1}{p} e^{i(p-1)\theta} = e^{-i\theta} \lim_{r \to 1^-} \sum_{p=1}^\infty \frac{1}{p} (re^{i\theta})^p. \]
Putting $z = re^{i\theta}$, we recognize the series on the right as the Taylor’s series expansion for the principal value of $-\log(1 - z)$, valid for $|z| < 1$.
Hence
\[ \sum_{p=1}^\infty \frac{1}{p} e^{i(p-1)\theta} = -e^{-i\theta} \log(1 - e^{i\theta}). \]
Now
\[ 1 - e^{i\theta} = 2 \sin \frac{1}{2}\theta e^{i(\pi/2 + \theta/2 - \pi)}, \]
for $0 < \theta < 2\pi$,
and here $2 \sin \frac{1}{2}\theta > 0$ and $\frac{1}{2}(\theta - \pi)$ is the principal value of the argument (since $-\pi < \frac{1}{2}(\theta - \pi) < \pi$). Therefore
\[ \log(1 - e^{i\theta}) = \log \left( 2 \sin \frac{1}{2}\theta \right) + \frac{1}{2}i(\theta - \pi). \]
Thus
\[ \sum_{p=1}^\infty \frac{1}{p} e^{i(p-1)\theta} = -e^{-i\theta} \left[ \log \left( 2 \sin \frac{1}{2}\theta \right) + \frac{1}{2}i(\theta - \pi) \right]. \]
Hence the limit point is at a distance
\[ \left| \sum_{p=1}^\infty \frac{1}{p} e^{i(p-1)\theta} \right| = \boxed{\sqrt{\left[ \log \left( 2 \sin \frac{1}{2}\theta \right) \right]^2 + \frac{1}{4}(\theta - \pi)^2}}, \]
from the origin; its argument is
\[ \pi - \theta + \arg \left[ \log \left( 2 \sin \frac{1}{2}\theta \right) + \frac{1}{2}i(\theta - \pi) \right]. \] | algebraic | putnam (modified boxing) | Complex Numbers | The lengths of successive segments of a broken line are represented by the successive terms of the harmonic progression $1, \frac{1}{2}, \frac{1}{3}, \dots, \frac{1}{n}, \dots$. Each segment makes with the preceding segment a given angle $\theta$. What is the distance and what is the direction of the limiting point (if there is one) from the initial point of the first segment? | We may identify the plane with the complex number plane in such a way that the first segment extends from $0$ to $1$. Then the next segment extends from $1$ to $1 + \frac{1}{2} e^{i\theta}$, since this segment represents the complex number $\frac{1}{2} e^{i\theta}$.
The $n$th segment of the path represents the complex number $\frac{1}{n} e^{i(n-1)\theta}$, and when added to the previous $(n-1)$ segments, the sum ends at the point $\sum_{p=1}^\infty \frac{1}{p} e^{i(p-1)\theta}$. Thus the question really concerns the convergence and evaluation of
\[ \sum_{p=1}^\infty \frac{1}{p} e^{i(p-1)\theta}. \]
This becomes the harmonic series if $\theta = 0$ (or a multiple of $2\pi$) and does not converge in this case. We shall show that the series $(1)$ converges in all other cases.
A theorem of Abel asserts: If $b_1, b_2, b_3, \dots$ is a real sequence that decreases monotonically to zero, and if the partial sums of the series $\sum_{p=1}^\infty a_p b_p$ are bounded, then $\sum_{p=1}^\infty a_p b_p$ converges. A proof of this theorem was given in the discussion of Problem A.M. 7 of the Eleventh Competition (p. 325). It is also proved in a number of advanced calculus texts, for example, Louis Brand, *Advanced Calculus: An Introduction to Classical Analysis,* Wiley, New York, 1955, pages 418–420; or T. M. Apostol, *Mathematical Analysis,* Addison-Wesley, Reading, Mass., 1957, page 365.
For the present case, take $b_p = \frac{1}{p}$ and $a_p = e^{i(p-1)\theta}$. Since the $a$'s form a geometric progression, and since we are assuming $e^{i\theta} \neq 1$, we have
\[ \left| \sum_{p=1}^N a_p \right| = \left| \frac{1 - e^{iN\theta}}{1 - e^{i\theta}} \right| \leq \frac{2}{|1 - e^{i\theta}|}. \]
Thus the partial sums are bounded, Abel's theorem applies, and the series $(1)$ converges when $e^{i\theta} \neq 1$.
To evaluate $(1)$ we can use another theorem of Abel. If $\sum_{p=1}^\infty c_p$ converges, then
\[ \lim_{r \to 1^-} \sum_{p=1}^\infty c_p r^p = \sum_{p=1}^\infty c_p, \]
where the limit is taken for $r$ increasing to $1$ through real values. See Brand, pages 423–424, or Apostol, page 421.
In the present case,
\[ \sum_{p=1}^\infty \frac{1}{p} e^{i(p-1)\theta} = e^{-i\theta} \lim_{r \to 1^-} \sum_{p=1}^\infty \frac{1}{p} (re^{i\theta})^p. \]
Putting $z = re^{i\theta}$, we recognize the series on the right as the Taylor’s series expansion for the principal value of $-\log(1 - z)$, valid for $|z| < 1$.
Hence
\[ \sum_{p=1}^\infty \frac{1}{p} e^{i(p-1)\theta} = -e^{-i\theta} \log(1 - e^{i\theta}). \]
Now
\[ 1 - e^{i\theta} = 2 \sin \frac{1}{2}\theta e^{i(\pi/2 + \theta/2 - \pi)}, \]
for $0 < \theta < 2\pi$,
and here $2 \sin \frac{1}{2}\theta > 0$ and $\frac{1}{2}(\theta - \pi)$ is the principal value of the argument (since $-\pi < \frac{1}{2}(\theta - \pi) < \pi$). Therefore
\[ \log(1 - e^{i\theta}) = \log \left( 2 \sin \frac{1}{2}\theta \right) + \frac{1}{2}i(\theta - \pi). \]
Thus
\[ \sum_{p=1}^\infty \frac{1}{p} e^{i(p-1)\theta} = -e^{-i\theta} \left[ \log \left( 2 \sin \frac{1}{2}\theta \right) + \frac{1}{2}i(\theta - \pi) \right]. \]
\boxed{Hence the limit point is at a distance
\[ \left| \sum_{p=1}^\infty \frac{1}{p} e^{i(p-1)\theta} \right| = \sqrt{\left[ \log \left( 2 \sin \frac{1}{2}\theta \right) \right]^2 + \frac{1}{4}(\theta - \pi)^2}, \]
from the origin; its argument is
\[ \pi - \theta + \arg \left[ \log \left( 2 \sin \frac{1}{2}\theta \right) + \frac{1}{2}i(\theta - \pi) \right]}. \] | 0 |
1958 | 1958_14_Sp | Let $a_1, a_2, \dots, a_n$ be a permutation of the integers $1, 2, \dots, n$. Call $a_i$ a "big" integer if $a_i > a_j$ for all $j > i$. Find the mean number of "big" integers over all permutations on the first $n$ positive integers. | If $\sigma$ is a permutation, let $N_i(\sigma)$ be the number of "big" integers occurring at position $i$. Then $N_i(\sigma) = 0$ or $1$. The average value of $N_i(\sigma)$ over all the permutations is $1/(n - i + 1)$ because after $a_1, a_2, \dots, a_{i-1}$ have been selected, the question of whether or not $a_i$ will be a big integer is whether or not $a_i$ is the greatest among the $(n - i + 1)$ integers left.
Let the number of big integers in $\sigma$ be $N(\sigma)$. Then $N(\sigma) = N_1(\sigma) + N_2(\sigma) + \dots + N_n(\sigma)$, and the average value of $N(\sigma)$ over all the $n!$ permutations will be the sum of the average values of the separate terms, $N_i(\sigma)$, for $i = 1, 2, \dots, n$. Hence this average is
\[ \boxed{1/n + 1/(n - 1) + 1/(n - 2) + \dots + 1}. \] | algebraic | putnam | Combinatorics | Let $a_1, a_2, \dots, a_n$ be a permutation of the integers $1, 2, \dots, n$. Call $a_i$ a "big" integer if $a_i > a_j$ for all $j > i$. Find the mean number of "big" integers over all permutations on the first $n$ positive integers. | If $\sigma$ is a permutation, let $N_i(\sigma)$ be the number of "big" integers occurring at position $i$. Then $N_i(\sigma) = 0$ or $1$. The average value of $N_i(\sigma)$ over all the permutations is $1/(n - i + 1)$ because after $a_1, a_2, \dots, a_{i-1}$ have been selected, the question of whether or not $a_i$ will be a big integer is whether or not $a_i$ is the greatest among the $(n - i + 1)$ integers left.
Let the number of big integers in $\sigma$ be $N(\sigma)$. Then $N(\sigma) = N_1(\sigma) + N_2(\sigma) + \dots + N_n(\sigma)$, and the average value of $N(\sigma)$ over all the $n!$ permutations will be the sum of the average values of the separate terms, $N_i(\sigma)$, for $i = 1, 2, \dots, n$. Hence this average is
\[ \boxed{1/n + 1/(n - 1) + 1/(n - 2) + \dots + 1}. \] | 0 |
1959 | 1959_5 | A sparrow, flying horizontally in a straight line, is 50 feet directly below an eagle and 100 feet directly above a hawk. Both hawk and eagle fly directly toward the sparrow, reaching it simultaneously. The hawk flies twice as fast as the sparrow. How far does the eagle fly? | Although not specifically stated, it is clearly intended that each of the three birds flies at a uniform speed. So we consider the general problem of a predator flying with speed $v$ in pursuit of a prey flying along a straight path with speed $rv$, $r < 1$.
Let the prey start at $(0, 0)$ and move along the $x$-axis so that his position at time $t$ is $(rvt, 0)$. Let the predator start at $(0, h)$, $h > 0$. If at time $t$ the predator is at $(x, y)$, then the condition that he always flies directly toward his prey is that
\[ \frac{dx/dt}{dy/dt} = \frac{rvt - x}{-y}. \]
As long as $y$ remains positive, $dy/dt < 0$, so we may choose $y$ as the independent variable. Then (1) becomes
\[ y \frac{dx}{dy} = x - rvt. \]
The condition that the predator has constant speed $v$ is
\[ \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = v, \]
which we can write as
\[ \sqrt{\left(\frac{dx}{dy}\right)^2 + 1} = v \frac{dt}{dy}, \]
where we have introduced a minus sign because $dt/dy$ is negative.
We differentiate (2) to get
\[ y \frac{d^2x}{dy^2} + \frac{dx}{dy} = \frac{dx}{dy} - rv \frac{dt}{dy}, \]
and using (3) we have
\[ y \frac{d^2x}{dy^2} = r \sqrt{\left(\frac{dx}{dy}\right)^2 + 1}. \]
We can write this as the first-order differential equation
\[ y \frac{dz}{dy} = r \sqrt{1 + z^2}, \]
where $z = dx/dy$. Separating the variables we have
\[ r \frac{dy}{y} = \frac{dz}{\sqrt{1 + z^2}}, \]
whence
\[ r \log y = \log (z + \sqrt{1 + z^2}) + C. \]
Since $z = dx/dy = 0$ when $y = h$, we find $C = r \log h$. Then (4) can be rewritten
\[ \left(\frac{y}{h}\right)^r = z + \sqrt{1 + z^2}. \]
and solved for $z (= dx/dy)$ to get
\[ 2 \frac{dx}{dy} = \left(\frac{y}{h}\right)^r - \left(\frac{y}{h}\right)^{-r}. \]
This can be integrated again using the initial condition $x = 0$ when $y = h$ to give
\[ 2x = \frac{h}{1 + r} \left(\frac{y}{h}\right)^{1+r} - \frac{h}{1 - r} \left(\frac{y}{h}\right)^{1-r} + \frac{2rh}{1 - r^2}. \]
This is valid only for $y > 0$. Recalling that $0 < r < 1$, we see from (6) that
\[ x \to \frac{-rh}{1 - r^2} \]
as $y \to 0$. It follows that the predator catches his prey at the point $(rh/(1 - r^2), 0)$ when $t = h/(1 - r^2)v$. At this time the prey has flown $rh/(1 - r^2)$ and the predator, $h/(1 - r^2)$.
Now for the hawk and the sparrow, $r = \frac{1}{2}$, $h = 100$ ft., so the sparrow flies $200/3$ ft., and the hawk, $400/3$ ft. [Note that since the sparrow is above the hawk we would have to choose a coordinate system with $y$ positive in the downward direction.]
In the case of the eagle, $h = 50$ ft. and $r$ is unknown, but
\[ \frac{r50}{1 - r^2} = \frac{200}{3}; \]
hence $4r^2 + 3r - 4 = 0$ and
\[ r = \frac{-3 + \sqrt{73}}{8} \]
since $r$ must be positive.
The speed of the eagle is $1/r$ times the speed of the sparrow, or
\[ \frac{v(3 + \sqrt{73})}{16} \approx .721v \]
where $v$ is the speed of the hawk.
Since the time of capture is $400/3v$, the eagle flies a distance of
\[ \boxed{\frac{400}{3} \left(\frac{3 + \sqrt{73}}{16}\right)} \approx 96.2 \text{ ft.} \] | numerical | putnam | Differential Equations Geometry | A sparrow, flying horizontally in a straight line, is 50 feet directly below an eagle and 100 feet directly above a hawk. Both hawk and eagle fly directly toward the sparrow, reaching it simultaneously. The hawk flies twice as fast as the sparrow. How far does each bird fly? At what rate does the eagle fly? | Although not specifically stated, it is clearly intended that each of the three birds flies at a uniform speed. So we consider the general problem of a predator flying with speed $v$ in pursuit of a prey flying along a straight path with speed $rv$, $r < 1$.
Let the prey start at $(0, 0)$ and move along the $x$-axis so that his position at time $t$ is $(rvt, 0)$. Let the predator start at $(0, h)$, $h > 0$. If at time $t$ the predator is at $(x, y)$, then the condition that he always flies directly toward his prey is that
\[ \frac{dx/dt}{dy/dt} = \frac{rvt - x}{-y}. \]
As long as $y$ remains positive, $dy/dt < 0$, so we may choose $y$ as the independent variable. Then (1) becomes
\[ y \frac{dx}{dy} = x - rvt. \]
The condition that the predator has constant speed $v$ is
\[ \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = v, \]
which we can write as
\[ \sqrt{\left(\frac{dx}{dy}\right)^2 + 1} = v \frac{dt}{dy}, \]
where we have introduced a minus sign because $dt/dy$ is negative.
We differentiate (2) to get
\[ y \frac{d^2x}{dy^2} + \frac{dx}{dy} = \frac{dx}{dy} - rv \frac{dt}{dy}, \]
and using (3) we have
\[ y \frac{d^2x}{dy^2} = r \sqrt{\left(\frac{dx}{dy}\right)^2 + 1}. \]
We can write this as the first-order differential equation
\[ y \frac{dz}{dy} = r \sqrt{1 + z^2}, \]
where $z = dx/dy$. Separating the variables we have
\[ r \frac{dy}{y} = \frac{dz}{\sqrt{1 + z^2}}, \]
whence
\[ r \log y = \log (z + \sqrt{1 + z^2}) + C. \]
Since $z = dx/dy = 0$ when $y = h$, we find $C = r \log h$. Then (4) can be rewritten
\[ \left(\frac{y}{h}\right)^r = z + \sqrt{1 + z^2}. \]
and solved for $z (= dx/dy)$ to get
\[ 2 \frac{dx}{dy} = \left(\frac{y}{h}\right)^r - \left(\frac{y}{h}\right)^{-r}. \]
This can be integrated again using the initial condition $x = 0$ when $y = h$ to give
\[ 2x = \frac{h}{1 + r} \left(\frac{y}{h}\right)^{1+r} - \frac{h}{1 - r} \left(\frac{y}{h}\right)^{1-r} + \frac{2rh}{1 - r^2}. \]
This is valid only for $y > 0$. Recalling that $0 < r < 1$, we see from (6) that
\[ x \to \frac{-rh}{1 - r^2} \]
as $y \to 0$. It follows that the predator catches his prey at the point $(rh/(1 - r^2), 0)$ when $t = h/(1 - r^2)v$. At this time the prey has flown $rh/(1 - r^2)$ and the predator, $h/(1 - r^2)$.
Now for the hawk and the sparrow, $r = \frac{1}{2}$, $h = 100$ ft., so the sparrow flies $200/3$ ft., and the hawk, $400/3$ ft. [Note that since the sparrow is above the hawk we would have to choose a coordinate system with $y$ positive in the downward direction.]
In the case of the eagle, $h = 50$ ft. and $r$ is unknown, but
\[ \frac{r50}{1 - r^2} = \frac{200}{3}; \]
hence $4r^2 + 3r - 4 = 0$ and
\[ r = \frac{-3 + \sqrt{73}}{8} \]
since $r$ must be positive.
The speed of the eagle is $1/r$ times the speed of the sparrow, or
\[ \frac{v(3 + \sqrt{73})}{16} \approx .721v \]
where $v$ is the speed of the hawk.
Since the time of capture is $400/3v$, the eagle flies a distance of
\[ \boxed{\frac{400}{3} \left(\frac{3 + \sqrt{73}}{16}\right)} \approx 96.2 \text{ ft.} \] | 0 |
1959 | 1959_8 | Let each of $m$ distinct points on the positive part of the $X$-axis be joined to $n$ distinct points on the positive part of the $Y$-axis. Obtain a formula for the number of intersection points of these segments (exclusive of endpoints), assuming that no three of the segments are concurrent. | Each pair of points on the $X$-axis together with each pair of points on the $Y$-axis determine a convex quadrilateral whose diagonals meet somewhere in the first quadrant. Conversely, each intersection point arises in this way. Since no three segments are concurrent, except at the endpoints, each intersection point arises uniquely. There are, therefore, \[ \binom{m}{2} \binom{n}{2} = \boxed{\frac{mn(m-1)(n-1)}{4}} \] points of intersection. (Compare Problem A.M. 4 of the Fifteenth Competition.) | algebraic | putnam | Combinatorics Geometry | Let each of $m$ distinct points on the positive part of the $X$-axis be joined to $n$ distinct points on the positive part of the $Y$-axis. Obtain a formula for the number of intersection points of these segments (exclusive of endpoints), assuming that no three of the segments are concurrent. | Each pair of points on the $X$-axis together with each pair of points on the $Y$-axis determine a convex quadrilateral whose diagonals meet somewhere in the first quadrant. Conversely, each intersection point arises in this way. Since no three segments are concurrent, except at the endpoints, each intersection point arises uniquely. There are, therefore, \[ \binom{m}{2} \binom{n}{2} = \boxed{\frac{mn(m-1)(n-1)}{4}} \] points of intersection. (Compare Problem A.M. 4 of the Fifteenth Competition.) | 0 |
1959 | 1959_11 | Given the following matrix of 25 elements \[ \begin{bmatrix} 11 & 17 & 25 & 19 & 16 \\ 24 & 10 & 13 & 15 & 3 \\ 12 & 5 & 14 & 2 & 18 \\ 23 & 4 & 1 & 8 & 22 \\ 6 & 20 & 7 & 21 & 9 \end{bmatrix}, \] choose five of these elements, no two coming from the same row or column, in such a way that the minimum of these five elements is as large as possible. Prove that your answer is correct while returning this minimum as the final answer. | Since the set of border elements of the matrix is the union of two rows and two columns, we may choose at most four elements from the border and must choose at least one element from the central $3 \times 3$ submatrix. Since the largest element in this central submatrix is 15, there is no admissible choice for which the minimum exceeds 15. But the choice $25, 15, 18, 23, 20$ satisfies the conditions and has minimum \boxed{15}. | numerical | putnam | Combinatorics Linear Algebra | Given the following matrix of 25 elements \[ \begin{bmatrix} 11 & 17 & 25 & 19 & 16 \\ 24 & 10 & 13 & 15 & 3 \\ 12 & 5 & 14 & 2 & 18 \\ 23 & 4 & 1 & 8 & 22 \\ 6 & 20 & 7 & 21 & 9 \end{bmatrix}, \] choose five of these elements, no two coming from the same row or column, in such a way that the minimum of these five elements is as large as possible. Prove that your answer is correct. | Since the set of border elements of the matrix is the union of two rows and two columns, we may choose at most four elements from the border and must choose at least one element from the central $3 \times 3$ submatrix. Since the largest element in this central submatrix is 15, there is no admissible choice for which the minimum exceeds 15. But the choice $25, 15, 18, 23, 20$ satisfies the conditions and has minimum 15. | 0 |
1959 | 1959_12 | Find the equation of the smallest sphere which is tangent to both of the lines: \begin{itemize} \item[(i)] $x = t + 1, y = 2t + 4, z = -3t + 5$, and \item[(ii)] $x = 4t - 12, y = -t + 8, z = t + 17$. \end{itemize} | Let the given lines be $l$ and $m$. Then there is a unique segment $PQ$ perpendicular to both lines with $P$ on $l$ and $Q$ on $m$. The required sphere has $PQ$ as its diameter. \par Suppose $l$ and $m$ are given in terms of a parameter $t$ by $a + tv$ and $b + tw$, respectively, where $a, b, v,$ and $w$ are vectors. If the lines are not parallel, $v$ and $w$ are linearly independent. Since $PQ$ is perpendicular to both lines, it has the direction of $v \times w$, say $PQ = \rho v \times w$, where $\rho$ is a scalar. Let $P$ and $Q$ be the points $a + \sigma v$ and $b + \tau w$, respectively. Then \[ PQ = b - a - \sigma v + \tau w \] and \[ a - b = - \rho (v \times w) - \sigma v + \tau w. \] Hence we can calculate $\rho, \sigma, \text{ and } \tau$ by expressing $a - b$ in terms of the independent vectors $v \times w, v, \text{ and } w$. Then the center of the required sphere is at \[ a + \sigma v + \frac{1}{2} \rho (v \times w) \] and its radius is \[ \frac{1}{2} |\rho| \|v \times w\|. \] \par For the example in question, $a = \langle 1, 4, 5 \rangle$, $b = \langle -12, 8, 17 \rangle$, $v = \langle 1, 2, -3 \rangle$, $w = \langle 4, -1, 1 \rangle$ and $v \times w = \langle -1, -13, -9 \rangle$. Then $\rho, \sigma, \text{ and } \tau$ are found from the equations \[ \begin{aligned} 13 & = \rho - \sigma + 4 \tau, \\ -4 & = 13 \rho - 2 \sigma - \tau, \\ -12 & = 9 \rho + 3 \sigma + \tau \end{aligned} \] which give \[ \rho = \frac{-147}{251}, \quad \sigma = \frac{-782}{251}, \quad \tau = \frac{657}{251}. \] \par The center of the sphere is therefore at \[ \langle 1, 4, 5 \rangle - \frac{782}{251} \langle 1, 2, -3 \rangle - \frac{147}{502} \langle -1, -13, -9 \rangle \] \[ = \frac{1}{502} \langle -915, 791, 8525 \rangle. \] \par The square of the radius is \[ \frac{1}{4} \rho^2 \|v \times w\|^2 = \left( \frac{147}{502} \right)^2 (251) = \frac{147^2}{1004}. \] \par The equation of the sphere is \[ \boxed{(502x + 915)^2 + (502y - 791)^2 + (502z - 8525)^2 = 251(147)^2}. \] | numerical | putnam | Geometry Linear Algebra | Find the equation of the smallest sphere which is tangent to both of the lines: \begin{itemize} \item[(i)] $x = t + 1, y = 2t + 4, z = -3t + 5$, and \item[(ii)] $x = 4t - 12, y = -t + 8, z = t + 17$. \end{itemize} | Let the given lines be $l$ and $m$. Then there is a unique segment $PQ$ perpendicular to both lines with $P$ on $l$ and $Q$ on $m$. The required sphere has $PQ$ as its diameter. \par Suppose $l$ and $m$ are given in terms of a parameter $t$ by $a + tv$ and $b + tw$, respectively, where $a, b, v,$ and $w$ are vectors. If the lines are not parallel, $v$ and $w$ are linearly independent. Since $PQ$ is perpendicular to both lines, it has the direction of $v \times w$, say $PQ = \rho v \times w$, where $\rho$ is a scalar. Let $P$ and $Q$ be the points $a + \sigma v$ and $b + \tau w$, respectively. Then \[ PQ = b - a - \sigma v + \tau w \] and \[ a - b = - \rho (v \times w) - \sigma v + \tau w. \] Hence we can calculate $\rho, \sigma, \text{ and } \tau$ by expressing $a - b$ in terms of the independent vectors $v \times w, v, \text{ and } w$. Then the center of the required sphere is at \[ a + \sigma v + \frac{1}{2} \rho (v \times w) \] and its radius is \[ \frac{1}{2} |\rho| \|v \times w\|. \] \par For the example in question, $a = \langle 1, 4, 5 \rangle$, $b = \langle -12, 8, 17 \rangle$, $v = \langle 1, 2, -3 \rangle$, $w = \langle 4, -1, 1 \rangle$ and $v \times w = \langle -1, -13, -9 \rangle$. Then $\rho, \sigma, \text{ and } \tau$ are found from the equations \[ \begin{aligned} 13 & = \rho - \sigma + 4 \tau, \\ -4 & = 13 \rho - 2 \sigma - \tau, \\ -12 & = 9 \rho + 3 \sigma + \tau \end{aligned} \] which give \[ \rho = \frac{-147}{251}, \quad \sigma = \frac{-782}{251}, \quad \tau = \frac{657}{251}. \] \par The center of the sphere is therefore at \[ \langle 1, 4, 5 \rangle - \frac{782}{251} \langle 1, 2, -3 \rangle - \frac{147}{502} \langle -1, -13, -9 \rangle \] \[ = \frac{1}{502} \langle -915, 791, 8525 \rangle. \] \par The square of the radius is \[ \frac{1}{4} \rho^2 \|v \times w\|^2 = \left( \frac{147}{502} \right)^2 (251) = \frac{147^2}{1004}. \] \par The equation of the sphere is \[ \boxed{(502x + 915)^2 + (502y - 791)^2 + (502z - 8525)^2 = 251(147)^2}. \] | 0 |
1960 | 1960_1 | Let $n$ be a given positive integer. How many solutions are there in ordered positive integer pairs $(x, y)$ to the equation \[ \frac{xy}{x+y} = n? \] | The given equation is equivalent to \[ (x - n)(y - n) = n^2. \] Evidently, either both $x > n$ and $y > n$, or $x < n$ and $y < n$. There are no solutions satisfying both $0 < x < n$, $0 < y < n$ because then $|x - n||y - n| < n \cdot n = n^2$. Hence we need only look for integral solutions of $(1)$ with $x - n > 0$, $y - n > 0$. It is clear that there are as many solutions as there are ordered factorizations of $n^2$ into two factors. \par If the prime factorization of $n$ is $p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$, then \[ n^2 = p_1^{2\alpha_1} p_2^{2\alpha_2} \cdots p_k^{2\alpha_k} \] and the number of ordered factorizations of $n^2$ is \[ \boxed{(2\alpha_1 + 1)(2\alpha_2 + 1) \cdots (2\alpha_k + 1)}. \] | algebraic | putnam | Number Theory | Let $n$ be a given positive integer. How many solutions are there in ordered positive integer pairs $(x, y)$ to the equation \[ \frac{xy}{x+y} = n? \] | The given equation is equivalent to \[ (x - n)(y - n) = n^2. \] Evidently, either both $x > n$ and $y > n$, or $x < n$ and $y < n$. There are no solutions satisfying both $0 < x < n$, $0 < y < n$ because then $|x - n||y - n| < n \cdot n = n^2$. Hence we need only look for integral solutions of $(1)$ with $x - n > 0$, $y - n > 0$. It is clear that there are as many solutions as there are ordered factorizations of $n^2$ into two factors. \par If the prime factorization of $n$ is $p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$, then \[ n^2 = p_1^{2\alpha_1} p_2^{2\alpha_2} \cdots p_k^{2\alpha_k} \] and the number of ordered factorizations of $n^2$ is \[ \boxed{(2\alpha_1 + 1)(2\alpha_2 + 1) \cdots (2\alpha_k + 1)}. \] | 0 |
1960 | 1960_3 | Given that $t_1, t_2, t_3, t_4, t_5$ are real numbers, find the upper bound of \[ \sum_{j=1}^5 (1-t_j) \exp \left( \sum_{k=1}^j t_k \right). \] | We first show that \[ (1-s+a)e^s \leq e^a \] for any choice of $s$. The derivative of the function on the left with respect to $s$ is \[ (a-s)e^s \] which vanishes only for $s = a$, and this critical point is easily seen to be a maximum point (since $\lim_{s \to -\infty} (1-s+a)e^s = 0$ and \[ \lim_{s \to +\infty} (1-s+a)e^s = -\infty \]). \par The claimed inequality $(1)$ follows. \par Then, taking $a=0$, $s=t_5$ in $(1)$, we have \[ (1-t_5)e^{t_5} \leq e^0 = 1. \] Multiplying by the positive number $e^{t_4}$ and adding $(1-t_4)e^{t_4}$, we find \[ (1-t_4)e^{t_4} + (1-t_5)e^{t_4+t_5} \leq (1-t_4)e^{t_4} + e^{t_4} \leq e, \] the last step by $(1)$ with $a=1$. \par We continue this process. \par \[ (1-t_3)e^{t_4} + (1-t_4)e^{t_4+t_4} + (1-t_5)e^{t_4+t_4+t_5} \leq (1-t_3)e^{t_4} + e \cdot e^{t_4} \leq e^e. \] Similarly, \[ (1-t_2)e^{t_4} + (1-t_3)e^{t_4+t_4} + (1-t_4)e^{t_4+t_4+t_4} + (1-t_5)e^{t_4+t_4+t_4+t_4} \leq (1-t_2)e^{t_4} + e^e \cdot e^{t_4} \leq e^{e^e}. \] And finally, \[ \sum_{j=1}^5 (1-t_j) \exp \left( \sum_{k=1}^j t_k \right) \leq \boxed{e^{e^{e^e}}}. \] | algebraic | putnam (modified boxing) | Analysis Algebra | Show that if $t_1, t_2, t_3, t_4, t_5$ are real numbers, then \[ \sum_{j=1}^5 (1-t_j) \exp \left( \sum_{k=1}^j t_k \right) \leq e^{e^{e^e}}. \] | We first show that \[ (1-s+a)e^s \leq e^a \] for any choice of $s$. The derivative of the function on the left with respect to $s$ is \[ (a-s)e^s \] which vanishes only for $s = a$, and this critical point is easily seen to be a maximum point (since $\lim_{s \to -\infty} (1-s+a)e^s = 0$ and \[ \lim_{s \to +\infty} (1-s+a)e^s = -\infty \]). \par The claimed inequality $(1)$ follows. \par Then, taking $a=0$, $s=t_5$ in $(1)$, we have \[ (1-t_5)e^{t_5} \leq e^0 = 1. \] Multiplying by the positive number $e^{t_4}$ and adding $(1-t_4)e^{t_4}$, we find \[ (1-t_4)e^{t_4} + (1-t_5)e^{t_4+t_5} \leq (1-t_4)e^{t_4} + e^{t_4} \leq e, \] the last step by $(1)$ with $a=1$. \par We continue this process. \par \[ (1-t_3)e^{t_4} + (1-t_4)e^{t_4+t_4} + (1-t_5)e^{t_4+t_4+t_5} \leq (1-t_3)e^{t_4} + e \cdot e^{t_4} \leq e^e. \] Similarly, \[ (1-t_2)e^{t_4} + (1-t_3)e^{t_4+t_4} + (1-t_4)e^{t_4+t_4+t_4} + (1-t_5)e^{t_4+t_4+t_4+t_4} \leq (1-t_2)e^{t_4} + e^e \cdot e^{t_4} \leq e^{e^e}. \] And finally, \[ \sum_{j=1}^5 (1-t_j) \exp \left( \sum_{k=1}^j t_k \right) \e^{e^{e^e}}. \] | 0 |
1960 | 1960_5 | Consider a polynomial $f(x)$ with real coefficients having the property $f(g(x)) = g(f(x))$ for every polynomial $g(x)$ with real coefficients. Determine $f(x)$. | Consider a constant function $g$, say $g(x) = a$. Then $f(g(x)) = g(f(x))$ becomes $f(a) = a$. Since this is true for all real $a$, $f$ is the identity function, i.e., $f(x) = \boxed{x}$. | algebraic | putnam | Algebra | Consider a polynomial $f(x)$ with real coefficients having the property $f(g(x)) = g(f(x))$ for every polynomial $g(x)$ with real coefficients. Determine and prove the nature of $f(x)$. | Consider a constant function $g$, say $g(x) = a$. Then $f(g(x)) = g(f(x))$ becomes $f(a) = a$. Since this is true for all real $a$, $f$ is the identity function, i.e., $f(x) = \boxed{x}$. | 0 |
1960 | 1960_6 | A player throwing a die scores as many points as on the top face of the die and is to play until his score reaches or passes a total $n$. Denote by $p(n)$ the probability of making exactly the total $n$, and find the value of $\lim_{n \to \infty} p(n)$. | If by definition $p(0) = 1$ and $p(n) = 0$, $n < 0$ then the following relations are easy to verify.
\[ p(0) = 1 \]
\[ p(1) = \frac{1}{6} p(0) \]
\[ p(2) = \frac{1}{6} [p(1) + p(0)] \]
\[ p(3) = \frac{1}{6} [p(2) + p(1) + p(0)] \]
\[ \vdots \]
\[ p(n) = \frac{1}{6} [p(n - 1) + p(n - 2) + \cdots + p(n - 6)], \quad n > 0. \]
Adding these equations and then canceling, we obtain
\[ p(n) + \frac{5}{6} p(n - 1) + \frac{4}{6} p(n - 2) + \cdots + \frac{1}{6} p(n - 5) = 1. \]
Now if it is assumed that $p(n) \to p$ as $n \to \infty$, then it follows that $(21/6)p = 1$ and hence that $p = \boxed{2/7}$. | numerical | putnam | Probability | A player throwing a die scores as many points as on the top face of the die and is to play until his score reaches or passes a total $n$. Denote by $p(n)$ the probability of making exactly the total $n$, and find the value of $\lim_{n \to \infty} p(n)$. | Remark. Since the score of a dice player will increase by an average of $3\frac{1}{2}$ per throw, it seems clear that his score will ultimately assume about $1/3\frac{1}{2}$ or $2/7$ of the possible values. Hence the probability that he achieves some large score $n$ precisely must be very nearly $2/7$. On the basis of this heuristic argument we expect that \[ \lim_{n \to \infty} p(n) = \frac{2}{7}. \] We shall show that this is indeed true.
First Solution. If by definition $p(0) = 1$ and $p(n) = 0$, $n < 0$ then the following relations are easy to verify.
\[ p(0) = 1 \]
\[ p(1) = \frac{1}{6} p(0) \]
\[ p(2) = \frac{1}{6} [p(1) + p(0)] \]
\[ p(3) = \frac{1}{6} [p(2) + p(1) + p(0)] \]
\[ \vdots \]
\[ p(n) = \frac{1}{6} [p(n - 1) + p(n - 2) + \cdots + p(n - 6)], \quad n > 0. \]
Adding these equations and then canceling, we obtain
\[ p(n) + \frac{5}{6} p(n - 1) + \frac{4}{6} p(n - 2) + \cdots + \frac{1}{6} p(n - 5) = 1. \]
Now if it is assumed that $p(n) \to p$ as $n \to \infty$, then it follows that $(21/6)p = 1$ and hence that $p = \boxed{2/7}$. | 0 |
1960 | 1960_8 | Find all solutions of $n^m = m^n$ in integers $n$ and $m$ ($n \neq m$) and return the sum of all ordered pairs by adding corresponding $m$ and $n$ values and give the final summed ordered pair.. | First we consider positive integer solutions. Then the given equation is equivalent to
\[
\frac{1}{m} \log m = \frac{1}{n} \log n.
\]
The function $(\log x)/x$ is strictly increasing for $0 < x \leq e$ and strictly decreasing for $e \leq x$, as we see by considering its derivative $(1 - \log x)/x^2$. Hence a solution of (1) with $m < n$ must have $m < e$, so $m = 1$ or $2$. If $m = 1$, there are no values of $n > 1$ which satisfy (1). If $m = 2$, then $n = 4$ is the unique solution of (1) with $n > 2$. Thus $m = 2$, $n = 4$ is the only solution of the given equation in positive integers with $m < n$. Clearly, there is just one other solution in positive integers with $m \neq n$, namely, $m = 4$, $n = 2$.
In the original equation neither $m$ nor $n$ can be zero, since $0^n = n^0$ has no non-zero solution $n$. If $m$ is negative, say $m = -k$, and $n$ is positive, the requirement becomes $(-k)^n = 1/n^k$. But this has no positive integral solution $n$, since the left member would be an integer but the right would not, unless $n = 1$, in which case the left member is negative.
If both $m$ and $n$ are negative, say $m = -k$, $n = -l$, then $(-k)^{-l} = (-l)^{-k}$ giving
\[
(-1)^l = (-1)^k \quad \text{and} \quad k^l = l^k
\]
with $k$ and $l$ unequal positive integers. As we have seen, this implies $k = 2$, $l = 4$, or vice versa. Both of these solutions satisfy the sign condition.
Thus there are four solutions to the original equation in unequal integers:
\[
(m, n) = (2, 4), (4, 2), (-2, -4), (-4, -2).
\] Thus the final answer is \boxed{(0, 0}. | numerical | putnam (modified boxing) | Algebra Number Theory | Find all solutions of $n^m = m^n$ in integers $n$ and $m$ ($n \neq m$). Prove that you have obtained all of them. | First we consider positive integer solutions. Then the given equation is equivalent to
\[
\frac{1}{m} \log m = \frac{1}{n} \log n.
\]
The function $(\log x)/x$ is strictly increasing for $0 < x \leq e$ and strictly decreasing for $e \leq x$, as we see by considering its derivative $(1 - \log x)/x^2$. Hence a solution of (1) with $m < n$ must have $m < e$, so $m = 1$ or $2$. If $m = 1$, there are no values of $n > 1$ which satisfy (1). If $m = 2$, then $n = 4$ is the unique solution of (1) with $n > 2$. Thus $m = 2$, $n = 4$ is the only solution of the given equation in positive integers with $m < n$. Clearly, there is just one other solution in positive integers with $m \neq n$, namely, $m = 4$, $n = 2$.
In the original equation neither $m$ nor $n$ can be zero, since $0^n = n^0$ has no non-zero solution $n$. If $m$ is negative, say $m = -k$, and $n$ is positive, the requirement becomes $(-k)^n = 1/n^k$. But this has no positive integral solution $n$, since the left member would be an integer but the right would not, unless $n = 1$, in which case the left member is negative.
If both $m$ and $n$ are negative, say $m = -k$, $n = -l$, then $(-k)^{-l} = (-l)^{-k}$ giving
\[
(-1)^l = (-1)^k \quad \text{and} \quad k^l = l^k
\]
with $k$ and $l$ unequal positive integers. As we have seen, this implies $k = 2$, $l = 4$, or vice versa. Both of these solutions satisfy the sign condition.
Thus there are four solutions to the original equation in unequal integers:
\[
\boxed{(m, n) = (2, 4), (4, 2), (-2, -4), (-4, -2)}.
\] | 0 |
1960 | 1960_9 | Evaluate the double series
\[
\sum_{j=0}^\infty \sum_{k=0}^\infty 2^{-3k-j-(k+j)^2}.
\] | If we write out a few terms of the double sequence of terms, we see that each of the terms $2^{-2m}$ occurs exactly once. Since all the terms are positive, the double series may be rearranged to make the simple series
\[
\sum_{m=0}^\infty 2^{-2m}
\]
which sums to $4/3$. The sum of the original double series is therefore $4/3$. To formalize this argument one must prove that
\[
(j, k) \mapsto (k+j)^2 + 3k + j
\]
is a bijection from the set of ordered pairs of non-negative integers to the set of non-negative even integers. This is straightforward but long.
A direct analytic argument can be given as follows: Sum the double series along the diagonals defined by $j + k = n$, and then sum on $n$.
\[
\sum_{j=0}^\infty \sum_{k=0}^\infty 2^{-3k-j-(k+j)^2} = \sum_{n=0}^\infty \sum_{k=0}^n 2^{-2k-n-n^2}
\]
\[
= \sum_{n=0}^\infty \left[ \frac{4}{3} (1 - 2^{-2n-2})2^{-n^2}\right]
\]
\[
= \frac{4}{3} \left[ \sum_{n=0}^\infty 2^{-n-n^2} - \sum_{n=0}^\infty 2^{-(n+1)(n+2)} \right]
\]
\[
= \frac{4}{3} \left[ \sum_{n=0}^\infty 2^{-n(n+1)} - \sum_{m=1}^\infty 2^{-m(m+1)} \right]
\]
\[
= \boxed{4/3}.
\]
At the third step, the separation of the sum into two sums is permissible because the new sums are convergent. | numerical | putnam | Analysis Algebra | Evaluate the double series
\[
\sum_{j=0}^\infty \sum_{k=0}^\infty 2^{-3k-j-(k+j)^2}.
\] | If we write out a few terms of the double sequence of terms, we see that each of the terms $2^{-2m}$ occurs exactly once. Since all the terms are positive, the double series may be rearranged to make the simple series
\[
\sum_{m=0}^\infty 2^{-2m}
\]
which sums to $4/3$. The sum of the original double series is therefore $4/3$. To formalize this argument one must prove that
\[
(j, k) \mapsto (k+j)^2 + 3k + j
\]
is a bijection from the set of ordered pairs of non-negative integers to the set of non-negative even integers. This is straightforward but long.
A direct analytic argument can be given as follows: Sum the double series along the diagonals defined by $j + k = n$, and then sum on $n$.
\[
\sum_{j=0}^\infty \sum_{k=0}^\infty 2^{-3k-j-(k+j)^2} = \sum_{n=0}^\infty \sum_{k=0}^n 2^{-2k-n-n^2}
\]
\[
= \sum_{n=0}^\infty \left[ \frac{4}{3} (1 - 2^{-2n-2})2^{-n^2}\right]
\]
\[
= \frac{4}{3} \left[ \sum_{n=0}^\infty 2^{-n-n^2} - \sum_{n=0}^\infty 2^{-(n+1)(n+2)} \right]
\]
\[
= \frac{4}{3} \left[ \sum_{n=0}^\infty 2^{-n(n+1)} - \sum_{m=1}^\infty 2^{-m(m+1)} \right]
\]
\[
= \boxed{4/3}.
\]
At the third step, the separation of the sum into two sums is permissible because the new sums are convergent. | 0 |
1960 | 1960_12 | Define a sequence as follows:
\[
a_0 = 0, \quad a_1 = 1 + \sin(-1), \quad \dots, \quad a_n = 1 + \sin(a_{n-1} - 1), \dots
\]
Evaluate
\[
\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n a_k.
\] | Let $b_n = a_n - 1$. Then $b_0 = -1$ and
\[
(1) \quad b_n = \sin b_{n-1}, \quad n = 1, 2, 3, \dots.
\]
\[
\text{(Graph of $y = \sin x$ and $y = x$ showing iterations converging to zero.)}
\]
The polygonal representation of this recursion (see page 223) suggests that $b_n$ increases to $0$ as $n \to \infty$. Analytically, we note that for $-\pi < x < 0$, $x < \sin x < 0$. So from $-1 \leq b_{n-1} < 0$ there follows $b_{n-1} < b_n < 0$. Hence
\[
-1 = b_0 < b_1 < b_2 < \cdots < 0.
\]
Therefore $\lim b_n = c$ exists. Passing to the limit in (1) we obtain $\sin c = c$, so $c = 0$.
We next prove that
\[
(2) \quad \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n b_k = 0.
\]
Let $\varepsilon > 0$ be given and choose $p$ so that $b_k > -\varepsilon$ for $k > p$. Choose $m$ so that $m \varepsilon > p$ (and $m > p$). Then if $n > m$, we have
\[
\sum_{k=1}^n (-b_k) = \sum_{k=1}^p (-b_k) + \sum_{k=p+1}^n (-b_k) < p + n \varepsilon,
\]
and hence
\[
0 > \frac{1}{n} \sum_{k=1}^n b_k > -2\varepsilon.
\]
Since $\varepsilon$ was arbitrary, (2) follows.
Finally, we have
\[
\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n a_k = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \sum_{k=1}^n b_k \right) = \boxed{1}.
\] | numerical | putnam | Analysis Algebra | Define a sequence as follows:
\[
a_0 = 0, \quad a_1 = 1 + \sin(-1), \quad \dots, \quad a_n = 1 + \sin(a_{n-1} - 1), \dots
\]
Evaluate
\[
\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n a_k.
\] | Let $b_n = a_n - 1$. Then $b_0 = -1$ and
\[
(1) \quad b_n = \sin b_{n-1}, \quad n = 1, 2, 3, \dots.
\]
\[
\text{(Graph of $y = \sin x$ and $y = x$ showing iterations converging to zero.)}
\]
The polygonal representation of this recursion (see page 223) suggests that $b_n$ increases to $0$ as $n \to \infty$. Analytically, we note that for $-\pi < x < 0$, $x < \sin x < 0$. So from $-1 \leq b_{n-1} < 0$ there follows $b_{n-1} < b_n < 0$. Hence
\[
-1 = b_0 < b_1 < b_2 < \cdots < 0.
\]
Therefore $\lim b_n = c$ exists. Passing to the limit in (1) we obtain $\sin c = c$, so $c = 0$.
We next prove that
\[
(2) \quad \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n b_k = 0.
\]
Let $\varepsilon > 0$ be given and choose $p$ so that $b_k > -\varepsilon$ for $k > p$. Choose $m$ so that $m \varepsilon > p$ (and $m > p$). Then if $n > m$, we have
\[
\sum_{k=1}^n (-b_k) = \sum_{k=1}^p (-b_k) + \sum_{k=p+1}^n (-b_k) < p + n \varepsilon,
\]
and hence
\[
0 > \frac{1}{n} \sum_{k=1}^n b_k > -2\varepsilon.
\]
Since $\varepsilon$ was arbitrary, (2) follows.
Finally, we have
\[
\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n a_k = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \sum_{k=1}^n b_k \right) = \boxed{1}.
\] | 0 |
1961 | 1961_1 | The graph of the equation $x^y = y^x$ in the first quadrant (i.e., the region where $x > 0$ and $y > 0$) consists of a straight line and a curve. Find the coordinates of the intersection point of the line and the curve. | In the first quadrant the given equation is equivalent to
\[
(1) \quad \frac{1}{y} \log y = \frac{1}{x} \log x.
\]
Consider the function given by
\[
f(t) = \frac{1}{t} \log t \quad \text{for } t > 0.
\]
\[
\text{(Graph of $f(t)$ showing maximum value at $t = e$.)}
\]
Since $f'(t) = (1 - \log t)/t^2$, it is clear that $f$ is strictly increasing for $t \leq e$, is strictly decreasing for $t \geq e$, and achieves its maximum value $e^{-1}$ for $t = e$. Moreover, $f(t) \to -\infty$ as $t \to 0$, and $f(t) \to 0$ as $t \to \infty$. It follows that for $\alpha \in (0, e^{-1})$ the equation $f(t) = \alpha$ has two solutions, one in $(1, e)$, the other in $(e, \infty)$. For $\alpha$ near $0$, the lower solution is just above $1$ and the upper solution is large. As $\alpha$ increases to $e^{-1}$, the lower solution increases to $e$ and the upper solution decreases to $e$. Therefore, the locus (1) consists of the line $y = x$ and a curve $M$ lying in the quadrant $x > 1$, $y > 1$ and asymptotic to the line $x = 1$ and $y = 1$, as shown. $M$ is evidently symmetric in the line $y = x$ and crosses that line at $(e, e)$.
\[
\text{(Diagram showing the curve $M$ and line $y = x$ intersecting at $(e, e)$.)}
\]
We can establish the smoothness of the curve $M$ analytically. If $F$ is a smooth function (say $C^\infty$) defined on an open set in $\mathbb{R}^2$ then the level sets (contour lines) of $F$ are smooth curves except possibly at critical points of $F$ (i.e., points where both partial derivatives of $F$ vanish). At a critical point the structure of the level sets is the same as that of the level sets of the second Taylor polynomial of $F$, provided the second degree terms are a non-degenerate quadratic form.
Let
\[
F = \frac{1}{y} \log y - \frac{1}{x} \log x \quad \text{for } x > 0, y > 0.
\]
Then
\[
F'_x = -\frac{1}{x^2}(1 - \log x), \quad F'_y = \frac{1}{y^2}(1 - \log y),
\]
and the only critical point of $F$ is at $(e, e)$. At this point the second Taylor polynomial of $F$ is
\[
\frac{1}{2e^2}((x - e)^2 - (y - e)^2).
\]
This is a non-degenerate quadratic form that vanishes along the lines of slopes $+1$ and $-1$ through the point $(e, e)$. It follows that the level sets of $F$ are everywhere smooth curves except the one through $(e, e)$ which is locally the union of two smooth curves having slopes $+1$ and $-1$ at that point. Since this level set is the required locus, this proves that the curve $M$ described above is smooth. Thus the final answer is $\boxed{(e, e)}$. | algebraic | putnam | Analysis Geometry | The graph of the equation $x^y = y^x$ in the first quadrant (i.e., the region where $x > 0$ and $y > 0$) consists of a straight line and a curve. Find the coordinates of the intersection point of the line and the curve. | In the first quadrant the given equation is equivalent to
\[
(1) \quad \frac{1}{y} \log y = \frac{1}{x} \log x.
\]
Consider the function given by
\[
f(t) = \frac{1}{t} \log t \quad \text{for } t > 0.
\]
\[
\text{(Graph of $f(t)$ showing maximum value at $t = e$.)}
\]
Since $f'(t) = (1 - \log t)/t^2$, it is clear that $f$ is strictly increasing for $t \leq e$, is strictly decreasing for $t \geq e$, and achieves its maximum value $e^{-1}$ for $t = e$. Moreover, $f(t) \to -\infty$ as $t \to 0$, and $f(t) \to 0$ as $t \to \infty$. It follows that for $\alpha \in (0, e^{-1})$ the equation $f(t) = \alpha$ has two solutions, one in $(1, e)$, the other in $(e, \infty)$. For $\alpha$ near $0$, the lower solution is just above $1$ and the upper solution is large. As $\alpha$ increases to $e^{-1}$, the lower solution increases to $e$ and the upper solution decreases to $e$. Therefore, the locus (1) consists of the line $y = x$ and a curve $M$ lying in the quadrant $x > 1$, $y > 1$ and asymptotic to the line $x = 1$ and $y = 1$, as shown. $M$ is evidently symmetric in the line $y = x$ and crosses that line at $(e, e)$.
\[
\text{(Diagram showing the curve $M$ and line $y = x$ intersecting at $\boxed{(e, e)}$.)}
\]
We can establish the smoothness of the curve $M$ analytically. If $F$ is a smooth function (say $C^\infty$) defined on an open set in $\mathbb{R}^2$ then the level sets (contour lines) of $F$ are smooth curves except possibly at critical points of $F$ (i.e., points where both partial derivatives of $F$ vanish). At a critical point the structure of the level sets is the same as that of the level sets of the second Taylor polynomial of $F$, provided the second degree terms are a non-degenerate quadratic form.
Let
\[
F = \frac{1}{y} \log y - \frac{1}{x} \log x \quad \text{for } x > 0, y > 0.
\]
Then
\[
F'_x = -\frac{1}{x^2}(1 - \log x), \quad F'_y = \frac{1}{y^2}(1 - \log y),
\]
and the only critical point of $F$ is at $(e, e)$. At this point the second Taylor polynomial of $F$ is
\[
\frac{1}{2e^2}((x - e)^2 - (y - e)^2).
\]
This is a non-degenerate quadratic form that vanishes along the lines of slopes $+1$ and $-1$ through the point $(e, e)$. It follows that the level sets of $F$ are everywhere smooth curves except the one through $(e, e)$ which is locally the union of two smooth curves having slopes $+1$ and $-1$ at that point. Since this level set is the required locus, this proves that the curve $M$ described above is smooth. | 0 |
1961 | 1961_3 | Evaluate
\[
\lim_{n \to \infty} \sum_{i=1}^n \frac{n}{n^2 + j^2}.
\] | We write the sum in the form
\[
S_n = \frac{1}{n} \sum_{i=1}^n \frac{1}{1 + \left( \frac{j}{n} \right)^2}.
\]
Since
\[
\int_{\frac{i-1}{n}}^{\frac{i}{n}} \frac{dx}{1+x^2} < \frac{1}{n} \cdot \frac{1}{1 + \left( \frac{j}{n} \right)^2} < \int_{\frac{i}{n}}^{\frac{i+1}{n}} \frac{dx}{1+x^2},
\]
we get
\[
(1) \quad \int_{\frac{1}{n}}^{\frac{n+1}{n}} \frac{dx}{1+x^2} < S_n < \int_{0}^n \frac{dx}{1+x^2}.
\]
Now
\[
\int_0^n \frac{dx}{1+x^2} = \arctan n, \quad \lim_{n \to \infty} \arctan n = \frac{\pi}{2},
\]
and
\[
\int_{n+\frac{1}{n}}^n \frac{dx}{1+x^2} = \arctan \left( n + \frac{1}{n} \right) - \arctan \left( \frac{1}{n} \right).
\]
\[
\lim_{n \to \infty} \left( \arctan \left( n + \frac{1}{n} \right) - \arctan \left( \frac{1}{n} \right) \right) = \frac{\pi}{2}.
\]
Thus both the left and right members of (1) have the limit $\pi/2$, so
\[
\lim_{n \to \infty} S_n = \boxed{\frac{\pi}{2}}.
\] | numerical | putnam | Calculus Analysis | Evaluate
\[
\lim_{n \to \infty} \sum_{i=1}^n \frac{n}{n^2 + j^2}.
\] | We write the sum in the form
\[
S_n = \frac{1}{n} \sum_{i=1}^n \frac{1}{1 + \left( \frac{j}{n} \right)^2}.
\]
Since
\[
\int_{\frac{i-1}{n}}^{\frac{i}{n}} \frac{dx}{1+x^2} < \frac{1}{n} \cdot \frac{1}{1 + \left( \frac{j}{n} \right)^2} < \int_{\frac{i}{n}}^{\frac{i+1}{n}} \frac{dx}{1+x^2},
\]
we get
\[
(1) \quad \int_{\frac{1}{n}}^{\frac{n+1}{n}} \frac{dx}{1+x^2} < S_n < \int_{0}^n \frac{dx}{1+x^2}.
\]
Now
\[
\int_0^n \frac{dx}{1+x^2} = \arctan n, \quad \lim_{n \to \infty} \arctan n = \frac{\pi}{2},
\]
and
\[
\int_{n+\frac{1}{n}}^n \frac{dx}{1+x^2} = \arctan \left( n + \frac{1}{n} \right) - \arctan \left( \frac{1}{n} \right).
\]
\[
\lim_{n \to \infty} \left( \arctan \left( n + \frac{1}{n} \right) - \arctan \left( \frac{1}{n} \right) \right) = \frac{\pi}{2}.
\]
Thus both the left and right members of (1) have the limit $\pi/2$, so
\[
\lim_{n \to \infty} S_n = \boxed{\frac{\pi}{2}}.
\] | 0 |
1961 | 1961_4 | Define a function $f$ over the domain of positive integers as follows: $f(1) = 1$, and for $n > 1$, $f(n) = (-1)^k$ where $k$ is the total number of prime factors of $n$. For example $f(9) = (-1)^2$, $f(20) = (-1)^3$. Define $F(n)$ as $\sum_{d | n} f(d)$ where the sum ranges over all positive integer divisors of $n$. Prove that for every positive integer $n$, $F(n) = 0$ or $F(n) = 1$. For which integers $n$ is $F(n) = 1$? Find the sum of the first 5 positive integers for which $F$ is 1 and return that as the final answer. | Suppose $m$ and $n$ are relatively prime positive integers. Then every divisor of $mn$ is uniquely a product $d_1d_2$ where $d_1 | m$, $d_2 | n$, and conversely. Also $f(d_1d_2) = f(d_1)f(d_2)$.
\[
F(mn) = \sum_{d | mn} f(d) = \sum_{d_1 | m} \sum_{d_2 | n} f(d_1)f(d_2) = F(m)F(n).
\]
Thus $F$ is a multiplicative numerical function and it suffices to evaluate $F$ on prime powers. Evidently for $p$ a prime
\[
F(p^\alpha) = f(1) + f(p) + f(p^2) + \cdots + f(p^\alpha)
= 1 - 1 + 1 - \cdots + (-1)^\alpha
= \begin{cases}
1 & \text{if } \alpha \text{ is even,} \\
0 & \text{if } \alpha \text{ is odd.}
\end{cases}
\]
Let $n$ be any positive integer and suppose
\[
n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}
\]
is its canonical factorization into primes. Then $F(n) = F(p_1^{\alpha_1}) F(p_2^{\alpha_2}) \cdots F(p_k^{\alpha_k})$ and we see that $F(n) = 0$ if some prime appears with odd exponent in the prime factorization of $n$, and $F(n) = 1$ if all primes appear with even exponents. In other words,
\[
F(n) = 1
\]
if $n$ is a perfect square, and
\[
F(n) = 0
\]
if $n$ is not a perfect square. Thus the sum becomes \boxed{55}. | numerical | putnam (modified boxing) | Number Theory Algebra | Define a function $f$ over the domain of positive integers as follows: $f(1) = 1$, and for $n > 1$, $f(n) = (-1)^k$ where $k$ is the total number of prime factors of $n$. For example $f(9) = (-1)^2$, $f(20) = (-1)^3$. Define $F(n)$ as $\sum_{d | n} f(d)$ where the sum ranges over all positive integer divisors of $n$. Prove that for every positive integer $n$, $F(n) = 0$ or $F(n) = 1$. For which integers $n$ is $F(n) = 1$? | Suppose $m$ and $n$ are relatively prime positive integers. Then every divisor of $mn$ is uniquely a product $d_1d_2$ where $d_1 | m$, $d_2 | n$, and conversely. Also $f(d_1d_2) = f(d_1)f(d_2)$.
\[
F(mn) = \sum_{d | mn} f(d) = \sum_{d_1 | m} \sum_{d_2 | n} f(d_1)f(d_2) = F(m)F(n).
\]
Thus $F$ is a multiplicative numerical function and it suffices to evaluate $F$ on prime powers. Evidently for $p$ a prime
\[
F(p^\alpha) = f(1) + f(p) + f(p^2) + \cdots + f(p^\alpha)
= 1 - 1 + 1 - \cdots + (-1)^\alpha
= \begin{cases}
1 & \text{if } \alpha \text{ is even,} \\
0 & \text{if } \alpha \text{ is odd.}
\end{cases}
\]
Let $n$ be any positive integer and suppose
\[
n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}
\]
is its canonical factorization into primes. Then $F(n) = F(p_1^{\alpha_1}) F(p_2^{\alpha_2}) \cdots F(p_k^{\alpha_k})$ and we see that $F(n) = 0$ if some prime appears with odd exponent in the prime factorization of $n$, and $F(n) = 1$ if all primes appear with even exponents. In other words,
\[
F(n) = 1
\]
\boxed{if $n$ is a perfect square}, and
\[
F(n) = 0
\]
if $n$ is not a perfect square. | 0 |
1961 | 1961_5 | Let $\Omega$ be a set of $n$ points, where $n > 2$. Let $\Sigma$ be a nonempty subcollection of the $2^n$ subsets of $\Omega$ that is closed with respect to unions, intersections, and complements (that is, if $A$ and $B$ are members of $\Sigma$, then so are $A \cup B$, $A \cap B$, $\Omega - A$, and $\Omega - B$, where $\Omega - B$ denotes all points in $\Omega$ but not in $B$). If $k$ is the number of members of $\Sigma$, what are the possible values of $k$? Give the sum of the first 5 possible values of $k$. | Since $\Sigma$ is not empty, say it contains $A$. Then $\Sigma$ contains also $\Omega - A$ and $A \cap (\Omega - A) = \emptyset$. Hence also $\Omega = \Omega - \emptyset \in \Sigma$.
Among the non-empty members of $\Sigma$ certain are minimal in the sense that they do not contain any other member of $\Sigma$ except $\emptyset$. Any non-empty member $A$ of $\Sigma$ contains a minimal element, for example, a set $B$ of least cardinal satisfying $B \subseteq A$, $B \in \Sigma$, $B \neq \emptyset$.
Let $B_1, B_2, \ldots, B_p$ be a proper enumeration of all the minimal elements of $\Sigma$. These elements are mutually disjoint. Suppose $i \neq j$, then $B_i \cap B_j \in \Sigma$ and $B_i \cap B_j \subseteq B_i$. Hence by the minimality of $B_i$ either $B_i \cap B_j = \emptyset$ or $B_i \cap B_j = B_i$. The latter is impossible since it implies $B_j = B_i$ or $B_j = \emptyset$, both contradictions.
Now if $\Omega - (B_1 \cup B_2 \cup \cdots \cup B_p) \neq \emptyset$, then it would contain some minimal element of $\Sigma$, say $B_i$, and we would have
\[
B_i = B_i \cap [\Omega - (B_1 \cup B_2 \cup \cdots \cup B_p)] = \emptyset,
\]
a contradiction. So $B_1 \cup B_2 \cup \cdots \cup B_p = \Omega$.
Suppose $C \in \Sigma$. Then
\[
C = C \cap \Omega = C \cap (B_1 \cup \cdots \cup B_p) = (C \cap B_1) \cup (C \cap B_2) \cup \cdots \cup (C \cap B_p).
\]
Now each of the sets $C \cap B_i$ is either $\emptyset$ or $B_i$ (by the minimality of $B_i$). Thus we have shown: Every element of $\Sigma$ is the union of some subcollection of the sets $\{B_i\}$. Conversely every such union is a member of $\Sigma$. Moreover, since the sets $\{B_i\}$ are non-empty and mutually disjoint, distinct subcollections of $\{B_i\}$ have distinct unions. There are therefore exactly $2^p$ elements of $\Sigma$, one for each subset of $\{B_1, B_2, \ldots, B_p\}$. Thus $k = 2^r$.
Every power of 2 from 2 to $2^n$ is possible. Thus the final answer is \boxed{62}. | numerical | putnam (modified boxing) | Number Theory Algebra | Let $\Omega$ be a set of $n$ points, where $n > 2$. Let $\Sigma$ be a nonempty subcollection of the $2^n$ subsets of $\Omega$ that is closed with respect to unions, intersections, and complements (that is, if $A$ and $B$ are members of $\Sigma$, then so are $A \cup B$, $A \cap B$, $\Omega - A$, and $\Omega - B$, where $\Omega - B$ denotes all points in $\Omega$ but not in $B$). If $k$ is the number of members of $\Sigma$, what are the possible values of $k$? Give a proof. | Since $\Sigma$ is not empty, say it contains $A$. Then $\Sigma$ contains also $\Omega - A$ and $A \cap (\Omega - A) = \emptyset$. Hence also $\Omega = \Omega - \emptyset \in \Sigma$.
Among the non-empty members of $\Sigma$ certain are minimal in the sense that they do not contain any other member of $\Sigma$ except $\emptyset$. Any non-empty member $A$ of $\Sigma$ contains a minimal element, for example, a set $B$ of least cardinal satisfying $B \subseteq A$, $B \in \Sigma$, $B \neq \emptyset$.
Let $B_1, B_2, \ldots, B_p$ be a proper enumeration of all the minimal elements of $\Sigma$. These elements are mutually disjoint. Suppose $i \neq j$, then $B_i \cap B_j \in \Sigma$ and $B_i \cap B_j \subseteq B_i$. Hence by the minimality of $B_i$ either $B_i \cap B_j = \emptyset$ or $B_i \cap B_j = B_i$. The latter is impossible since it implies $B_j = B_i$ or $B_j = \emptyset$, both contradictions.
Now if $\Omega - (B_1 \cup B_2 \cup \cdots \cup B_p) \neq \emptyset$, then it would contain some minimal element of $\Sigma$, say $B_i$, and we would have
\[
B_i = B_i \cap [\Omega - (B_1 \cup B_2 \cup \cdots \cup B_p)] = \emptyset,
\]
a contradiction. So $B_1 \cup B_2 \cup \cdots \cup B_p = \Omega$.
Suppose $C \in \Sigma$. Then
\[
C = C \cap \Omega = C \cap (B_1 \cup \cdots \cup B_p) = (C \cap B_1) \cup (C \cap B_2) \cup \cdots \cup (C \cap B_p).
\]
Now each of the sets $C \cap B_i$ is either $\emptyset$ or $B_i$ (by the minimality of $B_i$). Thus we have shown: Every element of $\Sigma$ is the union of some subcollection of the sets $\{B_i\}$. Conversely every such union is a member of $\Sigma$. Moreover, since the sets $\{B_i\}$ are non-empty and mutually disjoint, distinct subcollections of $\{B_i\}$ have distinct unions. There are therefore exactly $2^p$ elements of $\Sigma$, one for each subset of $\{B_1, B_2, \ldots, B_p\}$. Thus $k = 2^r$.
\boxed{Every power of 2 from 2 to $2^n$ is possible}. | 0 |
1961 | 1961_8 | Let $\alpha_1, \alpha_2, \alpha_3, \ldots$ be a sequence of positive real numbers; define $s_n$ as $(\alpha_1 + \alpha_2 + \cdots + \alpha_n)/n$ and $r_n$ as $(\alpha_1^{-1} + \alpha_2^{-1} + \cdots + \alpha_n^{-1})/n$. Given that $\lim s_n$ and $\lim r_n$ exist as $n \to \infty$, find the lower bound of the product of these limits. | It is clearly sufficient to prove that $r_ns_n \geq 1$ for all $n$. Let $\beta_i = \alpha_i^{1/2}$ and $\gamma_i = \alpha_i^{-1/2}$. Then by the Cauchy-Schwarz inequality,
\[
n^2 = \left( \sum_{i=1}^n \beta_i \gamma_i \right)^2 \leq \left( \sum_{i=1}^n \beta_i^2 \right) \left( \sum_{i=1}^n \gamma_i^2 \right).
\]
Substituting $\beta_i^2 = \alpha_i$ and $\gamma_i^2 = \alpha_i^{-1}$, we obtain
\[
n^2 \leq \left( \sum_{i=1}^n \alpha_i \right) \left( \sum_{i=1}^n \alpha_i^{-1} \right).
\]
Dividing both sides by $n^2$, we get
\[
r_ns_n \geq 1.
\]
Taking the limits as $n \to \infty$, it follows that
\[
\lim_{n \to \infty} s_n \cdot \lim_{n \to \infty} r_n \geq \boxed{1}.
\] | numerical | putnam (modified boxing) | Analysis | Let $\alpha_1, \alpha_2, \alpha_3, \ldots$ be a sequence of positive real numbers; define $s_n$ as $(\alpha_1 + \alpha_2 + \cdots + \alpha_n)/n$ and $r_n$ as $(\alpha_1^{-1} + \alpha_2^{-1} + \cdots + \alpha_n^{-1})/n$. Given that $\lim s_n$ and $\lim r_n$ exist as $n \to \infty$, prove that the product of these limits is not less than $1$. | It is clearly sufficient to prove that $r_ns_n \geq 1$ for all $n$. Let $\beta_i = \alpha_i^{1/2}$ and $\gamma_i = \alpha_i^{-1/2}$. Then by the Cauchy-Schwarz inequality,
\[
n^2 = \left( \sum_{i=1}^n \beta_i \gamma_i \right)^2 \leq \left( \sum_{i=1}^n \beta_i^2 \right) \left( \sum_{i=1}^n \gamma_i^2 \right).
\]
Substituting $\beta_i^2 = \alpha_i$ and $\gamma_i^2 = \alpha_i^{-1}$, we obtain
\[
n^2 \leq \left( \sum_{i=1}^n \alpha_i \right) \left( \sum_{i=1}^n \alpha_i^{-1} \right).
\]
Dividing both sides by $n^2$, we get
\[
r_ns_n \geq 1.
\]
Taking the limits as $n \to \infty$, it follows that
\[
\lim_{n \to \infty} s_n \cdot \lim_{n \to \infty} r_n \geq 1.
\] | 0 |
1961 | 1961_9 | Let $\alpha$ and $\beta$ be given positive real numbers, with $\alpha < \beta$. If two points are selected at random from a straight line segment of length $\beta$, what is the probability that the distance between them is at least $\alpha$? | We interpret "at random" to mean that the pair of points $x, y$ is chosen so that the probability that $(x, y)$ falls in any region in the square $[0, \beta] \times [0, \beta]$ is proportional to the area of that region. Then the favorable region is evidently the union of the two triangular regions shown, and the probability of a favorable outcome is
\[
\frac{(\beta - \alpha)^2}{\beta^2} = \boxed{\left(1 - \frac{\alpha}{\beta}\right)^2}.
\] | algebraic | putnam | Probability | Let $\alpha$ and $\beta$ be given positive real numbers, with $\alpha < \beta$. If two points are selected at random from a straight line segment of length $\beta$, what is the probability that the distance between them is at least $\alpha$? | We interpret "at random" to mean that the pair of points $x, y$ is chosen so that the probability that $(x, y)$ falls in any region in the square $[0, \beta] \times [0, \beta]$ is proportional to the area of that region. Then the favorable region is evidently the union of the two triangular regions shown, and the probability of a favorable outcome is
\[
\frac{(\beta - \alpha)^2}{\beta^2} = \boxed{\left(1 - \frac{\alpha}{\beta}\right)^2}.
\] | 0 |
1961 | 1961_11 | For a fixed positive odd integer $n$, let $x_1, x_2, \ldots, x_n$ be real numbers satisfying $0 \leq x_k \leq 1$ for $k = 1, 2, \ldots, n$. Determine the maximum value, as a function of $n$, of the sum of the $n(n-1)/2$ terms:
\[
\sum_{1 \leq i < j \leq n} |x_i - x_j|.
\] | If we keep all but one of the $x$'s, say $x_2, x_3, \ldots, x_n$, fixed, then the sum in question is a convex function of $x_1$, since it is a sum of convex functions. Hence the maximum value is achieved either for $x_1 = 0$ or for $x_1 = 1$; conceivably it could be achieved for both and even for all intermediate values.
Since the set of $n$-tuples being considered is a bounded closed set in $\mathbb{R}^n$, a maximum is achieved at some point. We can then change any one of the $x$'s not already either a $0$ or a $1$ into either a $0$ or a $1$ without decreasing the sum. Hence among the points where the maximum is achieved there is at least one for which every $x$ is either a $0$ or a $1$. Consider this point.
If $p$ of the $x$'s are zero and $(n - p)$ are one, then the sum is clearly
\[
p(n - p) = \binom{n}{2}^2 - \binom{n/2}{2} - p.
\]
and this will be largest when $p = n/2$ if $n$ is even, or $p = (n \pm 1)/2$ if $n$ is odd.
The maximum value of the sum is therefore
\[
\frac{1}{4} n^2 \text{ if } n \text{ is even},
\]
\[
\boxed{\frac{1}{4}(n^2 - 1)} \text{ if } n \text{ is odd}.
\] | algebraic | putnam (modified boxing) | Analysis | For a fixed positive integer $n$, let $x_1, x_2, \ldots, x_n$ be real numbers satisfying $0 \leq x_k \leq 1$ for $k = 1, 2, \ldots, n$. Determine the maximum value, as a function of $n$, of the sum of the $n(n-1)/2$ terms:
\[
\sum_{1 \leq i < j \leq n} |x_i - x_j|.
\] | If we keep all but one of the $x$'s, say $x_2, x_3, \ldots, x_n$, fixed, then the sum in question is a convex function of $x_1$, since it is a sum of convex functions. Hence the maximum value is achieved either for $x_1 = 0$ or for $x_1 = 1$; conceivably it could be achieved for both and even for all intermediate values.
Since the set of $n$-tuples being considered is a bounded closed set in $\mathbb{R}^n$, a maximum is achieved at some point. We can then change any one of the $x$'s not already either a $0$ or a $1$ into either a $0$ or a $1$ without decreasing the sum. Hence among the points where the maximum is achieved there is at least one for which every $x$ is either a $0$ or a $1$. Consider this point.
If $p$ of the $x$'s are zero and $(n - p)$ are one, then the sum is clearly
\[
p(n - p) = \binom{n}{2}^2 - \binom{n/2}{2} - p.
\]
and this will be largest when $p = n/2$ if $n$ is even, or $p = (n \pm 1)/2$ if $n$ is odd.
The maximum value of the sum is therefore
\[
\boxed{\frac{1}{4} n^2 \text{ if } n \text{ is even},
\]
\[
\frac{1}{4}(n^2 - 1) \text{ if } n \text{ is odd}}.
\] | 0 |
1961 | 1961_13 | Consider the function $y(x)$ satisfying the differential equation
\[ y'' = -(1 + \sqrt{x})y \]
with $y(0) = 1$ and $y'(0) = 0$. Prove that $y(x)$ vanishes exactly once on the interval $0 < x < \pi^2$, and find a positive lower bound for the zero. | We shall apply the Sturm comparison theorem to the three functions determined by the following differential equations with initial conditions:
\[
u'' + 3u = 0, \quad u(0) = 1, \quad u'(0) = 0,\]
\[y'' + (1 + \sqrt{x})y = 0, \quad y(0) = 1, \quad y'(0) = 0,\]
\[v'' + v = 0, \quad v(0) = 1, \quad v'(0) = 0.\]
We see that \( u(x) = \cos \sqrt{3}x \) and \( v(x) = \cos x.\)
For $0 < x < \pi/2$, we have
\[3 > 1 + \sqrt{x} > 1;\]
hence by the Sturm theorem, the first zero of $u$, namely $\pi/(2\sqrt{3})$, occurs before the first zero of $y$, say $\xi$, and the first zero of $y$ occurs before the first zero of $v$, namely $\pi/2$. So we have
\[\pi/(2\sqrt{3}) < \xi < \pi/2.\]
Suppose $y$ had a second zero, say $\eta$, in $[0, \pi/2]$. Then by the Sturm theorem, a zero of $u$ would appear in $(\xi, \eta) \subseteq (\pi/(2\sqrt{3}), \pi/2)$. But $u$ has no such zero, so $y$ has but one zero in $[0, \pi/2]$. Thus the lower bound is \boxed{\pi/(2\sqrt{3})}. | numerical | putnam | Differential Equations | Consider the function $y(x)$ satisfying the differential equation
\[ y'' = -(1 + \sqrt{x})y \]
with $y(0) = 1$ and $y'(0) = 0$. Prove that $y(x)$ vanishes exactly once on the interval $0 < x < \pi^2$, and find a positive lower bound for the zero. | We shall apply the Sturm comparison theorem to the three functions determined by the following differential equations with initial conditions:
\[
u'' + 3u = 0, \quad u(0) = 1, \quad u'(0) = 0,\]
\[y'' + (1 + \sqrt{x})y = 0, \quad y(0) = 1, \quad y'(0) = 0,\]
\[v'' + v = 0, \quad v(0) = 1, \quad v'(0) = 0.\]
We see that \( u(x) = \cos \sqrt{3}x \) and \( v(x) = \cos x.\)
For $0 < x < \pi/2$, we have
\[3 > 1 + \sqrt{x} > 1;\]
hence by the Sturm theorem, the first zero of $u$, namely $\pi/(2\sqrt{3})$, occurs before the first zero of $y$, say $\xi$, and the first zero of $y$ occurs before the first zero of $v$, namely $\pi/2$. So we have
\[\boxed{\pi/(2\sqrt{3})} < \xi < \pi/2.\]
Suppose $y$ had a second zero, say $\eta$, in $[0, \pi/2]$. Then by the Sturm theorem, a zero of $u$ would appear in $(\xi, \eta) \subseteq (\pi/(2\sqrt{3}), \pi/2)$. But $u$ has no such zero, so $y$ has but one zero in $[0, \pi/2]$. | 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 |
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 |
1962 | 1962_5 | Evaluate in closed form: \( \sum_{k=1}^n \binom{n}{k} k^2. \) | First, rewrite \( \sum_{k=1}^n \binom{n}{k} k^2 \) using the properties of binomial coefficients:
\[ \sum_{k=1}^n \binom{n}{k} k^2 = \sum_{k=1}^n \frac{n!}{(n-k)!k!} \big[k(k-1) + k\big]. \]
Split this into two parts:
\[ \sum_{k=1}^n \binom{n}{k} k^2 = \sum_{k=2}^n \frac{n(n-1)(n-2)!}{(n-k)!(k-2)!} + \sum_{k=1}^n \frac{n(n-1)}{(n-k)!(k-1)!}. \]
Recognizing the first part as related to \( \sum_{j=0}^{n-2} \binom{n-2}{j} \), and the second as related to \( \sum_{i=0}^{n-1} \binom{n-1}{i} \), simplify using the binomial theorem:
\[ \sum_{k=1}^n \binom{n}{k} k^2 = n(n-1)2^{n-2} + n2^{n-1} = \boxed{n(n+1)2^{n-2}}. \] | algebraic | putnam | Combinatorics | Evaluate in closed form: \( \sum_{k=1}^n \binom{n}{k} k^2. \) | First, rewrite \( \sum_{k=1}^n \binom{n}{k} k^2 \) using the properties of binomial coefficients:
\[ \sum_{k=1}^n \binom{n}{k} k^2 = \sum_{k=1}^n \frac{n!}{(n-k)!k!} \big[k(k-1) + k\big]. \]
Split this into two parts:
\[ \sum_{k=1}^n \binom{n}{k} k^2 = \sum_{k=2}^n \frac{n(n-1)(n-2)!}{(n-k)!(k-2)!} + \sum_{k=1}^n \frac{n(n-1)}{(n-k)!(k-1)!}. \]
Recognizing the first part as related to \( \sum_{j=0}^{n-2} \binom{n-2}{j} \), and the second as related to \( \sum_{i=0}^{n-1} \binom{n-1}{i} \), simplify using the binomial theorem:
\[ \sum_{k=1}^n \binom{n}{k} k^2 = n(n-1)2^{n-2} + n2^{n-1} = \boxed{n(n+1)2^{n-2}}. \] | 0 |
1962 | 1962_11 | For every integer $n$ greater than 1 find the upper and lower bound of: \binom{1}{n}^n + \binom{2}{n}^n + \cdots + \binom{n}{n}^n < 2 and return the sum of the two. | For $n > 1$ and $x > 0$ the function $x^n$ is convex (i.e., its graph is concave up), so its integral is less than the approximation given by the trapezoidal rule. Hence \[ \frac{1}{n + 1} = \int_0^1 x^n dx < \frac{1}{n} \left[ \frac{1}{2} \binom{0}{n}^n + \binom{1}{n}^n + \binom{2}{n}^n + \cdots + \binom{n-1}{n}^n + \frac{1}{2} \binom{n}{n}^n \right]. \] Multiply by $n$ and add $1/2$ to get \[ \frac{3n + 1}{2n + 2} = \frac{n}{n + 1} + \frac{1}{2} < \binom{1}{n}^n + \binom{2}{n}^n + \cdots + \binom{n}{n}^n. \] The required upper approximation can be obtained by considering the lower Riemann sum over the same subdivision (and adding 1 to both sides), but the argument below gives a much sharper estimate. Since $1 - x \leq e^{-x}$ for any $x$, we have \[ \left(1 - \frac{i}{n}\right)^n \leq e^{-i} \] for $0 \leq i \leq n$. Hence \[ \binom{n}{n}^n + \binom{n-1}{n}^n + \cdots + \binom{2}{n}^n + \binom{1}{n}^n \leq 1 + e^{-1} + \cdots + e^{-(n-1)} \leq \frac{1}{1 - e^{-1}} = \frac{e}{e - 1} < 2. \] Since $\lim_{n \to \infty} \left(1 - \frac{i}{n}\right)^n = e^{-i}$, it follows easily that \[ \lim_{n \to \infty} \left[ \binom{n}{n}^n + \binom{n-1}{n}^n + \cdots + \binom{2}{n}^n + \binom{1}{n}^n \right] = \frac{e}{e - 1}. \] Thus the final sum is \boxed{frac{7n + 5}{2n + 2}}. | algebraic | putnam (modified boxing) | Algebra Analysis | Prove that for every integer $n$ greater than 1: \[ \frac{3n + 1}{2n + 2} < \binom{1}{n}^n + \binom{2}{n}^n + \cdots + \binom{n}{n}^n < 2. \] | For $n > 1$ and $x > 0$ the function $x^n$ is convex (i.e., its graph is concave up), so its integral is less than the approximation given by the trapezoidal rule. Hence \[ \frac{1}{n + 1} = \int_0^1 x^n dx < \frac{1}{n} \left[ \frac{1}{2} \binom{0}{n}^n + \binom{1}{n}^n + \binom{2}{n}^n + \cdots + \binom{n-1}{n}^n + \frac{1}{2} \binom{n}{n}^n \right]. \] Multiply by $n$ and add $1/2$ to get \[ \frac{3n + 1}{2n + 2} = \frac{n}{n + 1} + \frac{1}{2} < \binom{1}{n}^n + \binom{2}{n}^n + \cdots + \binom{n}{n}^n. \] The required upper approximation can be obtained by considering the lower Riemann sum over the same subdivision (and adding 1 to both sides), but the argument below gives a much sharper estimate. Since $1 - x \leq e^{-x}$ for any $x$, we have \[ \left(1 - \frac{i}{n}\right)^n \leq e^{-i} \] for $0 \leq i \leq n$. Hence \[ \binom{n}{n}^n + \binom{n-1}{n}^n + \cdots + \binom{2}{n}^n + \binom{1}{n}^n \leq 1 + e^{-1} + \cdots + e^{-(n-1)} \leq \frac{1}{1 - e^{-1}} = \frac{e}{e - 1} < 2. \] Since $\lim_{n \to \infty} \left(1 - \frac{i}{n}\right)^n = e^{-i}$, it follows easily that \[ \lim_{n \to \infty} \left[ \binom{n}{n}^n + \binom{n-1}{n}^n + \cdots + \binom{2}{n}^n + \binom{1}{n}^n \right] = \frac{e}{e - 1}. \] | 0 |
1963 | 1963_3 | Find an integral formula for the solution of the differential equation \[ \delta(\delta - 1)(\delta - 2) \cdots (\delta - n + 1)y = f(x), \quad x \geq 1, \] for $y$ as a function of $x$ satisfying the initial conditions \[ y(1) = y'(1) = \cdots = y^{(n-1)}(1) = 0, \] where $f$ is continuous and \[ \delta \equiv x \frac{d}{dx}. \] | We first show that \[ \delta(\delta - 1)(\delta - 2) \cdots (\delta - n + 1)y = x^n \frac{d^n y}{dx^n}. \] We prove this by induction on $n$. It is true for $n = 1$. Assume (1) is true for $n = k$. Since polynomials in the operator $\delta$ commute with one another, we have \[ \delta(\delta - 1)(\delta - 2) \cdots (\delta - k + 1)(\delta - k)y = (\delta - k)\delta(\delta - 1)(\delta - 2) \cdots (\delta - k + 1)y \] \[ = \left(x \frac{d}{dx} - k\right)x^k \frac{d^k y}{dx^k} \] \[ = x^{k+1} \frac{d^{k+1}y}{dx^{k+1}}. \] Thus (1) is proved by induction for all $n$. The differential equation is thus \[ x^n y^{(n)} = f(x), \quad x \geq 1, \] and the solution can obviously be obtained by applying the integral operator $\int_1^x n$ times to the function $f(x)x^{-n}$. However, it is possible to collapse the $n$-fold integration to a single integration as follows: \[ y(x) = \boxed{\int_1^x \frac{(x-t)^{n-1}}{(n-1)!} \frac{f(t)}{t^n} dt}. \] | algebraic | putnam | Differential Equations Analysis | Find an integral formula for the solution of the differential equation \[ \delta(\delta - 1)(\delta - 2) \cdots (\delta - n + 1)y = f(x), \quad x \geq 1, \] for $y$ as a function of $x$ satisfying the initial conditions \[ y(1) = y'(1) = \cdots = y^{(n-1)}(1) = 0, \] where $f$ is continuous and \[ \delta \equiv x \frac{d}{dx}. \] | We first show that \[ \delta(\delta - 1)(\delta - 2) \cdots (\delta - n + 1)y = x^n \frac{d^n y}{dx^n}. \] We prove this by induction on $n$. It is true for $n = 1$. Assume (1) is true for $n = k$. Since polynomials in the operator $\delta$ commute with one another, we have \[ \delta(\delta - 1)(\delta - 2) \cdots (\delta - k + 1)(\delta - k)y = (\delta - k)\delta(\delta - 1)(\delta - 2) \cdots (\delta - k + 1)y \] \[ = \left(x \frac{d}{dx} - k\right)x^k \frac{d^k y}{dx^k} \] \[ = x^{k+1} \frac{d^{k+1}y}{dx^{k+1}}. \] Thus (1) is proved by induction for all $n$. The differential equation is thus \[ x^n y^{(n)} = f(x), \quad x \geq 1, \] and the solution can obviously be obtained by applying the integral operator $\int_1^x n$ times to the function $f(x)x^{-n}$. However, it is possible to collapse the $n$-fold integration to a single integration as follows: \[ y(x) = \boxed{\int_1^x \frac{(x-t)^{n-1}}{(n-1)!} \frac{f(t)}{t^n} dt}. \] | 0 |
1963 | 1963_4 | Let $\{a_n\}$ be a sequence of positive real numbers. Find the lower bound of \[ \limsup_{n \to \infty} n \left(1 + \frac{a_{n+1}}{a_n} - 1\right). \] (The symbol $\limsup$ is sometimes written $\lim$.) | Suppose that for some fixed integer $k$, \[ n \left(1 + \frac{a_{n+1}}{a_n} - 1\right) \leq 1 \] for all $n \geq k$. Then \[ 1 + \frac{a_{n+1}}{a_n} \leq \frac{n+1}{n}, \] \[ \frac{a_n}{n} \geq \frac{1}{n+1} + \frac{a_{n+1}}{n+1}, \] for $n \geq k$. Accordingly we have \[ \frac{a_k}{k} \geq \frac{1}{k+1} + \frac{a_{k+1}}{k+1} \geq \frac{1}{k+1} + \frac{1}{k+1} + \frac{a_{k+2}}{k+2} \geq \frac{1}{k+1} + \frac{1}{k+2} + \cdots + \frac{1}{k+p} \] for each $p$. However, this is impossible since the harmonic series diverges. Thus for any $k$ there exists an $n \geq k$ such that \[ n \left(1 + \frac{a_{n+1}}{a_n} - 1\right) > 1. \] Therefore \[ \limsup_{n \to \infty} n \left(1 + \frac{a_{n+1}}{a_n} - 1\right) \geq 1. \] If we take $a_n = n \log n$, then \[ n \left(1 + \frac{a_{n+1}}{a_n} - 1\right) = \frac{1 + (n+1) \log(n+1) - n \log n}{\log n} \] \[ = \frac{1}{\log n} \left[1 + n \log \left(1 + \frac{1}{n}\right) + \log(n+1)\right] \] \[ \leq \frac{1}{\log n} \left[2 + \log(n+1)\right], \] since $\log(1+x) \leq x$ for all $x > -1$. Since the right side has limit 1, we have, for the particular sequence, \[ \limsup_{n \to \infty} n \left(1 + \frac{a_{n+1}}{a_n} - 1\right) \leq 1. \] Thus we cannot increase the bound \boxed{1}. We could also take $a_n = n^{1+\epsilon}$ where $\epsilon > 0$ and obtain \[ \limsup_{n \to \infty} n \left(1 + \frac{a_{n+1}}{a_n} - 1\right) = 1 + \epsilon. \] | numerical | putnam (modified boxing) | Analysis | Let $\{a_n\}$ be a sequence of positive real numbers. Show that \[ \limsup_{n \to \infty} n \left(1 + \frac{a_{n+1}}{a_n} - 1\right) \geq 1. \] Show that the number 1 on the right-hand side of this inequality cannot be replaced by any larger number. (The symbol $\limsup$ is sometimes written $\lim$.) | Suppose that for some fixed integer $k$, \[ n \left(1 + \frac{a_{n+1}}{a_n} - 1\right) \leq 1 \] for all $n \geq k$. Then \[ 1 + \frac{a_{n+1}}{a_n} \leq \frac{n+1}{n}, \] \[ \frac{a_n}{n} \geq \frac{1}{n+1} + \frac{a_{n+1}}{n+1}, \] for $n \geq k$. Accordingly we have \[ \frac{a_k}{k} \geq \frac{1}{k+1} + \frac{a_{k+1}}{k+1} \geq \frac{1}{k+1} + \frac{1}{k+1} + \frac{a_{k+2}}{k+2} \geq \frac{1}{k+1} + \frac{1}{k+2} + \cdots + \frac{1}{k+p} \] for each $p$. However, this is impossible since the harmonic series diverges. Thus for any $k$ there exists an $n \geq k$ such that \[ n \left(1 + \frac{a_{n+1}}{a_n} - 1\right) > 1. \] Therefore \[ \limsup_{n \to \infty} n \left(1 + \frac{a_{n+1}}{a_n} - 1\right) \geq 1. \] If we take $a_n = n \log n$, then \[ n \left(1 + \frac{a_{n+1}}{a_n} - 1\right) = \frac{1 + (n+1) \log(n+1) - n \log n}{\log n} \] \[ = \frac{1}{\log n} \left[1 + n \log \left(1 + \frac{1}{n}\right) + \log(n+1)\right] \] \[ \leq \frac{1}{\log n} \left[2 + \log(n+1)\right], \] since $\log(1+x) \leq x$ for all $x > -1$. Since the right side has limit 1, we have, for the particular sequence, \[ \limsup_{n \to \infty} n \left(1 + \frac{a_{n+1}}{a_n} - 1\right) \leq 1. \] Thus we cannot increase the bound 1. We could also take $a_n = n^{1+\epsilon}$ where $\epsilon > 0$ and obtain \[ \limsup_{n \to \infty} n \left(1 + \frac{a_{n+1}}{a_n} - 1\right) = 1 + \epsilon. \] | 0 |
1963 | 1963_7 | For what integer $a$ does $x^2 - x + a$ divide $x^{13} + x + 90$? | Suppose $x^{13} + x + 90 = (x^2 - x + a)q(x)$ where $a$ is an integer. Then $q$ is a polynomial with integer coefficients. If $a \leq 0$, then $x^2 - x + a$, and hence also $x^{13} + x + 90$, would have a non-negative zero, which is impossible. So $a > 0$. \nSubstituting $x = -1, 0, 1$, we find \[ (a + 2)q(-1) = 88, \] \[ aq(0) = 90, \] \[ aq(1) = 92. \] The last two equations show that $a$ divides 2, so $a = 1$ or $2$. But if $a = 1$, then 3 would divide 88. So $a = \boxed{2}$ is the only possibility and, in fact, \[ x^{13} + x + 90 = (x^2 - x + 2)(x^{11} + x^{10} - x^9 - 3x^8 - x^7 + 5x^6 + 7x^5 - 3x^4 - 7x^3 - 11x^2 + 23x + 45). \] | numerical | putnam | Algebra | For what integer $a$ does $x^2 - x + a$ divide $x^{13} + x + 90$? | Suppose $x^{13} + x + 90 = (x^2 - x + a)q(x)$ where $a$ is an integer. Then $q$ is a polynomial with integer coefficients. If $a \leq 0$, then $x^2 - x + a$, and hence also $x^{13} + x + 90$, would have a non-negative zero, which is impossible. So $a > 0$. \nSubstituting $x = -1, 0, 1$, we find \[ (a + 2)q(-1) = 88, \] \[ aq(0) = 90, \] \[ aq(1) = 92. \] The last two equations show that $a$ divides 2, so $a = 1$ or $2$. But if $a = 1$, then 3 would divide 88. So $a = \boxed{2}$ is the only possibility and, in fact, \[ x^{13} + x + 90 = (x^2 - x + 2)(x^{11} + x^{10} - x^9 - 3x^8 - x^7 + 5x^6 + 7x^5 - 3x^4 - 7x^3 - 11x^2 + 23x + 45). \] | 0 |
1963 | 1963_11 | Let \{a_n\} be a sequence of real numbers satisfying the inequalities \[ 0 \leq a_k \leq 100a_n \quad \text{for} \quad n \leq k \leq 2n \quad \text{and} \quad n = 1, 2, \dots, \] and such that the series \[ \sum_{n=0}^\infty a_n \] converges. Find the value of \[ \lim_{n \to \infty} na_n. \] | For each positive integer $k$, let \[ S_k = \sum_{n \geq k/2}^k a_n. \] Since $\sum a_n$ converges, we have $S_k \to 0$ as $k \to \infty$.\nRewriting the given condition slightly we have \[ 0 \leq a_k \leq 100a_n \quad \text{for} \quad \frac{1}{2}k \leq n \leq k. \] For each $k$ there are at least $k/2$ integers $n$ satisfying this double inequality.\nAdding these inequalities we have, therefore, \[ \frac{1}{2}ka_k \leq 100S_k. \] Hence \[ \limsup ka_k \leq 200 \lim S_k = 0. \] Since $a_k \geq 0$, we conclude $\lim ka_k = \boxed{0}$. | numerical | putnam (modified boxing) | Analysis | Let \{a_n\} be a sequence of real numbers satisfying the inequalities \[ 0 \leq a_k \leq 100a_n \quad \text{for} \quad n \leq k \leq 2n \quad \text{and} \quad n = 1, 2, \dots, \] and such that the series \[ \sum_{n=0}^\infty a_n \] converges. Prove that \[ \lim_{n \to \infty} na_n = 0. \] | For each positive integer $k$, let \[ S_k = \sum_{n \geq k/2}^k a_n. \] Since $\sum a_n$ converges, we have $S_k \to 0$ as $k \to \infty$.\nRewriting the given condition slightly we have \[ 0 \leq a_k \leq 100a_n \quad \text{for} \quad \frac{1}{2}k \leq n \leq k. \] For each $k$ there are at least $k/2$ integers $n$ satisfying this double inequality.\nAdding these inequalities we have, therefore, \[ \frac{1}{2}ka_k \leq 100S_k. \] Hence \[ \limsup ka_k \leq 200 \lim S_k = 0. \] Since $a_k \geq 0$, we conclude $\lim ka_k = 0$. | 0 |
1964 | 1964_1 | Given a set of 6 points in the plane, find the lower bound of the ratio of the longest distance between any pair to the shortest$. | Suppose three of the points, say $A, B, C$, are collinear in that order. Then either $|AC| \geq 2|BC|$ or $|AC| \geq 2|AB|$. Hence a ratio of at least two occurs in this case, so we assume from now on that no three of the given points are collinear.\n\nNext we prove that some three of the points form a triangle with one angle at least $120^\circ$.\n\nIf the convex hull of the given points is a hexagon, then the angle sum is $720^\circ$, and one of the angles must be at least $120^\circ$, so some three consecutive vertices of the hexagon form the required triangle.\n\nIf the convex hull of the given points has fewer than six vertices, then one of the points, say $P$, is in the interior of the convex hull of the others. $P$ must be in the interior of a triangle spanned by some other three of the given points (since we have disposed of the case where three points are collinear). Say $P$ lies inside the triangle $QRS$, where $Q, R, S$ are among the given points. Then one of the angles $QPR, RPS, SPQ$ must be at least $120^\circ$, since their sum is $360^\circ$.\n\nFinally, we show that in a triangle in which one angle is at least $120^\circ$, the ratio of the longest side to the shortest side is at least $\sqrt{3}$. Let $ABC$ be a triangle with $\angle A \geq 120^\circ$. Then $-\cos A \geq \frac{1}{2}$. Suppose $b \geq c$.\n\nBy the law of cosines\n\[ a^2 = b^2 + c^2 - 2bc \cos A \geq b^2 + c^2 + bc \geq 3c^2, \]\nso $a \geq \boxed{\sqrt{3}}c$, as claimed. | numerical | putnam (modified boxing) | Geometry | Given a set of 6 points in the plane, prove that the ratio of the longest distance between any pair to the shortest is at least $\sqrt{3}$. | Suppose three of the points, say $A, B, C$, are collinear in that order. Then either $|AC| \geq 2|BC|$ or $|AC| \geq 2|AB|$. Hence a ratio of at least two occurs in this case, so we assume from now on that no three of the given points are collinear.\n\nNext we prove that some three of the points form a triangle with one angle at least $120^\circ$.\n\nIf the convex hull of the given points is a hexagon, then the angle sum is $720^\circ$, and one of the angles must be at least $120^\circ$, so some three consecutive vertices of the hexagon form the required triangle.\n\nIf the convex hull of the given points has fewer than six vertices, then one of the points, say $P$, is in the interior of the convex hull of the others. $P$ must be in the interior of a triangle spanned by some other three of the given points (since we have disposed of the case where three points are collinear). Say $P$ lies inside the triangle $QRS$, where $Q, R, S$ are among the given points. Then one of the angles $QPR, RPS, SPQ$ must be at least $120^\circ$, since their sum is $360^\circ$.\n\nFinally, we show that in a triangle in which one angle is at least $120^\circ$, the ratio of the longest side to the shortest side is at least $\sqrt{3}$. Let $ABC$ be a triangle with $\angle A \geq 120^\circ$. Then $-\cos A \geq \frac{1}{2}$. Suppose $b \geq c$.\n\nBy the law of cosines\n\[ a^2 = b^2 + c^2 - 2bc \cos A \geq b^2 + c^2 + bc \geq 3c^2, \]\nso $a \geq \sqrt{3}c$, as claimed. | 0 |
1964 | 1964_3 | Let $P_1, P_2, \dots$ be a sequence of distinct points which is dense in the interval $(0, 1)$. The points $P_1, P_2, \dots, P_{n-1}$ decompose the interval into $n$ parts, and $P_n$ decomposes one of these into two parts. Let $a_n$ and $b_n$ be the lengths of these two intervals. Find the value of \[ \sum_{n=1}^\infty a_nb_n(a_n + b_n). \] (A sequence of points in an interval is said to be dense when every subinterval contains at least one point of the sequence.) | Let $S_n$ be the sum of the cubes of the lengths of the segments formed by the points $P_1, P_2, \dots, P_{n-1}, P_n$. Take $S_0 = 1$. We can obtain $S_n$ from $S_{n-1}$ by removing the term $(a_n + b_n)^3$ and replacing it by $a_n^3 + b_n^3$, leaving all other terms fixed.\n\nHence\n\[ S_{n-1} - S_n = (a_n + b_n)^3 - a_n^3 - b_n^3 = 3a_nb_n(a_n + b_n). \]\n\nTherefore\n\[ 1 - S_k = S_0 - S_k = \sum_{n=1}^k 3a_nb_n(a_n + b_n). \]\n\nSince, as we shall prove below, $\lim_{k \to \infty} S_k = 0$,\n\[ \sum_{n=1}^\infty 3a_nb_n(a_n + b_n) = 1, \]\nand the required result follows immediately.\n\nWe now show that $\lim_{k \to \infty} S_k = 0$. Let $t$ be a positive integer. Because the set $\{P_i\}$ is dense, we can choose an integer $q$ so large that $\{P_1, P_2, \dots, P_q\}$ meets each of the intervals\n\[ \left[0, \frac{1}{t}\right], \left[\frac{1}{t}, \frac{2}{t}\right], \dots, \left[\frac{t-1}{t}, 1\right]. \]\n\nSuppose $k \geq q$, and let $l_0, l_1, \dots, l_k$ be the lengths of the intervals determined by the division points $P_1, P_2, \dots, P_k$. Then each of these intervals has length at most $2/t$, so\n\[ S_k = \sum_{i=0}^k l_i^3 \leq \left(\frac{2}{t}\right)^2 \sum_{i=0}^k l_i = \frac{4}{t^2}. \]\n\nSince $t$ was arbitrary, this proves\n\[ \lim S_k = 0. \] Thus the value is \boxed{\frac{1}{3}}. | numerical | putnam (modified boxing) | Analysis | Let $P_1, P_2, \dots$ be a sequence of distinct points which is dense in the interval $(0, 1)$. The points $P_1, P_2, \dots, P_{n-1}$ decompose the interval into $n$ parts, and $P_n$ decomposes one of these into two parts. Let $a_n$ and $b_n$ be the lengths of these two intervals. Prove that \[ \sum_{n=1}^\infty a_nb_n(a_n + b_n) = \frac{1}{3}. \] (A sequence of points in an interval is said to be dense when every subinterval contains at least one point of the sequence.) | Let $S_n$ be the sum of the cubes of the lengths of the segments formed by the points $P_1, P_2, \dots, P_{n-1}, P_n$. Take $S_0 = 1$. We can obtain $S_n$ from $S_{n-1}$ by removing the term $(a_n + b_n)^3$ and replacing it by $a_n^3 + b_n^3$, leaving all other terms fixed.\n\nHence\n\[ S_{n-1} - S_n = (a_n + b_n)^3 - a_n^3 - b_n^3 = 3a_nb_n(a_n + b_n). \]\n\nTherefore\n\[ 1 - S_k = S_0 - S_k = \sum_{n=1}^k 3a_nb_n(a_n + b_n). \]\n\nSince, as we shall prove below, $\lim_{k \to \infty} S_k = 0$,\n\[ \sum_{n=1}^\infty 3a_nb_n(a_n + b_n) = 1, \]\nand the required result follows immediately.\n\nWe now show that $\lim_{k \to \infty} S_k = 0$. Let $t$ be a positive integer. Because the set $\{P_i\}$ is dense, we can choose an integer $q$ so large that $\{P_1, P_2, \dots, P_q\}$ meets each of the intervals\n\[ \left[0, \frac{1}{t}\right], \left[\frac{1}{t}, \frac{2}{t}\right], \dots, \left[\frac{t-1}{t}, 1\right]. \]\n\nSuppose $k \geq q$, and let $l_0, l_1, \dots, l_k$ be the lengths of the intervals determined by the division points $P_1, P_2, \dots, P_k$. Then each of these intervals has length at most $2/t$, so\n\[ S_k = \sum_{i=0}^k l_i^3 \leq \left(\frac{2}{t}\right)^2 \sum_{i=0}^k l_i = \frac{4}{t^2}. \]\n\nSince $t$ was arbitrary, this proves\n\[ \lim S_k = 0. \] | 0 |
1964 | 1964_5 | Prove that there is a constant $K$ such that the following inequality holds for any sequence of positive numbers $a_1, a_2, a_3, \dots$: \[ \sum_{n=1}^\infty \frac{n}{a_1 + a_2 + \cdots + a_n} \leq K \sum_{n=1}^\infty \frac{1}{a_n}. \] Find the value of K. | Let $k = 2t$ be some fixed even positive integer. We shall prove \[ \sum_{n=1}^k \frac{n}{a_1 + a_2 + \cdots + a_n} \leq 4 \sum_{n=1}^k \frac{1}{a_n}. \] From this inequality, the required inequality for infinite sums, with $K = 4$, follows immediately.\n\nLet $b_1, b_2, \dots, b_k$ be the terms $a_1, a_2, \dots, a_k$ enumerated in increasing order. For $1 \leq p \leq t$ we have \[ a_1 + a_2 + \cdots + a_{2p} \geq a_1 + a_2 + \cdots + a_{2p-1} \geq b_1 + b_2 + \cdots + b_{p-1} \geq pb_p, \] since all terms are positive and since the last $p$ terms are at least $b_p$. Therefore \[ \frac{2p-1}{a_1 + a_2 + \cdots + a_{2p-1}} \leq \frac{2p-1}{pb_p} < \frac{2}{b_p}. \] Also \[ \frac{2p}{a_1 + a_2 + \cdots + a_{2p}} \leq \frac{2p}{pb_p} = \frac{2}{b_p}. \]\n\nHence \[ \frac{2p-1}{a_1 + a_2 + \cdots + a_{2p-1}} + \frac{2p}{a_1 + a_2 + \cdots + a_{2p}} < \frac{4}{b_p}. \]\n\nThus \[ \sum_{p=1}^t \left[ \frac{2p-1}{a_1 + a_2 + \cdots + a_{2p-1}} + \frac{2p}{a_1 + a_2 + \cdots + a_{2p}} \right] < \sum_{p=1}^t \frac{4}{b_p}. \]\n\nRewriting the left-hand sum, we have \[ \sum_{n=1}^k \frac{n}{a_1 + \cdots + a_n} \leq \sum_{p=1}^t 4b_p < \sum_{p=1}^k 4b_p = 4 \sum_{n=1}^k \frac{1}{a_n}. \]\n\nIf the series $\sum_{n=1} (1/a_n)$ diverges, the given inequality is deemed to be satisfied automatically. Otherwise we have \[ \sum_{n=1}^k \frac{n}{a_1 + \cdots + a_n} \leq 4 \sum_{n=1}^\infty \frac{1}{a_n}. \]\n\nAnd hence \[ \sum_{n=1}^\infty \frac{n}{a_1 + a_2 + \cdots + a_n} \leq \boxed{4} \sum_{n=1}^\infty \frac{1}{a_n}. \] | numerical | putnam (modified boxing) | Analysis | Prove that there is a constant $K$ such that the following inequality holds for any sequence of positive numbers $a_1, a_2, a_3, \dots$: \[ \sum_{n=1}^\infty \frac{n}{a_1 + a_2 + \cdots + a_n} \leq K \sum_{n=1}^\infty \frac{1}{a_n}. \] | Let $k = 2t$ be some fixed even positive integer. We shall prove \[ \sum_{n=1}^k \frac{n}{a_1 + a_2 + \cdots + a_n} \leq 4 \sum_{n=1}^k \frac{1}{a_n}. \] From this inequality, the required inequality for infinite sums, with $K = 4$, follows immediately.\n\nLet $b_1, b_2, \dots, b_k$ be the terms $a_1, a_2, \dots, a_k$ enumerated in increasing order. For $1 \leq p \leq t$ we have \[ a_1 + a_2 + \cdots + a_{2p} \geq a_1 + a_2 + \cdots + a_{2p-1} \geq b_1 + b_2 + \cdots + b_{p-1} \geq pb_p, \] since all terms are positive and since the last $p$ terms are at least $b_p$. Therefore \[ \frac{2p-1}{a_1 + a_2 + \cdots + a_{2p-1}} \leq \frac{2p-1}{pb_p} < \frac{2}{b_p}. \] Also \[ \frac{2p}{a_1 + a_2 + \cdots + a_{2p}} \leq \frac{2p}{pb_p} = \frac{2}{b_p}. \]\n\nHence \[ \frac{2p-1}{a_1 + a_2 + \cdots + a_{2p-1}} + \frac{2p}{a_1 + a_2 + \cdots + a_{2p}} < \frac{4}{b_p}. \]\n\nThus \[ \sum_{p=1}^t \left[ \frac{2p-1}{a_1 + a_2 + \cdots + a_{2p-1}} + \frac{2p}{a_1 + a_2 + \cdots + a_{2p}} \right] < \sum_{p=1}^t \frac{4}{b_p}. \]\n\nRewriting the left-hand sum, we have \[ \sum_{n=1}^k \frac{n}{a_1 + \cdots + a_n} \leq \sum_{p=1}^t 4b_p < \sum_{p=1}^k 4b_p = 4 \sum_{n=1}^k \frac{1}{a_n}. \]\n\nIf the series $\sum_{n=1} (1/a_n)$ diverges, the given inequality is deemed to be satisfied automatically. Otherwise we have \[ \sum_{n=1}^k \frac{n}{a_1 + \cdots + a_n} \leq 4 \sum_{n=1}^\infty \frac{1}{a_n}. \]\n\nAnd hence \[ \sum_{n=1}^\infty \frac{n}{a_1 + a_2 + \cdots + a_n} \leq 4 \sum_{n=1}^\infty \frac{1}{a_n}. \] | 0 |
1964 | 1964_7 | Let $u_k$ $(k = 1, 2, \dots)$ be a sequence of integers, and let $V_n$ be the number of those which are less than or equal to $n$. If \[ \sum_{k=1}^\infty \frac{1}{u_k} < \infty, \] then find the value of \[ \lim_{n \to \infty} \frac{V_n}{n}. \] | It must have been intended that the $u$'s be positive, since otherwise we could have $\sum 1/u_k$ convergent but all the $V$'s infinite. Hence we assume that $\sum 1/u_k$ is a convergent series of positive terms. Then $u_k \to \infty$, and the $V$'s are finite. Since the convergence of a series of positive terms is not affected by rearrangement, we can assume that $u_1 \leq u_2 \leq u_3 \leq \cdots$. Then, for a fixed $p$ and any $n$ so large that $V_n > p$ we have \[ \sum_{k=p+1}^{V_n} \frac{1}{u_k} \geq \frac{V_n - p}{n}, \] because there are $V_n - p$ terms in the sum and each is at least $1/n$. Therefore \[ \limsup_{n \to \infty} \frac{V_n}{n} \leq \sum_{k=p+1}^\infty \frac{1}{u_k}. \] Since this is true for any $p$ we have \[ \limsup_{n \to \infty} \frac{V_n}{n} \leq \lim_{p \to \infty} \sum_{k=p+1}^\infty \frac{1}{u_k} = 0. \] But $\frac{V_n}{n} \geq 0$ for all $n$, and therefore $\lim_{n \to \infty} \frac{V_n}{n} = \boxed{0}.$ | numerical | putnam (modified boxing) | Analysis | Let $u_k$ $(k = 1, 2, \dots)$ be a sequence of integers, and let $V_n$ be the number of those which are less than or equal to $n$. Show that if \[ \sum_{k=1}^\infty \frac{1}{u_k} < \infty, \] then \[ \lim_{n \to \infty} \frac{V_n}{n} = 0. \] | It must have been intended that the $u$'s be positive, since otherwise we could have $\sum 1/u_k$ convergent but all the $V$'s infinite. Hence we assume that $\sum 1/u_k$ is a convergent series of positive terms. Then $u_k \to \infty$, and the $V$'s are finite. Since the convergence of a series of positive terms is not affected by rearrangement, we can assume that $u_1 \leq u_2 \leq u_3 \leq \cdots$. Then, for a fixed $p$ and any $n$ so large that $V_n > p$ we have \[ \sum_{k=p+1}^{V_n} \frac{1}{u_k} \geq \frac{V_n - p}{n}, \] because there are $V_n - p$ terms in the sum and each is at least $1/n$. Therefore \[ \limsup_{n \to \infty} \frac{V_n}{n} \leq \sum_{k=p+1}^\infty \frac{1}{u_k}. \] Since this is true for any $p$ we have \[ \limsup_{n \to \infty} \frac{V_n}{n} \leq \lim_{p \to \infty} \sum_{k=p+1}^\infty \frac{1}{u_k} = 0. \] But $\frac{V_n}{n} \geq 0$ for all $n$, and therefore $\lim_{n \to \infty} \frac{V_n}{n} = 0.$ | 0 |
1964 | 1964_8 | Let $S$ be a set of $n > 0$ elements, and let $A_1, A_2, \dots, A_k$ be a family of distinct subsets, with the property that any two of these subsets meet. Assume that no other subset of $S$ meets all of the $A_i$. Find the value of $k$. | There are $2^n$ distinct subsets of $S$ which we may consider as being arranged in $2^{n-1}$ complementary pairs. If $k$, the number of subsets in the family, is greater than $2^{n-1}$, then two of these subsets would be complementary, and these two subsets would not intersect. On the other hand, if $k < 2^{n-1}$, we can find two complementary sets $X$ and $Y$, neither of which is among the given sets $A_1, A_2, \dots, A_k$. But no set not among the $A$'s meets all the $A$'s, so there is a set $A_i$ with $A_i \cap X = \emptyset$, i.e., $A_i \subseteq Y$. Similarly there is a set $A_j$ with $A_j \subseteq X$. But then $A_i \cap A_j \subseteq X \cap Y = \emptyset$, which contradicts the fact that any two of the $A$'s meet. So the only possibility is $k = \boxed{2^{n-1}}$. | algebraic | putnam (modified boxing) | Combinatorics | Let $S$ be a set of $n > 0$ elements, and let $A_1, A_2, \dots, A_k$ be a family of distinct subsets, with the property that any two of these subsets meet. Assume that no other subset of $S$ meets all of the $A_i$. Prove that $k = 2^{n-1}$. | There are $2^n$ distinct subsets of $S$ which we may consider as being arranged in $2^{n-1}$ complementary pairs. If $k$, the number of subsets in the family, is greater than $2^{n-1}$, then two of these subsets would be complementary, and these two subsets would not intersect. On the other hand, if $k < 2^{n-1}$, we can find two complementary sets $X$ and $Y$, neither of which is among the given sets $A_1, A_2, \dots, A_k$. But no set not among the $A$'s meets all the $A$'s, so there is a set $A_i$ with $A_i \cap X = \emptyset$, i.e., $A_i \subseteq Y$. Similarly there is a set $A_j$ with $A_j \subseteq X$. But then $A_i \cap A_j \subseteq X \cap Y = \emptyset$, which contradicts the fact that any two of the $A$'s meet. So the only possibility is $k = 2^{n-1}$. | 0 |
1964 | 1964_9 | Let $f(x)$ be a real continuous function defined for all real $x$. Assume that for every $\epsilon > 0$ \[ \lim_{n \to \infty} f(n\epsilon) = 0, \] (where $n$ is a positive integer). Find the value of \[ \lim_{x \to \infty} f(x). \] | We begin by proving the following fact. \\ \textbf{Lemma.} If $0 < a < b$ and $k$ is any positive integer, then \[ \bigcup_{n=k}^\infty [na, nb] \] contains the ray $[c, \infty)$ for some $c$. \\ \textbf{Proof.} Suppose $t$ is an integer such that $t \geq k$, $t \geq a/(b-a)$. If $n \geq t$, then $n(b-a) \geq a$ so $nb \geq (n+1)a$. Hence the intervals $[na, nb]$ and $[(n+1)a, (n+1)b]$ overlap. Therefore \[ \bigcup_{n=k}^\infty [na, nb] \supseteq [ta, \infty). \] \(\text{\textbullet}\)\\ We return to the problem. Let $\alpha > 0$ be given. Define \\ \[ F_n = \{x : |f(nx)| \leq \alpha \} \] and \\ \[ E_k = \{x : (\forall n \geq k) |f(nx)| \leq \alpha \}. \] Then $E_k = \bigcap_{n \geq k} F_n$. Because $f$ is continuous, each $F_n$ is closed, and therefore each $E_k$ is closed. If $y \in (0, \infty)$, then $\lim_{n \to \infty} f(ny) = 0$; hence for some $k$ and all $n \geq k$, $|f(ny)| \leq \alpha$; that is, $y \in E_k$. Thus $(0, \infty) \subseteq \bigcup E_k$. By the Baire category theorem (proof below) one of the $E$'s, say $E_m$, contains an interval $[a, b]$. This means \\ \[ (\forall x \in [a, b])(\forall n \geq m) |f(nx)| \leq \alpha. \] Therefore \\ \[ (\forall y \in \bigcup_{n=m} [na, nb]) |f(y)| \leq \alpha. \] Choose $c$ so that \[ \bigcup_{n=m} [na, nb] \supseteq [c, \infty). \] Then \\ \[ (\forall y \geq c) |f(y)| \leq \alpha. \] Since $\alpha$ was arbitrary, this proves that \[ \lim_{y \to \infty} f(y) = \boxed{0}, \] as required. | numerical | putnam (modified boxing) | Analysis | Let $f(x)$ be a real continuous function defined for all real $x$. Assume that for every $\epsilon > 0$ \[ \lim_{n \to \infty} f(n\epsilon) = 0, \] (where $n$ is a positive integer). Prove that \[ \lim_{x \to \infty} f(x) = 0. \] | We begin by proving the following fact. \\ \textbf{Lemma.} If $0 < a < b$ and $k$ is any positive integer, then \[ \bigcup_{n=k}^\infty [na, nb] \] contains the ray $[c, \infty)$ for some $c$. \\ \textbf{Proof.} Suppose $t$ is an integer such that $t \geq k$, $t \geq a/(b-a)$. If $n \geq t$, then $n(b-a) \geq a$ so $nb \geq (n+1)a$. Hence the intervals $[na, nb]$ and $[(n+1)a, (n+1)b]$ overlap. Therefore \[ \bigcup_{n=k}^\infty [na, nb] \supseteq [ta, \infty). \] \(\text{\textbullet}\)\\ We return to the problem. Let $\alpha > 0$ be given. Define \\ \[ F_n = \{x : |f(nx)| \leq \alpha \} \] and \\ \[ E_k = \{x : (\forall n \geq k) |f(nx)| \leq \alpha \}. \] Then $E_k = \bigcap_{n \geq k} F_n$. Because $f$ is continuous, each $F_n$ is closed, and therefore each $E_k$ is closed. If $y \in (0, \infty)$, then $\lim_{n \to \infty} f(ny) = 0$; hence for some $k$ and all $n \geq k$, $|f(ny)| \leq \alpha$; that is, $y \in E_k$. Thus $(0, \infty) \subseteq \bigcup E_k$. By the Baire category theorem (proof below) one of the $E$'s, say $E_m$, contains an interval $[a, b]$. This means \\ \[ (\forall x \in [a, b])(\forall n \geq m) |f(nx)| \leq \alpha. \] Therefore \\ \[ (\forall y \in \bigcup_{n=m} [na, nb]) |f(y)| \leq \alpha. \] Choose $c$ so that \[ \bigcup_{n=m} [na, nb] \supseteq [c, \infty). \] Then \\ \[ (\forall y \geq c) |f(y)| \leq \alpha. \] Since $\alpha$ was arbitrary, this proves that \[ \lim_{y \to \infty} f(y) = 0, \] as required. | 0 |
1964 | 1964_10 | Into how many regions do $n$ great circles (no three concurrent) decompose the surface of the sphere on which they lie? | Let $f(n)$ be the number of regions on the surface of a sphere formed by $n$ great circles of which no three are concurrent. Clearly $f(1) = 2, f(2) = 4$. Suppose $n$ circles have been drawn and an $(n + 1)$st circle is added. The new circle meets each of the old ones in two points, making $2n$ points of intersection, and these $2n$ points are all different since no three circles are concurrent. The $2n$ points divide the new circle into $2n$ arcs. Each of these arcs divides one of the old regions into two parts. Thus there are $2n + f(n)$ regions formed by the $(n + 1)$ circles. Hence we have \\ \[ f(n + 1) = 2n + f(n), \quad \text{for } n \geq 1. \] \\ It follows easily by induction that \\ \[ f(n) = \boxed{n^2 - n + 2} \quad \text{for } n \geq 1. \] \\ Obviously, $f(0) = 1$. Note that the argument leading to (1) breaks down if $n = 0.$ | algebraic | putnam | Geometry | Into how many regions do $n$ great circles (no three concurrent) decompose the surface of the sphere on which they lie? | Let $f(n)$ be the number of regions on the surface of a sphere formed by $n$ great circles of which no three are concurrent. Clearly $f(1) = 2, f(2) = 4$. Suppose $n$ circles have been drawn and an $(n + 1)$st circle is added. The new circle meets each of the old ones in two points, making $2n$ points of intersection, and these $2n$ points are all different since no three circles are concurrent. The $2n$ points divide the new circle into $2n$ arcs. Each of these arcs divides one of the old regions into two parts. Thus there are $2n + f(n)$ regions formed by the $(n + 1)$ circles. Hence we have \\ \[ f(n + 1) = 2n + f(n), \quad \text{for } n \geq 1. \] \\ It follows easily by induction that \\ \[ f(n) = \boxed{n^2 - n + 2} \quad \text{for } n \geq 1. \] \\ Obviously, $f(0) = 1$. Note that the argument leading to (1) breaks down if $n = 0.$ | 0 |
1965 | 1965_A1 | Let $ABC$ be a triangle with angle $A < \angle C < 90^\circ < \angle B$. Consider the bisectors of the external angles at $A$ and $B$, each measured from the vertex to the opposite side (extended). Suppose both of these line-segments are equal to $AB$. Compute the angle $A$. | Suppose the bisector of the exterior angle at $A$ intersects line $BC$ at $X$ and the bisector of the exterior angle at $B$ meets the line $AC$ at $Y$. The assumption that $C$ is between $B$ and $X$ contradicts the fact that $\angle B > \angle C$, so we may assume that $B$ is between $X$ and $C$. Similarly, we conclude that $C$ is between $A$ and $Y$ because $\angle A < \angle C$.
If $Z$ is a point on line $AB$ with $B$ between $A$ and $Z$, we have from triangle $ABY$ that $\angle ZBY = 2A$. Hence, $\angle BXA = \angle ABX = \angle ZBC = 2 \angle ZBY = 4A$, and the angle sum of triangle $ABX$ is $90^\circ - \frac{1}{2}A + 8A$. Thus, $A = \boxed{12}^\circ$. | numerical | putnam | Geometry | Let $ABC$ be a triangle with angle $A < \angle C < 90^\circ < \angle B$. Consider the bisectors of the external angles at $A$ and $B$, each measured from the vertex to the opposite side (extended). Suppose both of these line-segments are equal to $AB$. Compute the angle $A$. | Suppose the bisector of the exterior angle at $A$ intersects line $BC$ at $X$ and the bisector of the exterior angle at $B$ meets the line $AC$ at $Y$. The assumption that $C$ is between $B$ and $X$ contradicts the fact that $\angle B > \angle C$, so we may assume that $B$ is between $X$ and $C$. Similarly, we conclude that $C$ is between $A$ and $Y$ because $\angle A < \angle C$.
If $Z$ is a point on line $AB$ with $B$ between $A$ and $Z$, we have from triangle $ABY$ that $\angle ZBY = 2A$. Hence, $\angle BXA = \angle ABX = \angle ZBC = 2 \angle ZBY = 4A$, and the angle sum of triangle $ABX$ is $90^\circ - \frac{1}{2}A + 8A$. Thus, $A = \boxed{12}^\circ$. | 0 |
1965 | 1965_A2 | Given any positive integer $n$ find the value of \n\[ \sum_{r=0}^{\lfloor (n-1)/2 \rfloor} \left\{\frac{n - 2r}{n}\binom{n}{r}\right\}^2, \]\n where $\lfloor x \rfloor$ means the greatest integer not exceeding $x$, and $\binom{n}{r}$ is the binomial coefficient "$n$ choose $r$," with the convention $\binom{0}{0} = 1$. Return your final answer in binomial coefficient form with all other multiplicants reduced to the lowest form. | Substituting $s=n-r$ in the given summation reveals that twice this sum is equal to:
\[
\sum_{r=0}^n \left(\frac{n-2r}{n} \binom{n}{r}\right)^2 = \sum \left(1 - 2\frac{r}{n}\right)^2 \binom{n}{r}^2 = \sum \binom{n}{r}^2 - 4\sum \frac{r}{n} \binom{n}{r}^2 + 4\sum \left(\frac{r}{n}\right)^2 \binom{n}{r}^2.
\]
\[
= \binom{2n}{n} - 4 \sum_{r=1}^n \binom{n-1}{r-1} \binom{n}{r} + 4 \sum_{r=1}^n \binom{n-1}{r-1}^2.
\]
\[
= \binom{2n}{n} - 4\binom{2n-1}{n-1} + 4\binom{2n-2}{n-1}.
\]
\[
= \frac{2n(2n-1)}{n^2} - \frac{4(n-1)}{n}\binom{2n-2}{n-1} = \boxed{\frac{1}{n}\binom{2n-2}{n-1}}.\n\] | algebraic | putnam (modified boxing) | Combinatorics Algebra | Show that, for any positive integer $n$, \n\[ \sum_{r=0}^{\lfloor (n-1)/2 \rfloor} \left\{\frac{n - 2r}{n}\binom{n}{r}\right\}^2 = \frac{1}{n}\binom{2n-2}{n-1}, \]
where $\lfloor x \rfloor$ means the greatest integer not exceeding $x$, and $\binom{n}{r}$ is the binomial coefficient "$n$ choose $r$," with the convention $\binom{0}{0} = 1$. | Substituting $s=n-r$ in the given summation reveals that twice this sum is equal to:
\[
\sum_{r=0}^n \left(\frac{n-2r}{n} \binom{n}{r}\right)^2 = \sum \left(1 - 2\frac{r}{n}\right)^2 \binom{n}{r}^2 = \sum \binom{n}{r}^2 - 4\sum \frac{r}{n} \binom{n}{r}^2 + 4\sum \left(\frac{r}{n}\right)^2 \binom{n}{r}^2.
\]
\[
= \binom{2n}{n} - 4 \sum_{r=1}^n \binom{n-1}{r-1} \binom{n}{r} + 4 \sum_{r=1}^n \binom{n-1}{r-1}^2.
\]
\[
= \binom{2n}{n} - 4\binom{2n-1}{n-1} + 4\binom{2n-2}{n-1}.
\]
\[
= \frac{2n(2n-1)}{n^2} - \frac{4(n-1)}{n}\binom{2n-2}{n-1} = \frac{1}{n}\binom{2n-2}{n-1}.\n\]
Comment: This solution assumes the well-known identities:\n\[\n\sum_r \binom{n}{r}^2 = \binom{2n}{n} \quad \text{and} \quad \sum_{r=0}^k \binom{m}{k-r}\binom{n}{r} = \binom{m+n}{k},\n\]\n which may be proved by comparing coefficients in the expansion of\n\[\n(1+x)^m \cdot (1+x)^n = (1+x)^{m+n}.\n\] | 0 |
1965 | 1965_A5 | In how many ways can the integers from 1 to $n$ be ordered subject to the condition that, except for the first integer on the left, every integer differs by 1 from some integer to the left of it? | On the basis of the first few cases we conjecture the answer is $2^{n-1}$ and proceed by induction.
We first show (also by induction) that an $n$-arrangement ends in 1 or $n$. Note that when $n$ is deleted from an $n$-arrangement, the result is an $(n-1)$-arrangement. If an $n$-arrangement does not end in 1 or $n$, deletion of $n$ produces an $(n-1)$-arrangement ending (by induction) in $(n-1)$. This implies the $n$-arrangement ended in $n$ because $n$ cannot precede $(n-1)$.
For any $n$-arrangement $(a_1, a_2, \cdots, a_n)$ there is another $n$-arrangement $(a_1', a_2', \cdots, a_n')$, where $a_i' = n+1-a_i$. If one of these ends in $n$, the other ends in 1 and consequently exactly half of the $n$-arrangements end in $n$.
All of the $n$-arrangements which end in $n$ can be obtained by adjoining an $n$ to the end of all $(n-1)$-arrangements, and by induction there are $2^{n-2}$ of these. Hence, there are $\boxed{2^{n-1}}$ $n$-arrangements. | numerical | putnam | Combinatorics | In how many ways can the integers from 1 to $n$ be ordered subject to the condition that, except for the first integer on the left, every integer differs by 1 from some integer to the left of it? | On the basis of the first few cases we conjecture the answer is $2^{n-1}$ and proceed by induction.
We first show (also by induction) that an $n$-arrangement ends in 1 or $n$. Note that when $n$ is deleted from an $n$-arrangement, the result is an $(n-1)$-arrangement. If an $n$-arrangement does not end in 1 or $n$, deletion of $n$ produces an $(n-1)$-arrangement ending (by induction) in $(n-1)$. This implies the $n$-arrangement ended in $n$ because $n$ cannot precede $(n-1)$.
For any $n$-arrangement $(a_1, a_2, \cdots, a_n)$ there is another $n$-arrangement $(a_1', a_2', \cdots, a_n')$, where $a_i' = n+1-a_i$. If one of these ends in $n$, the other ends in 1 and consequently exactly half of the $n$-arrangements end in $n$.
All of the $n$-arrangements which end in $n$ can be obtained by adjoining an $n$ to the end of all $(n-1)$-arrangements, and by induction there are $2^{n-2}$ of these. Hence, there are $\boxed{2^{n-1}}$ $n$-arrangements. | 0 |
1965 | 1965_B1 | Evaluate
\[
\lim_{n \to \infty} \int_0^1 \int_0^1 \cdots \int_0^1 \cos^2 \left(\frac{\pi}{2n}(x_1 + x_2 + \cdots + x_n)\right) dx_1 dx_2 \cdots dx_n.
\] | The change of variables $x_k \to 1 - x_k$ yields
\[
\int_0^1 \int_0^1 \cdots \int_0^1 \cos^2 \left(\frac{\pi}{2n}(x_1 + x_2 + \cdots + x_n)\right) dx_1 dx_2 \cdots dx_n \\
= \int_0^1 \int_0^1 \cdots \int_0^1 \sin^2 \left(\frac{\pi}{2n}(x_1 + x_2 + \cdots + x_n)\right) dx_1 dx_2 \cdots dx_n.
\]
Each of these expressions, being equal to half their sum, must equal $\frac{1}{2}$. The limit is also $\boxed{\frac{1}{2}}$. | numerical | putnam | Calculus | Evaluate
\[
\lim_{n \to \infty} \int_0^1 \int_0^1 \cdots \int_0^1 \cos^2 \left(\frac{\pi}{2n}(x_1 + x_2 + \cdots + x_n)\right) dx_1 dx_2 \cdots dx_n.
\] | The change of variables $x_k \to 1 - x_k$ yields
\[
\int_0^1 \int_0^1 \cdots \int_0^1 \cos^2 \left(\frac{\pi}{2n}(x_1 + x_2 + \cdots + x_n)\right) dx_1 dx_2 \cdots dx_n \\
= \int_0^1 \int_0^1 \cdots \int_0^1 \sin^2 \left(\frac{\pi}{2n}(x_1 + x_2 + \cdots + x_n)\right) dx_1 dx_2 \cdots dx_n.
\]
Each of these expressions, being equal to half their sum, must equal $\frac{1}{2}$. The limit is also $\boxed{\frac{1}{2}}$. | 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 |
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 |
1966 | 1966_A1 | Let $f(n)$ be the sum of the first $n$ terms of the sequence $0, 1, 1, 2, 2, 3, 3, 4, \ldots$, where the $n$th term is given by
\[
a_n =
\begin{cases}
\frac{n}{2}, & \text{if } n \text{ is even}, \\
\frac{n-1}{2}, & \text{if } n \text{ is odd}.
\end{cases}
\]
Show that if $x$ and $y$ are positive integers and $x > y$, then find the value of $f(x+y) - f(x-y)$. | It is easily verified by induction that
\[
f(n) =
\begin{cases}
\frac{n^2}{4}, & \text{when } n \text{ is even}, \\
\frac{n^2 - 1}{4}, & \text{when } n \text{ is odd}.
\end{cases}
\]
Therefore, since $x+y$ and $x-y$ always have the same parity, in any case we must have
\[
f(x+y) - f(x-y) = \frac{(x+y)^2 - (x-y)^2}{4} = \boxed{xy}.
\] | algebraic | putnam (modified boxing) | Algebra | Let $f(n)$ be the sum of the first $n$ terms of the sequence $0, 1, 1, 2, 2, 3, 3, 4, \ldots$, where the $n$th term is given by
\[
a_n =
\begin{cases}
\frac{n}{2}, & \text{if } n \text{ is even}, \\
\frac{n-1}{2}, & \text{if } n \text{ is odd}.
\end{cases}
\]
Show that if $x$ and $y$ are positive integers and $x > y$, then
\[
xy = f(x+y) - f(x-y).
\] | It is easily verified by induction that
\[
f(n) =
\begin{cases}
\frac{n^2}{4}, & \text{when } n \text{ is even}, \\
\frac{n^2 - 1}{4}, & \text{when } n \text{ is odd}.
\end{cases}
\]
Therefore, since $x+y$ and $x-y$ always have the same parity, in any case we must have
\[
f(x+y) - f(x-y) = \frac{(x+y)^2 - (x-y)^2}{4} = xy.
\] | 0 |
1966 | 1966_A2 | Let $a, b, c$ be the lengths of the sides of a triangle, let $p = \frac{a+b+c}{2}$, and $r$ be the radius of the inscribed circle. Find the upper bound of [
\frac{1}{(p-a)^2} + \frac{1}{(p-b)^2} + \frac{1}{(p-c)^2}
\] in terms of the radius. | The area of the given triangle can be calculated in two ways, $A = pr$ and $A = \sqrt{p(p-a)(p-b)(p-c)}$. Squaring and equating we get
\[
p^2r^2 = p(p-a)(p-b)(p-c).
\]
Setting $x = 1/(p-a)$, $y = 1/(p-b)$, $z = 1/(p-c)$, we can write this equation in the form
\[
\frac{1}{r^2} = pxyz = xyz \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right).
\]
Thus we need only show that $yz + xz + xy \leq x^2 + y^2 + z^2$. However, this follows from the trivial inequalities $y^2 + z^2 \geq 2yz$, $x^2 + z^2 \geq 2xz$, $x^2 + y^2 \geq 2xy$. Thus the lowerbound is \boxed{\frac{1}{r^2}} | algebraic | putnam (modified boxing) | Geometry | Let $a, b, c$ be the lengths of the sides of a triangle, let $p = \frac{a+b+c}{2}$, and $r$ be the radius of the inscribed circle. Show that
\[
\frac{1}{(p-a)^2} + \frac{1}{(p-b)^2} + \frac{1}{(p-c)^2} \geq \frac{1}{r^2}.
\] | The area of the given triangle can be calculated in two ways, $A = pr$ and $A = \sqrt{p(p-a)(p-b)(p-c)}$. Squaring and equating we get
\[
p^2r^2 = p(p-a)(p-b)(p-c).
\]
Setting $x = 1/(p-a)$, $y = 1/(p-b)$, $z = 1/(p-c)$, we can write this equation in the form
\[
\frac{1}{r^2} = pxyz = xyz \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right).
\]
Thus we need only show that $yz + xz + xy \leq x^2 + y^2 + z^2$. However, this follows from the trivial inequalities $y^2 + z^2 \geq 2yz$, $x^2 + z^2 \geq 2xz$, $x^2 + y^2 \geq 2xy$. | 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 |
1966 | 1966_A6 | Find the upper bound of \sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + 4\sqrt{1 + 5\sqrt{1 + \cdots}}}}}. | We understand the statement to mean that
\[
3 = \lim_{n \to \infty} \sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \cdots \sqrt{1 + (n-1)\sqrt{1 + n}}}}}.
\]
We see that
\[
3 = \sqrt{1 + 2 \cdot 4} = \sqrt{1 + 2\sqrt{16}} = \sqrt{1 + 2\sqrt{1 + 3\sqrt{25}}} = \sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + 4\sqrt{36}}}}.
\]
This leads us to conjecture the relation
\[
3 = \sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \cdots + \sqrt{1 + n\sqrt{(n+2)^2}}}}}
\]
for all $n \geq 1$. Proceeding by induction, we verify that $(n+2)^2 = n^2 + 4n + 4 = 1 + (n+1)\sqrt{(n+3)^2}$. This given, we must have
\[
3 \geq \sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \cdots + \sqrt{1 + (n-1)\sqrt{1 + n}}}}}.
\]
To set an inequality in the other direction, observe that for any $\alpha > 1$
\[
\sqrt{1 + n\alpha} \leq \sqrt{\alpha\sqrt{1 + n}}.
\]
A repetition of this inequality gives then
\[
3 \leq (n+2)^\alpha \sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \cdots + \sqrt{1 + (n-1)\sqrt{1 + n}}}}}, \quad \text{where } \alpha = 2^{1-n}.
\]This makes the upper bound \boxed{3}. | numerical | putnam (modified boxing) | Analysis | Justify the statement that
\[
3 = \sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + 4\sqrt{1 + 5\sqrt{1 + \cdots}}}}}.
\] | We understand the statement to mean that
\[
3 = \lim_{n \to \infty} \sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \cdots \sqrt{1 + (n-1)\sqrt{1 + n}}}}}.
\]
We see that
\[
3 = \sqrt{1 + 2 \cdot 4} = \sqrt{1 + 2\sqrt{16}} = \sqrt{1 + 2\sqrt{1 + 3\sqrt{25}}} = \sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + 4\sqrt{36}}}}.
\]
This leads us to conjecture the relation
\[
3 = \sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \cdots + \sqrt{1 + n\sqrt{(n+2)^2}}}}}
\]
for all $n \geq 1$. Proceeding by induction, we verify that $(n+2)^2 = n^2 + 4n + 4 = 1 + (n+1)\sqrt{(n+3)^2}$. This given, we must have
\[
3 \geq \sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \cdots + \sqrt{1 + (n-1)\sqrt{1 + n}}}}}.
\]
To set an inequality in the other direction, observe that for any $\alpha > 1$
\[
\sqrt{1 + n\alpha} \leq \sqrt{\alpha\sqrt{1 + n}}.
\]
A repetition of this inequality gives then
\[
3 \leq (n+2)^\alpha \sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \cdots + \sqrt{1 + (n-1)\sqrt{1 + n}}}}}, \quad \text{where } \alpha = 2^{1-n}.
\] | 0 |
1966 | 1966_B1 | Let a convex polygon $P$ be contained in a square of side 1. Find the upper bound of the sum of the squares of the sides of $P$. | Let $P_1, P_2, \dots, P_n$ be the vertices of $P$. Let $P'_1, P'_2, \dots, P'_n$ be the projections of $P_1, P_2, \dots, P_n$ upon one of the sides of the squares, and let $P''_1, P''_2, \dots, P''_n$ be the projections of $P_1, P_2, \dots, P_n$ upon a side that is orthogonal to the previous one. Since $P$ is convex, the first side will be covered at most twice by the segments $P'_1P'_2, \dots, P'_{n-1}P'_n, P'_nP'_1$. We thus deduce the inequality \[ P'_1P'_2^2 + \dots + P'_nP'_1^2 \leq 2. \] Similarly \[ P''_1P''_2^2 + \dots + P''_nP''_1^2 \leq 2. \] Adding these two inequalities and using the Pythagorean theorem the assertion follows. Thus the upper bound is \boxed{4}. | numerical | putnam (modified boxing) | Geometry | Let a convex polygon $P$ be contained in a square of side one. Show that the sum of the squares of the sides of $P$ is less than or equal to $4$. | Let $P_1, P_2, \dots, P_n$ be the vertices of $P$. Let $P'_1, P'_2, \dots, P'_n$ be the projections of $P_1, P_2, \dots, P_n$ upon one of the sides of the squares, and let $P''_1, P''_2, \dots, P''_n$ be the projections of $P_1, P_2, \dots, P_n$ upon a side that is orthogonal to the previous one. Since $P$ is convex, the first side will be covered at most twice by the segments $P'_1P'_2, \dots, P'_{n-1}P'_n, P'_nP'_1$. We thus deduce the inequality \[ P'_1P'_2^2 + \dots + P'_nP'_1^2 \leq 2. \] Similarly \[ P''_1P''_2^2 + \dots + P''_nP''_1^2 \leq 2. \] Adding these two inequalities and using the Pythagorean theorem the assertion follows. | 0 |
1967 | 1967_A1 | Let $f(x)=a_1\sin x + a_2\sin 2x + \cdots + a_n\sin nx$, where $a_1, a_2, \cdots, a_n$ are real numbers and $n$ is a positive integer. Given that $|f(x)| \leq |\sin x|$ for all real $x$, what is the minimum value of \[ |a_1 + 2a_2 + \cdots + na_n|. \] | We have \[ |a_1 + 2a_2 + \cdots + na_n| = |f'(0)| = \lim_{x \to 0} \left| \frac{f(x) - f(0)}{x} \right| = \lim_{x \to 0} \left| \frac{f(x)}{\sin x} \cdot \frac{\sin x}{x} \right| = \lim_{x \to 0} \left| \frac{f(x)}{\sin x} \right| \cdot \lim_{x \to 0} \left| \frac{\sin x}{x} \right| \leq 1 making the minimum value \boxed{1}.\] | numerical | putnam (modified boxing) | Trigonometry Analysis | Let $f(x)=a_1\sin x + a_2\sin 2x + \cdots + a_n\sin nx$, where $a_1, a_2, \cdots, a_n$ are real numbers and $n$ is a positive integer. Given that $|f(x)| \leq |\sin x|$ for all real $x$, prove that \[ |a_1 + 2a_2 + \cdots + na_n| \leq 1. \] | We have \[ |a_1 + 2a_2 + \cdots + na_n| = |f'(0)| = \lim_{x \to 0} \left| \frac{f(x) - f(0)}{x} \right| = \lim_{x \to 0} \left| \frac{f(x)}{\sin x} \cdot \frac{\sin x}{x} \right| = \lim_{x \to 0} \left| \frac{f(x)}{\sin x} \right| \cdot \lim_{x \to 0} \left| \frac{\sin x}{x} \right| \leq 1. \] | 0 |
1967 | 1967_A2 | Define $S_0$ to be 1. For $n \geq 1$, let $S_n$ be the number of $n \times n$ matrices whose elements are nonnegative integers with the property that $a_{ij} = a_{ji}, \ (i,j = 1, 2, \dots, n)$ and where $\sum_{i=1}^n a_{ij} = 1, \ (j = 1, 2, \dots, n)$. Find the $\sum_{n=0}^\infty \frac{S_n x^n}{n!}. | $S_n$ is the number of symmetric $n \times n$ permutation matrices (a permutation matrix has exactly one 1 in each row and column with 0's elsewhere). Let the 1 in the first row be in the $k$th column. If $k = 1$, then there are $S_{n-1}$ ways to complete the matrix. If $k \neq 1$ then $a_{1k} = a_{k1} = 1$ and the deletion of the 1st and $k$th rows and columns leaves a symmetric $(n-2) \times (n-2)$ permutation matrix. Consequently $S_n = S_{n-1} + (n-1)S_{n-2}$.\n\nFor part (b), let\n\[ F(x) = \sum_{n=0}^\infty \frac{S_n x^n}{n!}. \] Then\n\[ F'(x) = \sum_{n=1}^\infty \frac{S_n x^{n-1}}{(n-1)!} = \sum_{n=1}^\infty \frac{S_{n-1} x^{n-1}}{(n-1)!} + (n-1)\frac{S_{n-2} x^{n-1}}{(n-1)!}. \]\n\[ F'(x) = \sum_{n=0}^\infty \frac{S_n x^n}{n!} + \sum_{n=2}^\infty \frac{S_{n-2} x^{n-1}}{(n-2)!} = F(x) + xF(x). \]\n\nHence $\frac{F'(x)}{F(x)} = 1 + x$. Integration and use of $F(0) = S_0 = 1$ yields $F(x) = \boxed{e^(x + x^2/2)}$. Now the series for $F(x)$ is uniformly convergent for all $x$, so all the operations are legal. | algebraic | putnam (modified boxing) | Combinatorics Algebra | Define $S_0$ to be 1. For $n \geq 1$, let $S_n$ be the number of $n \times n$ matrices whose elements are nonnegative integers with the property that $a_{ij} = a_{ji}, \ (i,j = 1, 2, \dots, n)$ and where $\sum_{i=1}^n a_{ij} = 1, \ (j = 1, 2, \dots, n)$. Prove\n\n(a) $S_{n+1} = S_n + n S_{n-1}$,\n\n(b) $\sum_{n=0}^\infty \frac{S_n x^n}{n!} = \exp(x + x^2/2)$, where $\exp(x) = e^x$. | $S_n$ is the number of symmetric $n \times n$ permutation matrices (a permutation matrix has exactly one 1 in each row and column with 0's elsewhere). Let the 1 in the first row be in the $k$th column. If $k = 1$, then there are $S_{n-1}$ ways to complete the matrix. If $k \neq 1$ then $a_{1k} = a_{k1} = 1$ and the deletion of the 1st and $k$th rows and columns leaves a symmetric $(n-2) \times (n-2)$ permutation matrix. Consequently $S_n = S_{n-1} + (n-1)S_{n-2}$.\n\nFor part (b), let\n\[ F(x) = \sum_{n=0}^\infty \frac{S_n x^n}{n!}. \] Then\n\[ F'(x) = \sum_{n=1}^\infty \frac{S_n x^{n-1}}{(n-1)!} = \sum_{n=1}^\infty \frac{S_{n-1} x^{n-1}}{(n-1)!} + (n-1)\frac{S_{n-2} x^{n-1}}{(n-1)!}. \]\n\[ F'(x) = \sum_{n=0}^\infty \frac{S_n x^n}{n!} + \sum_{n=2}^\infty \frac{S_{n-2} x^{n-1}}{(n-2)!} = F(x) + xF(x). \]\n\nHence $\frac{F'(x)}{F(x)} = 1 + x$. Integration and use of $F(0) = S_0 = 1$ yields $F(x) = \exp(x + x^2/2)$. Now the series for $F(x)$ is uniformly convergent for all $x$, so all the operations are legal. | 0 |
1967 | 1967_A3 | Consider polynomial forms $ax^2 - bx + c$ with integer coefficients which have two distinct zeros in the open interval $0 < x < 1$. Exhibit with a proof the least positive integer value of $a$ for which such a polynomial exists. | Let $f(x) = ax^2 - bx + c = a(x - r)(x - s)$. Then $f(0) \cdot f(1) = a^2 r(r - 1)s(s - 1)$. The graph of $r(r - 1)$ shows that $0 < r < 1$ implies $0 < r(r - 1) \leq \frac{1}{4}$, with equality if and only if $r = \frac{1}{2}$. Similarly, $0 < s(s - 1) \leq \frac{1}{4}$. Since $r \neq s$, $r(r - 1)s(s - 1) < \frac{1}{16}$ and $0 < f(0) \cdot f(1) < a^2 / 16$. The coefficients $a, b, c$ are integers and thus $1 \leq f(0) \cdot f(1)$. Consequently $a^2 > 16$, i.e., $a \geq 5$.\n\nThe discriminant $b^2 - 4ac$ shows that the minimum possible value for $b$ is $5$. Furthermore, $5x^2 - 5x + 1$ has two distinct roots between 0 and 1. Thus the minimum value of a is \boxed{5}. | numerical | putnam | Algebra Analysis | Consider polynomial forms $ax^2 - bx + c$ with integer coefficients which have two distinct zeros in the open interval $0 < x < 1$. Exhibit with a proof the least positive integer value of $a$ for which such a polynomial exists. | Let $f(x) = ax^2 - bx + c = a(x - r)(x - s)$. Then $f(0) \cdot f(1) = a^2 r(r - 1)s(s - 1)$. The graph of $r(r - 1)$ shows that $0 < r < 1$ implies $0 < r(r - 1) \leq \frac{1}{4}$, with equality if and only if $r = \frac{1}{2}$. Similarly, $0 < s(s - 1) \leq \frac{1}{4}$. Since $r \neq s$, $r(r - 1)s(s - 1) < \frac{1}{16}$ and $0 < f(0) \cdot f(1) < a^2 / 16$. The coefficients $a, b, c$ are integers and thus $1 \leq f(0) \cdot f(1)$. Consequently $a^2 > 16$, i.e., $a \geq \boxed{5}$.\n\nThe discriminant $b^2 - 4ac$ shows that the minimum possible value for $b$ is $5$. Furthermore, $5x^2 - 5x + 1$ has two distinct roots between 0 and 1. | 0 |
1967 | 1967_A4 | Find the lower bound of $\lambda$ such that there does not exist a real-valued function $u$ such that for all $x$ in the closed interval $0 \leq x \leq 1$, \[ u(x) = 1 + \lambda \int_0^x u(y) u(y-x)\,dy. \] | Assuming that there is a solution $u$, then integrating with respect to $x$ from 0 to 1, one obtains \[ \int_0^1 u(x)\,dx = \int_0^1 1\,dx + \lambda \int_0^1 \int_0^x u(y)u(y-x)\,dy\,dx. \] In the iterated integral, one can interchange the order of integration, and letting \( \int_0^1 u(x)\,dx = \alpha \), one gets \[ \alpha = 1 + \lambda \int_0^1 u(y)\int_0^{1-y} u(y+z)\,dz\,dy. \] Now, holding $y$ fixed, let $z = y-x$ to get \[ \alpha = 1 + \lambda \int_0^1 u(y)\int_0^y u(z)\,dz\,dy. \] Set $f(y) = \int_0^y u(z)\,dz$. Then \[ \alpha = 1 + \lambda \int_0^1 f'(y)f(y)\,dy. \] Set $f^2(y) = \alpha$. Then \[ \alpha = 1 + \lambda \frac{1}{2}\big[f^2(1) - f^2(0)\big] = 1 + \lambda \frac{1}{2} \alpha^2. \] This results in the quadratic equation \( \lambda \alpha^2 - 2\alpha + 2 = 0 \). The discriminant of this quadratic equation shows that if $\lambda > \boxed{\frac{1}{2}}$ then the roots are imaginary, and hence no such function $u$ can exist. | numerical | putnam (modified boxing) | Analysis | Show that if $\lambda > \frac{1}{2}$ there does not exist a real-valued function $u$ such that for all $x$ in the closed interval $0 \leq x \leq 1$, \[ u(x) = 1 + \lambda \int_0^x u(y) u(y-x)\,dy. \] | Assuming that there is a solution $u$, then integrating with respect to $x$ from 0 to 1, one obtains \[ \int_0^1 u(x)\,dx = \int_0^1 1\,dx + \lambda \int_0^1 \int_0^x u(y)u(y-x)\,dy\,dx. \] In the iterated integral, one can interchange the order of integration, and letting \( \int_0^1 u(x)\,dx = \alpha \), one gets \[ \alpha = 1 + \lambda \int_0^1 u(y)\int_0^{1-y} u(y+z)\,dz\,dy. \] Now, holding $y$ fixed, let $z = y-x$ to get \[ \alpha = 1 + \lambda \int_0^1 u(y)\int_0^y u(z)\,dz\,dy. \] Set $f(y) = \int_0^y u(z)\,dz$. Then \[ \alpha = 1 + \lambda \int_0^1 f'(y)f(y)\,dy. \] Set $f^2(y) = \alpha$. Then \[ \alpha = 1 + \lambda \frac{1}{2}\big[f^2(1) - f^2(0)\big] = 1 + \lambda \frac{1}{2} \alpha^2. \] This results in the quadratic equation \( \lambda \alpha^2 - 2\alpha + 2 = 0 \). The discriminant of this quadratic equation shows that if $\lambda > \frac{1}{2}$ then the roots are imaginary, and hence no such function $u$ can exist. | 0 |
1967 | 1967_A6 | Given real numbers $\{a_i\}$ and $\{b_i\}$, $(i = 1, 2, 3, 4)$, such that $a_1b_2 - a_2b_1 \neq 0$. Consider the set of all solutions $(x_1, x_2, x_3, x_4)$ of the simultaneous equations \[ a_1x_1 + a_2x_2 + a_3x_3 + a_4x_4 = 0 \] and \[ b_1x_1 + b_2x_2 + b_3x_3 + b_4x_4 = 0, \] for which no $x_i$ $(i = 1, 2, 3, 4)$ is zero. Each such solution generates a 4-tuple of plus and minus signs $(\text{signum } x_1, \text{signum } x_2, \text{signum } x_3, \text{signum } x_4)$. Determine, with a proof, the maximum number of distinct 4-tuples possible. | Solving the given equations in terms of $x_3$ and $x_4$, leads to the equivalent system: \[ x_1 = A_1x_3 + B_1x_4, \quad x_2 = A_2x_3 + B_2x_4, \quad x_3 = x_3, \quad x_4 = x_4, \] where \[ A_1 = \frac{a_2b_3 - a_3b_2}{a_1b_2 - a_2b_1}, \quad B_1 = \frac{a_3b_1 - a_1b_3}{a_1b_2 - a_2b_1}, \] etc.\n\nEach point in the $x_3, x_4$-plane corresponds uniquely to a solution $(x_1, x_2, x_3, x_4)$. Signum $x_1$ is positive for $(x_3, x_4)$ on one side of $A_1x_3 + B_1x_4 = 0$ and negative on the other side. Similarly for signum $x_2$, signum $x_3$, and signum $x_4$, using $A_2x_3 + B_2x_3 = 0$, $x_3 = 0$, and $x_4 = 0$, respectively. These four lines through the origin, in general, divide the plane into eight regions, each having a different 4-tuple of signum values. Hence the maximum number of distinct 4-tuples is \boxed{8}. | numerical | putnam | Algebra Linear Algebra | Given real numbers $\{a_i\}$ and $\{b_i\}$, $(i = 1, 2, 3, 4)$, such that $a_1b_2 - a_2b_1 \neq 0$. Consider the set of all solutions $(x_1, x_2, x_3, x_4)$ of the simultaneous equations \[ a_1x_1 + a_2x_2 + a_3x_3 + a_4x_4 = 0 \] and \[ b_1x_1 + b_2x_2 + b_3x_3 + b_4x_4 = 0, \] for which no $x_i$ $(i = 1, 2, 3, 4)$ is zero. Each such solution generates a 4-tuple of plus and minus signs $(\text{signum } x_1, \text{signum } x_2, \text{signum } x_3, \text{signum } x_4)$.\n(a) Determine, with a proof, the maximum number of distinct 4-tuples possible.\n(b) Investigate necessary and sufficient conditions on the real numbers $\{a_i\}$ and $\{b_i\}$ such that the above maximum number of 4-tuples is attained. | Solving the given equations in terms of $x_3$ and $x_4$, leads to the equivalent system: \[ x_1 = A_1x_3 + B_1x_4, \quad x_2 = A_2x_3 + B_2x_4, \quad x_3 = x_3, \quad x_4 = x_4, \] where \[ A_1 = \frac{a_2b_3 - a_3b_2}{a_1b_2 - a_2b_1}, \quad B_1 = \frac{a_3b_1 - a_1b_3}{a_1b_2 - a_2b_1}, \] etc.\n\nEach point in the $x_3, x_4$-plane corresponds uniquely to a solution $(x_1, x_2, x_3, x_4)$. Signum $x_1$ is positive for $(x_3, x_4)$ on one side of $A_1x_3 + B_1x_4 = 0$ and negative on the other side. Similarly for signum $x_2$, signum $x_3$, and signum $x_4$, using $A_2x_3 + B_2x_3 = 0$, $x_3 = 0$, and $x_4 = 0$, respectively. These four lines through the origin, in general, divide the plane into eight regions, each having a different 4-tuple of signum values. Hence the maximum number of distinct 4-tuples is \boxed{8}.\n\nThe maximum number of eight will occur if and only if there are actually four distinct lines. This is equivalent to the conditions $A_1 \neq 0$, $A_2 \neq 0$, $B_1 \neq 0$, $B_2 \neq 0$, and $A_1B_2 - A_2B_1 \neq 0$, or, simply, $a_ib_j - a_jb_i \neq 0$ for $i, j = 1, 2, 3, 4$ and $i < j$. | 0 |
1967 | 1967_B2 | Let $0 \leq p \leq 1$ and $0 \leq r \leq 1$ and consider the identities \begin{align*} (a) \quad (px + (1-p)y)^2 &= Ax^2 + Bxy + Cy^2, \\ (b) \quad (px + (1-p)y)(rx + (1-r)y) &= \alpha x^2 + \beta xy + \gamma y^2. \end{align*} Find the lower bound of (with respect to $p$ and $r$) \max\{A, B, C\}, and \max\{\alpha, \beta, \gamma\} and return the sum of both. | For part (a), one has immediately that $A = p^2$, $B = 2p(1-p)$, and $C = (1-p)^2$. The result follows by examination of the graphs for $A$, $B$, and $C$ on $0 \leq p \leq 1$.\n\nFor part (b), $\alpha = pr$, $\beta = p(1-r) + r(1-p)$, $\gamma = (1-p)(1-r)$. Consider the region $R$ in the $p, r$-plane defined by $0 \leq p \leq 1$ and $0 \leq r \leq 1$. We will show that there is no point in $R$ with $\alpha < \frac{4}{9}$, $\beta < \frac{4}{9}$, and $\gamma < \frac{4}{9}$. If $\alpha < \frac{4}{9}$ and $\gamma < \frac{4}{9}$, then $(p, r)$ is between the hyperbolas $pr = \frac{4}{9}$ and $(1-p)(1-r) = \frac{4}{9}$. These hyperbolas have vertices in $R$ at $(2/3, 2/3)$ and $(1/3, 1/3)$, respectively. The symmetry about $(1/2, 1/2)$ suggests setting $p' = p - \frac{1}{2}$ and $r' = r - \frac{1}{2}$. Then $\beta = \frac{1}{2} - 2p'r'$ and thus $\beta < \frac{4}{9}$ if and only if $p'r' > 1/36$. Note that the vertices for the hyperbola $p'r' = 1/36$ are at $(p, r) = (1/3, 1/3)$ and $(2/3, 2/3)$. By looking at asymptotes, we see graphically that the region $\beta < \frac{4}{9}$ does not overlap the region in $R$ where $\alpha < \frac{4}{9}$ and $\gamma < \frac{4}{9}$ thus making the final sum \boxed{\frac{8}{9}}. | numerical | putnam (modified boxing) | Algebra Analysis | Let $0 \leq p \leq 1$ and $0 \leq r \leq 1$ and consider the identities \begin{align*} (a) \quad (px + (1-p)y)^2 &= Ax^2 + Bxy + Cy^2, \\ (b) \quad (px + (1-p)y)(rx + (1-r)y) &= \alpha x^2 + \beta xy + \gamma y^2. \end{align*} Show that (with respect to $p$ and $r$): \begin{align*} (a) \quad \max\{A, B, C\} \geq \frac{4}{9}, \\ (b) \quad \max\{\alpha, \beta, \gamma\} \geq \frac{4}{9}. \end{align*} | For part (a), one has immediately that $A = p^2$, $B = 2p(1-p)$, and $C = (1-p)^2$. The result follows by examination of the graphs for $A$, $B$, and $C$ on $0 \leq p \leq 1$.\n\nFor part (b), $\alpha = pr$, $\beta = p(1-r) + r(1-p)$, $\gamma = (1-p)(1-r)$. Consider the region $R$ in the $p, r$-plane defined by $0 \leq p \leq 1$ and $0 \leq r \leq 1$. We will show that there is no point in $R$ with $\alpha < \frac{4}{9}$, $\beta < \frac{4}{9}$, and $\gamma < \frac{4}{9}$. If $\alpha < \frac{4}{9}$ and $\gamma < \frac{4}{9}$, then $(p, r)$ is between the hyperbolas $pr = \frac{4}{9}$ and $(1-p)(1-r) = \frac{4}{9}$. These hyperbolas have vertices in $R$ at $(2/3, 2/3)$ and $(1/3, 1/3)$, respectively. The symmetry about $(1/2, 1/2)$ suggests setting $p' = p - \frac{1}{2}$ and $r' = r - \frac{1}{2}$. Then $\beta = \frac{1}{2} - 2p'r'$ and thus $\beta < \frac{4}{9}$ if and only if $p'r' > 1/36$. Note that the vertices for the hyperbola $p'r' = 1/36$ are at $(p, r) = (1/3, 1/3)$ and $(2/3, 2/3)$. By looking at asymptotes, we see graphically that the region $\beta < \frac{4}{9}$ does not overlap the region in $R$ where $\alpha < \frac{4}{9}$ and $\gamma < \frac{4}{9}$. | 0 |
1967 | 1967_B5 | Find the sum of the first $n$ terms in the binomial expansion of $(2 - 1)^{-n}$ where $n$ is a positive integer. | Let $A_n$ be the sum of the first $n$ terms in the binomial expansion of $(2 - 1)^{-n}$.\n\[ A_n = \sum_{i=0}^{n-1} \binom{n+i-1}{i} 2^{-n-i}. \]\n\nRewriting, we use recursive summation to evaluate:\n\[ A_n = 2^{-n} + \sum_{i=1}^{n-1} \left( \binom{n+i-2}{i} + \binom{n+i-2}{i-1} \right) 2^{-n-i}. \]\n\nAfter manipulation and combining terms:\n\[ A_n = \frac{1}{2} A_{n-1} + \frac{1}{2} A_{n-1} + 2^{-2n}, \]\nwhich simplifies to:\n\[ A_n = \frac{1}{2} A_n + \frac{1}{2}, \]\nso \( A_n = \boxed{\frac{1}{2}} \). | numerical | putnam (modified boxing) | Combinatorics | Show that the sum of the first $n$ terms in the binomial expansion of $(2 - 1)^{-n}$ is $\frac{1}{2}$, where $n$ is a positive integer. | Let $A_n$ be the sum of the first $n$ terms in the binomial expansion of $(2 - 1)^{-n}$.\n\[ A_n = \sum_{i=0}^{n-1} \binom{n+i-1}{i} 2^{-n-i}. \]\n\nRewriting, we use recursive summation to evaluate:\n\[ A_n = 2^{-n} + \sum_{i=1}^{n-1} \left( \binom{n+i-2}{i} + \binom{n+i-2}{i-1} \right) 2^{-n-i}. \]\n\nAfter manipulation and combining terms:\n\[ A_n = \frac{1}{2} A_{n-1} + \frac{1}{2} A_{n-1} + 2^{-2n}, \]\nwhich simplifies to:\n\[ A_n = \frac{1}{2} A_n + \frac{1}{2}, \]\nso \( A_n = \frac{1}{2} \).\n\nAlternate solution: Consider a random walk starting at $(0, 0)$, where moving to $(x, y+1)$ or $(x+1, y)$ each occurs with probability $\frac{1}{2}$. Let $S_n$ be the square with vertices $(0, 0)$, $(n, 0)$, $(0, n)$, and $(n, n)$. By symmetry, the probability of first touching $S_n$ at $(n, i)$ or $(i, n)$ equals $\frac{1}{2}$ for all $i$. Hence, $\sum_{i=0}^n R_i(n) = \frac{1}{2}$. Therefore, \( A_n = \frac{1}{2} \). | 0 |
1968 | 1968_A4 | Given $n$ points on the sphere \( \{(x, y, z): x^2 + y^2 + z^2 = 1\} \), find the upper bound of the sum of the squares of the distances between them in terms of $n$. | The \(n\) points can be represented by vectors \(\mathbf{v}_i\); \(i = 1, \dots, n\) with \(\|\mathbf{v}_i\| = 1\). Expanding "the sum of the squares of the distances between them" in the case \(n = 3\) suggests the following general identities:
\[
\sum_{1 \leq i < j \leq n} \|\mathbf{v}_i - \mathbf{v}_j\|^2 = \sum_{1 \leq i < j \leq n} (\mathbf{v}_i - \mathbf{v}_j) \cdot (\mathbf{v}_i - \mathbf{v}_j)
\]
\[
= (n-1) \sum_{1 \leq i \leq n} \mathbf{v}_i \cdot \mathbf{v}_i - 2 \sum_{1 \leq i < j \leq n} \mathbf{v}_i \cdot \mathbf{v}_j
\]
\[
= n \sum_{1 \leq i \leq n} \mathbf{v}_i \cdot \mathbf{v}_i - \left[ \sum_{1 \leq i \leq n} \mathbf{v}_i \cdot \mathbf{v}_i + 2 \sum_{1 \leq i < j \leq n} \mathbf{v}_i \cdot \mathbf{v}_j \right]
\]
\[
= n^2 - \left( \sum_{1 \leq i \leq n} \mathbf{v}_i \right) \cdot \left( \sum_{1 \leq i \leq n} \mathbf{v}_i \right).
\]
Thus the result follows and, in addition, equality exists if and only if \(\sum_{1 \leq i \leq n} \mathbf{v}_i = 0\). Thus the uper bound is \boxed{n^2}. | algebraic | putnam (modified boxing) | Geometry Algebra | Given $n$ points on the sphere \( \{(x, y, z): x^2 + y^2 + z^2 = 1\} \), demonstrate that the sum of the squares of the distances between them does not exceed \(n^2\). | The \(n\) points can be represented by vectors \(\mathbf{v}_i\); \(i = 1, \dots, n\) with \(\|\mathbf{v}_i\| = 1\). Expanding "the sum of the squares of the distances between them" in the case \(n = 3\) suggests the following general identities:
\[
\sum_{1 \leq i < j \leq n} \|\mathbf{v}_i - \mathbf{v}_j\|^2 = \sum_{1 \leq i < j \leq n} (\mathbf{v}_i - \mathbf{v}_j) \cdot (\mathbf{v}_i - \mathbf{v}_j)
\]
\[
= (n-1) \sum_{1 \leq i \leq n} \mathbf{v}_i \cdot \mathbf{v}_i - 2 \sum_{1 \leq i < j \leq n} \mathbf{v}_i \cdot \mathbf{v}_j
\]
\[
= n \sum_{1 \leq i \leq n} \mathbf{v}_i \cdot \mathbf{v}_i - \left[ \sum_{1 \leq i \leq n} \mathbf{v}_i \cdot \mathbf{v}_i + 2 \sum_{1 \leq i < j \leq n} \mathbf{v}_i \cdot \mathbf{v}_j \right]
\]
\[
= n^2 - \left( \sum_{1 \leq i \leq n} \mathbf{v}_i \right) \cdot \left( \sum_{1 \leq i \leq n} \mathbf{v}_i \right).
\]
Thus the result follows and, in addition, equality exists if and only if \(\sum_{1 \leq i \leq n} \mathbf{v}_i = 0\). | 0 |
1968 | 1968_A5 | Let $V$ be the collection of all quadratic polynomials $P$ with real coefficients such that $|P(x)| \leq 1$ for all $x$ on the closed interval $[0, 1]$. Determine \[ \sup\{|P'(0)| : P \in V\}. \] | Let $f(x) = ax^2 + bx + c$ be an arbitrary quadratic polynomial. Then $f(0) = c$, $f(\frac{1}{2}) = \frac{1}{4}a + \frac{1}{2}b + c$, and $f(1) = a + b + c$. $f'(0) = b = 4f(\frac{1}{2}) - 3f(0) - f(1)$. Using the given conditions, \[|P'(0)| \leq 4|P(\frac{1}{2})| + 3|P(0)| + |P(1)| \leq 8. \] Furthermore, $P(x) = 8x^2 - 8x + 1$ satisfies the given conditions and has $|P'(0)| = \boxed{8}$. | numerical | putnam | Analysis Algebra | Let $V$ be the collection of all quadratic polynomials $P$ with real coefficients such that $|P(x)| \leq 1$ for all $x$ on the closed interval $[0, 1]$. Determine \[ \sup\{|P'(0)| : P \in V\}. \] | Let $f(x) = ax^2 + bx + c$ be an arbitrary quadratic polynomial. Then $f(0) = c$, $f(\frac{1}{2}) = \frac{1}{4}a + \frac{1}{2}b + c$, and $f(1) = a + b + c$. $f'(0) = b = 4f(\frac{1}{2}) - 3f(0) - f(1)$. Using the given conditions, \[|P'(0)| \leq 4|P(\frac{1}{2})| + 3|P(0)| + |P(1)| \leq 8. \] Furthermore, $P(x) = 8x^2 - 8x + 1$ satisfies the given conditions and has $|P'(0)| = \boxed{8}$. | 0 |
1968 | 1968_A6 | Determine all polynomials of the form $\sum_{i=0}^n a_i x^{n-i}$ with $a_i = \pm 1$ $(0 \leq i \leq n, 1 \leq n < \infty)$ such that each has only real zeros and return the sum of the absolute values of the polynomials. | The desired polynomials with $a_0 = -1$ are the negatives of those with $a_0 = 1$, so consider $a_0 = 1$. The sum of the squares of the zeros of $x^n + a_1x^{n-1} + \cdots + a_n$ is $a_1^2 - 2a_2$. The product of the squares of these zeros is $a_n^2$. If all the zeros are real, we can apply the arithmetic-geometric mean inequality to obtain \[ \frac{a_1^2 - 2a_2}{n} \geq (a_n^2)^{1/n}, \] with equality only if the zeros are numerically equal. In our case, this inequality becomes \[ \frac{1 \pm 2}{n} \geq 1 \quad \text{or} \quad n \leq 3. \] Note that $n > 1$ implies $a_2 = -1$ and $n = 3$ implies all zeros are $\pm 1$. Thus, the list of polynomials is: \[ \pm(x-1), \quad \pm(x+1), \quad \pm(x^2+x-1), \quad \pm(x^2-x-1), \quad \pm(x^3+x^2-x-1), \quad \pm(x^3-x^2-x+1). \] Thus the sum of the absolute values of the polynomials is $\boxed{2x^3 + 2x^2 - 2}$. | algebraic | putnam (modified boxing) | Algebra | Determine all polynomials of the form $\sum_{i=0}^n a_i x^{n-i}$ with $a_i = \pm 1$ $(0 \leq i \leq n, 1 \leq n < \infty)$ such that each has only real zeros. | The desired polynomials with $a_0 = -1$ are the negatives of those with $a_0 = 1$, so consider $a_0 = 1$. The sum of the squares of the zeros of $x^n + a_1x^{n-1} + \cdots + a_n$ is $a_1^2 - 2a_2$. The product of the squares of these zeros is $a_n^2$. If all the zeros are real, we can apply the arithmetic-geometric mean inequality to obtain \[ \frac{a_1^2 - 2a_2}{n} \geq (a_n^2)^{1/n}, \] with equality only if the zeros are numerically equal. In our case, this inequality becomes \[ \frac{1 \pm 2}{n} \geq 1 \quad \text{or} \quad n \leq 3. \] Note that $n > 1$ implies $a_2 = -1$ and $n = 3$ implies all zeros are $\pm 1$. Thus, the list of polynomials is: \[ \pm(x-1), \quad \pm(x+1), \quad \pm(x^2+x-1), \quad \pm(x^2-x-1), \quad \pm(x^3+x^2-x-1), \quad \pm(x^3-x^2-x+1). \] | 0 |
1968 | 1968_B1 | The temperatures in Chicago and Detroit are $x^\circ$ and $y^\circ$, respectively. These temperatures are not assumed to be independent; namely, we are given:
\begin{itemize}
\item[(i)] $P(x^\circ = 70^\circ)$, the probability that the temperature in Chicago is $70^\circ$,
\item[(ii)] $P(y^\circ = 70^\circ)$, and
\item[(iii)] $P(\max(x^\circ, y^\circ) = 70^\circ)$.
\end{itemize}
Determine $P(\min(x^\circ, y^\circ) = 70^\circ)$. | Denote the four events $x^\circ = 70^\circ$, $y^\circ = 70^\circ$, $\max(x^\circ, y^\circ) = 70^\circ$, and $\min(x^\circ, y^\circ) = 70^\circ$ by $A$, $B$, $C$, $D$, respectively. Then $A \cup B = C \cup D$, and $A \cap B = C \cap D$. Hence
\[ P(A) + P(B) = P(A \cup B) + P(A \cap B) = P(C \cup D) + P(C \cap D) = P(C) + P(D), \]
and
\[ P(\min(x^\circ, y^\circ) = 70^\circ) = \boxed{P(x^\circ = 70^\circ) + P(y^\circ = 70^\circ) - P(\max(x^\circ, y^\circ) = 70^\circ)}. \] | algebraic | putnam | Probability | The temperatures in Chicago and Detroit are $x^\circ$ and $y^\circ$, respectively. These temperatures are not assumed to be independent; namely, we are given:
\begin{itemize}
\item[(i)] $P(x^\circ = 70^\circ)$, the probability that the temperature in Chicago is $70^\circ$,
\item[(ii)] $P(y^\circ = 70^\circ)$, and
\item[(iii)] $P(\max(x^\circ, y^\circ) = 70^\circ)$.
\end{itemize}
Determine $P(\min(x^\circ, y^\circ) = 70^\circ)$. | Denote the four events $x^\circ = 70^\circ$, $y^\circ = 70^\circ$, $\max(x^\circ, y^\circ) = 70^\circ$, and $\min(x^\circ, y^\circ) = 70^\circ$ by $A$, $B$, $C$, $D$, respectively. Then $A \cup B = C \cup D$, and $A \cap B = C \cap D$. Hence
\[ P(A) + P(B) = P(A \cup B) + P(A \cap B) = P(C \cup D) + P(C \cap D) = P(C) + P(D), \]
and
\[ P(\min(x^\circ, y^\circ) = 70^\circ) = \boxed{P(x^\circ = 70^\circ) + P(y^\circ = 70^\circ) - P(\max(x^\circ, y^\circ) = 70^\circ)}. \] | 0 |
1968 | 1968_B5 | Let $p$ be a prime number. Let $J$ be the set of all $2 \times 2$ matrices \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) whose entries are chosen from $\{0, 1, 2, \dots, p-1\}$ and satisfy the conditions $a+d \equiv 1 \pmod{p}$ and $ad-bc \equiv 0 \pmod{p}$. Determine how many members $J$ has. | If $a=0$ then $d=1$, and if $a=1$ then $d=0$. In either case $bc \equiv 0 \pmod{p}$ and $b$ or $c$ is $0$, while the other is arbitrary. There are $2p-1$ distinct solutions to $bc \equiv 0 \pmod{p}$, and thus the case $a=0$ or $a=1$ accounts for a total of $4p-2$ solutions. If $a \neq 0$ or $1$, then $d$ is uniquely determined and $bc \equiv ad \pmod{p}$ implies that for each $b \neq 0$, there is a unique $c$, since the integers mod $p$ form a field. Hence for each $a$ in this case, there are $p-1$ solutions. The total number of solutions is $4p-2+(p-2)(p-1)=\boxed{p^2+p}$. | algebraic | putnam | Algebra Linear Algebra | Let $p$ be a prime number. Let $J$ be the set of all $2 \times 2$ matrices \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) whose entries are chosen from $\{0, 1, 2, \dots, p-1\}$ and satisfy the conditions $a+d \equiv 1 \pmod{p}$ and $ad-bc \equiv 0 \pmod{p}$. Determine how many members $J$ has. | If $a=0$ then $d=1$, and if $a=1$ then $d=0$. In either case $bc \equiv 0 \pmod{p}$ and $b$ or $c$ is $0$, while the other is arbitrary. There are $2p-1$ distinct solutions to $bc \equiv 0 \pmod{p}$, and thus the case $a=0$ or $a=1$ accounts for a total of $4p-2$ solutions. If $a \neq 0$ or $1$, then $d$ is uniquely determined and $bc \equiv ad \pmod{p}$ implies that for each $b \neq 0$, there is a unique $c$, since the integers mod $p$ form a field. Hence for each $a$ in this case, there are $p-1$ solutions. The total number of solutions is $4p-2+(p-2)(p-1)=\boxed{p^2+p}$. | 0 |
1969 | 1969_A2 | Let $D_n$ be the determinant of order $n$ of which the element in the $i$th row and the $j$th column is the absolute value of the difference of $i$ and $j$. Find the value of $D_n$. | Subtract the first column from every other column. Then add the first row to every other row. The last row now has all zeros except for $(n-1)$ in the first column. $D_n$ is $(-1)^{n-1}(n-1)$ times the minor formed by deleting the first column and last row from the transformed determinant. This minor has only zeros below the main diagonal and thus is equal to the product of its diagonal elements. Hence the minor has value $2^{n-2}$ and $D_n = \boxed{(-1)^{n-1}(n-1)2^{n-2}}.$ | algebraic | putnam (modified boxing) | Linear Algebra | Let $D_n$ be the determinant of order $n$ of which the element in the $i$th row and the $j$th column is the absolute value of the difference of $i$ and $j$. Show that $D_n$ is equal to $(-1)^{n-1}(n-1)2^{n-2}.$ | Subtract the first column from every other column. Then add the first row to every other row. The last row now has all zeros except for $(n-1)$ in the first column. $D_n$ is $(-1)^{n-1}(n-1)$ times the minor formed by deleting the first column and last row from the transformed determinant. This minor has only zeros below the main diagonal and thus is equal to the product of its diagonal elements. Hence the minor has value $2^{n-2}$ and $D_n = (-1)^{n-1}(n-1)2^{n-2}.$ | 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 |
1969 | 1969_B3 | The terms of a sequence $T_n$ satisfy
\[ T_n T_{n+1} = n \quad (n = 1, 2, 3, \dots) \text{ and } \lim_{n \to \infty} \frac{T_n}{T_{n+1}} = 1. \]
Find the value of $\pi T_1^2$. | The first relation implies that
\[
T_n = \frac{(n-1)(n-3)(n-5)\cdots}{(n-2)(n-4)(n-6)\cdots} \cdot T_1 \quad \text{if $n$ is even}, \quad \text{and}
\]
\[
T_n = \frac{(n-1)(n-3)(n-5)\cdots}{(n-2)(n-4)(n-6)\cdots} \cdot T_1 \quad \text{if $n$ is odd}.
\]
If $n$ is odd, we have
\[ \frac{T_n}{T_{n+1}} = (T_1)^2 \cdot \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \cdots \cdot \frac{(n-1)}{(n-2)} \cdot \frac{(n-1)}{n}. \]
The Wallis product gives
\[ \frac{\pi}{2} = \prod \left( \frac{2 \cdot 4 \cdot 6 \cdot \cdots}{1 \cdot 3 \cdot 5 \cdot \cdots} \right). \]
Thus, after an even number of factors, the partial product is less than $\pi/2$, and after an odd number of factors, the partial product is greater than $\pi/2$. Therefore, $T_n/T_{n+1} < \frac{1}{2} \pi T_1^2$. Similarly, when $n$ is even, $T_n/T_{n+1} < \frac{2}{\pi T_1^2}$. Since the limit of $T_n/T_{n+1} \to 1$, we conclude that
\[ \pi T_1^2 = \boxed{2}. \] | numerical | putnam (modified boxing) | Algebra Calculus | The terms of a sequence $T_n$ satisfy
\[ T_n T_{n+1} = n \quad (n = 1, 2, 3, \dots) \text{ and } \lim_{n \to \infty} \frac{T_n}{T_{n+1}} = 1. \]
Show that $\pi T_1^2 = 2$. | The first relation implies that
\[
T_n = \frac{(n-1)(n-3)(n-5)\cdots}{(n-2)(n-4)(n-6)\cdots} \cdot T_1 \quad \text{if $n$ is even}, \quad \text{and}
\]
\[
T_n = \frac{(n-1)(n-3)(n-5)\cdots}{(n-2)(n-4)(n-6)\cdots} \cdot T_1 \quad \text{if $n$ is odd}.
\]
If $n$ is odd, we have
\[ \frac{T_n}{T_{n+1}} = (T_1)^2 \cdot \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \cdots \cdot \frac{(n-1)}{(n-2)} \cdot \frac{(n-1)}{n}. \]
The Wallis product gives
\[ \frac{\pi}{2} = \prod \left( \frac{2 \cdot 4 \cdot 6 \cdot \cdots}{1 \cdot 3 \cdot 5 \cdot \cdots} \right). \]
Thus, after an even number of factors, the partial product is less than $\pi/2$, and after an odd number of factors, the partial product is greater than $\pi/2$. Therefore, $T_n/T_{n+1} < \frac{1}{2} \pi T_1^2$. Similarly, when $n$ is even, $T_n/T_{n+1} < \frac{2}{\pi T_1^2}$. Since the limit of $T_n/T_{n+1} \to 1$, we conclude that
\[ \pi T_1^2 = 2. \] | 0 |
1969 | 1969_B4 | Find the maximum area covered by a closed rectangle such that any curve of unit length can be covered by it. | Place the curve so that its endpoints lie on the $x$-axis. Then take the smallest rectangle with sides parallel to the axes which covers the curve. Let its horizontal and vertical dimensions be $a$ and $b$ respectively. Let $P_0$ and $P_5$ be the endpoints of the curve, and let $P_1, P_2, P_3, P_4$ be the points on the curve, in the order named, which lie one on each of the four sides of the rectangle. Draw the broken line $P_0P_1P_2P_3P_4P_5$. This line has length at most one. The horizontal components of the segments of this broken line add up at least to $a$, since one of the vertices of the broken line lies on the left end of the rectangle and one on the right end. The vertical segments add to at least $2b$ since we start and finish on the $x$-axis and go to both the top and bottom sides. This implies that the total length of the broken line is at least $(a^2 + 4b^2)^{1/2}$.
We now have that $a$ and $b$ both lie between $0$ and $1$ and that $a^2 + 4b^2 \leq 1$. Under these conditions the product $ab$ is a maximum for $a = \frac{1}{2}\sqrt{2}, b = \frac{1}{4}\sqrt{2}$ and so the maximum of $ab$ is $\frac{1}{4}$. Thus the area of the rectangle we have constructed is at most $\boxed{\frac{1}{4}}$. | numerical | putnam (modified boxing) | Geometry Analysis | Show that any curve of unit length can be covered by a closed rectangle of area $\frac{1}{4}$. | Place the curve so that its endpoints lie on the $x$-axis. Then take the smallest rectangle with sides parallel to the axes which covers the curve. Let its horizontal and vertical dimensions be $a$ and $b$ respectively. Let $P_0$ and $P_5$ be the endpoints of the curve, and let $P_1, P_2, P_3, P_4$ be the points on the curve, in the order named, which lie one on each of the four sides of the rectangle. Draw the broken line $P_0P_1P_2P_3P_4P_5$. This line has length at most one. The horizontal components of the segments of this broken line add up at least to $a$, since one of the vertices of the broken line lies on the left end of the rectangle and one on the right end. The vertical segments add to at least $2b$ since we start and finish on the $x$-axis and go to both the top and bottom sides. This implies that the total length of the broken line is at least $(a^2 + 4b^2)^{1/2}$.
We now have that $a$ and $b$ both lie between $0$ and $1$ and that $a^2 + 4b^2 \leq 1$. Under these conditions the product $ab$ is a maximum for $a = \frac{1}{2}\sqrt{2}, b = \frac{1}{4}\sqrt{2}$ and so the maximum of $ab$ is $\frac{1}{4}$. Thus the area of the rectangle we have constructed is at most $\frac{1}{4}$. | 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 |
1970 | 1970_A3 | Find the length of the longest sequence of equal nonzero digits in which an integral square can terminate (in base 10) and find the smallest square which terminates in such a sequence. | If $x$ is an integer, then $x^2 \equiv 0, 1, 4, 6,$ or $9 \pmod{10}$. The case $x^2 \equiv 0 \pmod{10}$ is eliminated by the statement of the problem. If $x^2 \equiv 11, 55,$ or $99 \pmod{100}$, then $x^2 \equiv 3 \pmod{4}$ which is impossible. Similarly, $x^2 \equiv 66 \pmod{100}$ implies $x^2 \equiv 2 \pmod{4}$ which is also impossible. Therefore, $x^2 \equiv 44 \pmod{100}$. If $x^2 \equiv 4444 \pmod{10000}$, then $x^2 \equiv 12 \pmod{16}$, but a simple check shows that this is impossible. Finally, note that $38^2 = 1444$. Thus, the longest sequence of equal nonzero digits is $4$, and the smallest square terminating in such a sequence is $38^2 = \boxed{1444}$. | numerical | putnam | Number Theory | Find the length of the longest sequence of equal nonzero digits in which an integral square can terminate (in base 10) and find the smallest square which terminates in such a sequence. | If $x$ is an integer, then $x^2 \equiv 0, 1, 4, 6,$ or $9 \pmod{10}$. The case $x^2 \equiv 0 \pmod{10}$ is eliminated by the statement of the problem. If $x^2 \equiv 11, 55,$ or $99 \pmod{100}$, then $x^2 \equiv 3 \pmod{4}$ which is impossible. Similarly, $x^2 \equiv 66 \pmod{100}$ implies $x^2 \equiv 2 \pmod{4}$ which is also impossible. Therefore, $x^2 \equiv 44 \pmod{100}$. If $x^2 \equiv 4444 \pmod{10000}$, then $x^2 \equiv 12 \pmod{16}$, but a simple check shows that this is impossible. Finally, note that $38^2 = 1444$. Thus, the longest sequence of equal nonzero digits is $4$, and the smallest square terminating in such a sequence is $38^2 = \boxed{1444}$. | 0 |
1970 | 1970_A4 | Given a sequence $\{x_n\}$, $n=1, 2, \dots$, such that $\lim_{n \to \infty} \{x_n - x_{n-2}\} = 0$, find the value of \lim_{n \to \infty} \frac{x_n - x_{n-1}}{n}. | For $\varepsilon > 0$, let $N$ be sufficiently large so that $|x_n - x_{n-2}| < \varepsilon$ for all $n \geq N$. Note that for any $n > N$, \[ x_n - x_{n-1} = (x_n - x_{n-2}) - (x_{n-1} - x_{n-3}) + (x_{n-2} - x_{n-4}) - \dots \pm (x_{N+1} - x_{N-1}) \pm (x_N - x_{N-1}). \] Thus \[ |x_n - x_{n-1}| \leq (n-N)\varepsilon + |x_N - x_{N-1}|, \] and $\lim_{n \to \infty} \frac{x_n - x_{n-1}}{n} = \boxed{0}.$ | numerical | putnam (modified boxing) | Analysis | Given a sequence $\{x_n\}$, $n=1, 2, \dots$, such that $\lim_{n \to \infty} \{x_n - x_{n-2}\} = 0$, prove that \[ \lim_{n \to \infty} \frac{x_n - x_{n-1}}{n} = 0. \] | For $\varepsilon > 0$, let $N$ be sufficiently large so that $|x_n - x_{n-2}| < \varepsilon$ for all $n \geq N$. Note that for any $n > N$, \[ x_n - x_{n-1} = (x_n - x_{n-2}) - (x_{n-1} - x_{n-3}) + (x_{n-2} - x_{n-4}) - \dots \pm (x_{N+1} - x_{N-1}) \pm (x_N - x_{N-1}). \] Thus \[ |x_n - x_{n-1}| \leq (n-N)\varepsilon + |x_N - x_{N-1}|, \] and $\lim_{n \to \infty} \frac{x_n - x_{n-1}}{n} = 0.$ | 0 |
1970 | 1970_A5 | Determine the radius of the largest circle which can lie on the ellipsoid \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \] \( (a > b > c) \). | Since parallel cross sections of the ellipsoid are always similar ellipses, any circular cross section can be increased in size by taking a parallel cutting plane passing through the center. Every plane through \( (0, 0, 0) \) which makes a circular cross section must intersect the \( y-z \) plane. But this means that a diameter of the circular cross section must be a diameter of the ellipse \( x = 0, \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \). Hence the radius of the circle is at most \( b \). Similar reasoning with the \( x-y \) plane shows that the radius of the circle is at least \( b \), so that any circular cross section formed by a plane through \( (0, 0, 0) \) must have radius \( b \), and this will be the required maximum radius. To show that circular cross sections of radius \( b \) actually exist, consider all planes through the \( y \)-axis. It can be verified that the two planes given by \( a^2(b^2 - c^2)z^2 = c^2(a^2 - b^2)x^2 \) give circular cross sections of radius \( \boxed{b} \). | algebraic | putnam | Geometry | Determine the radius of the largest circle which can lie on the ellipsoid \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \] \( (a > b > c) \). | Since parallel cross sections of the ellipsoid are always similar ellipses, any circular cross section can be increased in size by taking a parallel cutting plane passing through the center. Every plane through \( (0, 0, 0) \) which makes a circular cross section must intersect the \( y-z \) plane. But this means that a diameter of the circular cross section must be a diameter of the ellipse \( x = 0, \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \). Hence the radius of the circle is at most \( b \). Similar reasoning with the \( x-y \) plane shows that the radius of the circle is at least \( b \), so that any circular cross section formed by a plane through \( (0, 0, 0) \) must have radius \( b \), and this will be the required maximum radius. To show that circular cross sections of radius \( b \) actually exist, consider all planes through the \( y \)-axis. It can be verified that the two planes given by \( a^2(b^2 - c^2)z^2 = c^2(a^2 - b^2)x^2 \) give circular cross sections of radius \( \boxed{b} \). | 0 |
1970 | 1970_A6 | Three numbers are chosen independently at random, one from each of the three intervals $[0, L_i]$ $(i=1, 2, 3)$. If the distribution of each random number is uniform with respect to length in the interval it is chosen from, determine the expected value of the smallest of the three numbers chosen. | Let $x$ be selected from $[0, L_1]$, $y$ from $[0, L_2]$, and $z$ from $[0, L_3]$, and assume $L_3 \geq L_2 \geq L_1$. Let $X = \min(x, y, z)$. The expected value is given by \[ E[X] = \int_0^{L_1} \int_0^{L_2} \int_0^{L_3} X \; dz \; dy \; dx. \] Breaking this into components, we compute the integral \[ L_1 L_2 L_3 E[X] = \int_0^{L_1} \int_0^{L_2} \left( \int_0^\mu z \; dz + \int_\mu^{L_3} \mu \; dz \right) dy \; dx, \] where $\mu = \min(x, y)$. This evaluates to \[ \int_0^{L_1} \int_0^{L_2} \{ L_3 \mu - \frac{1}{2} \mu^2 \} dy \; dx. \] Further, we split the integral and calculate the contributions for $x$ and $y$, leading to \[ \frac{1}{2} L_1^2 L_2 L_3 - \frac{1}{6} L_1^3 (L_2 + L_3) + \frac{1}{12} L_1^4. \] Hence, the expected value is \[ \boxed{\frac{1}{2} L_1^2 L_2 L_3 - \frac{1}{6} L_1^3 (L_2 + L_3) + \frac{1}{12} L_1^4}. \] | algebraic | putnam | Probability Calculus | Three numbers are chosen independently at random, one from each of the three intervals $[0, L_i]$ $(i=1, 2, 3)$. If the distribution of each random number is uniform with respect to length in the interval it is chosen from, determine the expected value of the smallest of the three numbers chosen. | Let $x$ be selected from $[0, L_1]$, $y$ from $[0, L_2]$, and $z$ from $[0, L_3]$, and assume $L_3 \geq L_2 \geq L_1$. Let $X = \min(x, y, z)$. The expected value is given by \[ E[X] = \int_0^{L_1} \int_0^{L_2} \int_0^{L_3} X \; dz \; dy \; dx. \] Breaking this into components, we compute the integral \[ L_1 L_2 L_3 E[X] = \int_0^{L_1} \int_0^{L_2} \left( \int_0^\mu z \; dz + \int_\mu^{L_3} \mu \; dz \right) dy \; dx, \] where $\mu = \min(x, y)$. This evaluates to \[ \int_0^{L_1} \int_0^{L_2} \{ L_3 \mu - \frac{1}{2} \mu^2 \} dy \; dx. \] Further, we split the integral and calculate the contributions for $x$ and $y$, leading to \[ \frac{1}{2} L_1^2 L_2 L_3 - \frac{1}{6} L_1^3 (L_2 + L_3) + \frac{1}{12} L_1^4. \] Hence, the expected value is \[ \boxed{\frac{1}{2} L_1^2 L_2 L_3 - \frac{1}{6} L_1^3 (L_2 + L_3) + \frac{1}{12} L_1^4}. \] | 0 |
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