\documentclass[amssymb,twocolumn]{revtex4}
\usepackage{times,amsmath,latexsym,epsfig}

\def\mymod#1{\,\,({\rm mod}\, #1)}

\begin{document}
\title{Solutions to the 57th William Lowell Putnam Mathematical Competition \\ 
    Saturday, December 7, 1996}
\author{Manjul Bhargava and Kiran Kedlaya}
\maketitle

\begin{itemize}
\item[A-1]
If $x$ and $y$ are the sides of two squares with combined area 1, then
$x^2 + y^2 = 1$. Suppose without loss of generality that $x \geq y$.
Then the shorter side of a rectangle containing both squares without
overlap must be at least $x$, and the longer side must be at least
$x+y$. Hence the desired value of $A$ is the maximum of $x(x+y)$.

To find this maximum, we let $x = \cos \theta, y = \sin \theta$ with
$\theta \in [0, \pi/4]$. Then we are to maximize
\begin{align*}
\cos^2 \theta + \sin \theta \cos \theta
&=& \frac 12 (1 + \cos 2\theta + \sin 2\theta) \\
&=& \frac 12 + \frac{\sqrt{2}}{2} \cos (2\theta - \pi/4) \\
&\leq& \frac{1 + \sqrt{2}}{2},
\end{align*}
with equality for $\theta = \pi/8$. Hence this value is the desired
value of $A$.

\item[A-2]
Let $O_1$ and $O_2$ be the centers of $C_1$ and $C_2$, respectively.
(We are assuming $C_1$ has radius 1 and $C_2$ has radius 3.)
Then the
desired locus is an annulus centered at the midpoint of $O_1O_2$, with
inner radius 1 and outer radius 2.

For a fixed point $Q$ on $C_2$, the locus of the midpoints of the
segments $PQ$ for $P$ lying on $C_1$ is the image of $C_1$ under a
homothety centered at $Q$ of radius $1/2$, which is a circle of radius
$1/2$. As $Q$ varies, the center of this smaller circle traces out a
circle $C_3$ of radius $3/2$ (again by homothety). By considering the two
positions of $Q$ on the line of centers of the circles, one sees that
$C_3$ is centered at the midpoint of $O_1O_2$, and the locus is now
clearly the specified annulus.

\item[A-3]
The claim is false. There are $\binom 63 = 20$ ways to choose 3 of the
6 courses; have each student choose a different set of 3 courses. Then
each pair of courses is chosen by 4 students (corresponding to the
four ways to complete this pair to a set of 3 courses) and is not
chosen by 4 students (corresponding to the 3-element subsets of the
remaining 4 courses).

Note: Assuming that no two students choose the same courses,
the above counterexample is unique (up to permuting students).
This may be seen as follows: Given a group of students, suppose that
for any pair of courses (among the six) there are at most 4 students
taking both, and at most 4 taking neither.  Then there are at most
$120=(4+4){6\choose 2}$ pairs $(s,p)$, where $s$ is a student, and $p$
is a set of two courses of which $s$ is taking either both or none.
On the other hand, if a student $s$ is taking $k$ courses, then he/she
occurs in $f(k)={k\choose 2}+{{6-k}\choose 2}$ such pairs $(s,p)$.  As
$f(k)$ is minimized for $k=3$, it follows that every student occurs in
at least $6={3\choose 2}+{3\choose 2}$ such pairs $(s,p)$.  Hence
there can be at most $120/6=20$ students, with equality only if each
student takes 3 courses, and for each set of two courses, there are
exactly 4 students who take both and exactly 4 who take neither.
Since there are only 4 ways to complete a given pair of courses to a
set of 3, and only 4 ways to choose 3 courses not containing the given
pair, the only way for there to be 20 students (under our hypotheses) 
is if all sets of 3 courses are in fact taken.  This is the desired conclusion.

However, Robin Chapman has pointed out that the solution is not unique
in the problem as stated, because a given selection of courses may be
made by more than one student. One alternate solution is to identify
the 6 courses with pairs of antipodal vertices of an icosahedron, and
have each student pick a different face and choose the three vertices
touching that face. In this example, each of 10 selections is made by
a pair of students.

\item[A-4]
In fact, we will show that such a function $g$ exists with the
property that $(a,b,c) \in S$ if and only if $g(d) < g(e) < g(f)$ for
some cyclic permutation $(d,e,f)$ of $(a,b,c)$. We proceed by
induction on the number of elements in $A$. If $A =
\{a,b,c\}$ and $(a,b,c) \in S$, then choose $g$ with $g(a) < g(b) <
g(c)$, otherwise choose $g$ with $g(a) > g(b) > g(c)$.

Now let $z$ be an element of $A$ and $B = A - \{z\}$.
Let $a_{1}, \dots, a_{n}$ be the elements of $B$ labeled such that 
$g(a_{1}) < g(a_{2}) < \cdots < g(a_{n})$. We claim that there exists 
a unique $i \in \{1, \dots, n\}$ such that $(a_{i}, z, a_{i+1}) 
\in S$, where hereafter $a_{n+k} = a_{k}$.

We show existence first. Suppose no such $i$ exists; then for all 
$i,k \in \{1, \dots, n\}$, we have $(a_{i+k}, z, a_{i}) \notin S$. 
This holds by property 1 for $k=1$ and by induction on $k$ in 
general, noting that
\begin{align*}
(a_{i+k+1}, z, &a_{i+k}), (a_{i+k}, z, a_{i})  \in S \\
&\Rightarrow
(a_{i+k}, a_{i+k+1}, z), (z, a_{i}, a_{i+k}) \in S  \\
&\Rightarrow (a_{i+k+1},z,a_{i}) \in S.
\end{align*}
Applying this when $k=n$, we get $(a_{i-1}, z, a_{i}) \in S$, 
contradicting the fact that $(a_{i}, z, a_{i-1}) \in S$. Hence 
existence follows.

Now we show uniqueness. Suppose $(a_{i}, z, a_{i+1}) \in S$; then for 
any $j \neq i-1, i, i+1$, we have $(a_{i}, a_{i+1}, a_{j}), (a_{j}, 
a_{j+1}, a_{i}) \in S$ by the 
assumption on $G$. Therefore
\begin{eqnarray*}
(a_{i}, z, a_{i+1}), (a_{i+1}, a_{j}, a_{i}) \in S &\Rightarrow&
(a_{j}, a_{i}, z) \in S \\
(a_{i}, z, a_{j}), (a_{j}, a_{j+1}, a_{i}) \in S &\Rightarrow& (z, 
a_{j}, a_{j+1}),
\end{eqnarray*}
so $(a_{j}, z, a_{j+1}) \notin S$. The case $j =i+1$ is ruled out by
\[
(a_{i}, z, a_{i+1}), (a_{i+1}, a_{i+2}, a_{i}) \in S \Rightarrow (z, 
a_{i+1}, a_{i+2}) \in S
\]
and the case $j=i-1$ is similar.

Finally, we put $g(z)$ in $(g(a_{n}), + \infty)$ if $i = n$, and 
$(g(a_{i}), g(a_{i+1}))$ otherwise; an analysis similar to that above 
shows that $g$ has the desired property.

\item[A-5] (due to Lenny Ng)
For $1 \leq n \leq p-1$, $p$ divides $\binom pn$ and
\begin{align*}
\frac{1}{p} \binom pn &= \frac{1}{n} \frac{p-1}{1} \frac{p-2}{2} \cdots 
\frac{p-n+1}{n-1} \\
&\equiv \frac{(-1)^{n-1}}{n} \mymod{p},
\end{align*}
where the congruence $x \equiv y \mymod{p}$ means that $x-y$ is a
rational number whose numerator, in reduced form, is divisible by $p$.
Hence it suffices to show that
\[
\sum_{n=1}^k \frac{(-1)^{n-1}}{n} \equiv 0 \mymod{p}.
\]
We distinguish two cases based on $p \mymod{6}$. First suppose $p = 
6r+1$, so that $k = 4r$. Then
\begin{align*}
\sum_{n=1}^{4r}\ &\frac{(-1)^{n-1}}{n}  \\
&= \sum_{n=1}^{4r} \frac{1}{n} - 2 \sum_{n=1}^{2r} \frac{1}{2n} \\
&= \sum_{n=1}^{2r} \left( \frac{1}{n} - \frac{1}{n} \right)
+ \sum_{n=2r+1}^{3r} \left( \frac{1}{n} + \frac{1}{6r+1-n} \right) \\
&= \sum_{n=2r+1}^{3r} \frac{p}{n(p-n)} \equiv 0 \mymod{p},
\end{align*}
since $p = 6r+1$.

Now suppose $p = 6r+5$, so that $k = 4r + 3$. A similar argument gives
\begin{align*}
\sum_{n=1}^{4r+3}\ &\frac{(-1)^{n-1}}{n} \\
&= \sum_{n=1}^{4r+3} \frac{1}{n} + 2 \sum_{n=1}^{2r+1} \frac{1}{2n} \\
&= \sum_{n=1}^{2r+1} \left( \frac{1}{n} - \frac{1}{n} \right) 
+ \sum_{n=2r+2}^{3r+2} \left( \frac{1}{n} + \frac{1}{6r+5-n} \right) 
\\ &= \sum_{n=2r+2}^{3r+2} \frac{p}{n(p-n)} \equiv 0 \mymod{p}.
\end{align*}

\item[A-6]
We first consider the case $c \leq 1/4$; we shall show in this case
$f$ must be constant. The relation
\[
f(x) = f(x^2 + c) = f((-x)^2 + c) = f(-x)
\]
proves that $f$ is an even function. Let $r_1 \leq r_2$ be the roots of
$x^2 + c - x$, both of which are real. If $x > r_{2}$, define $x_{0} = 
x$ and $x_{n+1} = \sqrt{x_{n} - c}$ for each positive integer $x$. By 
induction on $n$, $r_{2} < x_{n+1} < x_{n}$ for all $n$, so the 
sequence $\{x_{n}\}$ tends to a limit $L$ which is a root of $x^{2} + 
c = x$ not less than $r_{2}$. Of course this means $L = r_{2}$. 
Since $f(x) = f(x_{n})$ for all $n$ and $x_{n} \to r_{2}$, we 
conclude $f(x) = f(r_{2})$, so $f$ is constant on $x \geq r_{2}$.

If $r_{1} < x < r_{2}$ and $x_{n}$ is defined as before, then by 
induction, $x_{n} < x_{n+1} < r_{2}$. Note that the 
sequence can be defined because $r_{1} > c$; the latter follows by 
noting that the polynomial $x^{2} - x + c$ is positive at $x = c$ and 
has its minimum at $1/2 > c$, so both roots are greater than $c$. In 
any case, we deduce that $f(x)$ is also constant on $r_{1} \leq x \leq 
r_{2}$.

Finally, suppose $x < r_{1}$. Now define $x_{0} = x, x_{n+1} = 
x_{n}^{2} + c$. Given that $x_{n} < r_{1}$, we have $x_{n+1} > 
x_{n}$. Thus if we had $x_{n} < r_{1}$ for all $n$, by the same argument as 
in the first case we deduce $x_{n} \to r_{1}$ and so $f(x) = 
f(r_{1})$. Actually, this doesn't happen; eventually we have $x_{n} > 
r_{1}$, in which case $f(x) = f(x_{n}) = f(r_{1})$ by what we have 
already shown. We conclude that $f$ is a constant function. (Thanks 
to Marshall Buck for catching an inaccuracy in a previous version of 
this solution.)

Now suppose $c > 1/4$. Then the sequence $x_n$ defined by $x_0 = 0$
and $x_{n+1} = x_n^2 + c$ is strictly increasing and has no limit
point. Thus if we define $f$ on $[x_0, x_1]$ as any continuous
function with equal values on the endpoints, and extend the definition
from $[x_n, x_{n+1}]$ to $[x_{n+1}, x_{n+2}]$ by the relation $f(x) =
f(x^2 + c)$, and extend the definition further to $x < 0$ by the
relation $f(x) = f(-x)$, the resulting function has the desired
property. Moreover, any function with that property clearly has this form.

\item[B-1]
Let $[n]$ denote the set $\{1,2,\ldots,n\}$, and let $f_n$ denote the
number of minimal selfish subsets of $[n]$.  Then the number of
minimal selfish subsets of $[n]$ not containing $n$ is equal to
$f_{n-1}$.  On the other hand, for any minimal selfish subset of $[n]$
containing $n$, by subtracting 1 from each element, and then taking
away the element $n-1$ from the set, we obtain a minimal selfish
subset of $[n-2]$ (since $1$ and $n$ cannot both occur in a selfish
set).  Conversely, any minimal selfish subset of $[n-2]$ gives rise to
a minimal selfish subset of $[n]$ containing $n$ by the inverse
procedure.  Hence the number of minimal selfish subsets of $[n]$
containing $n$ is $f_{n-2}$.  Thus we obtain $f_n=f_{n-1}+f_{n-2}$.
Since $f_1=f_2=1$, we have $f_n=F_n$, where $F_n$ denotes the $n$th
term of the Fibonacci sequence.

\item[B-2]
By estimating the area under the graph of $\ln x$ using upper and
lower rectangles of width 2, we get
\begin{align*}
\int_1^{2n-1} \ln x\,dx &\leq 2(\ln(3) + \cdots + \ln(2n-1)) \\
&\leq \int_3^{2n+1} \ln x\,dx.
\end{align*}
Since $\int \ln x\,dx = x \ln x - x + C$, we have, upon exponentiating
and taking square roots,
%\begin{eqnarray*}
%\left( \frac{2n-1}{e} \right)^{\frac{2n-1}{2}}
%< (2n-1)^{\frac{2n-1}{2}} e^{-n+1}
%&\leq& 1 \cdot 3 \cdots (2n-1) \\
%&\leq& (2n+1)^{\frac{2n+1}{2}} \frac{e^{-n+1}}{3^{3/2}} 
%< \left( \frac{2n+1}{e} \right)^{\frac{2n+1}{2}},
%\end{eqnarray*}
\begin{align*}
\left( \frac{2n-1}{e} \right)^{\frac{2n-1}{2}}
&< (2n-1)^{\frac{2n-1}{2}} e^{-n+1} \\
& \leq 1 \cdot 3 \cdots (2n-1) \\
& \leq (2n+1)^{\frac{2n+1}{2}} \frac{e^{-n+1}}{3^{3/2}} \\
& < \left( \frac{2n+1}{e} \right)^{\frac{2n+1}{2}},
\end{align*}
using the fact that $1 < e < 3$.

\item[B-3]
View $x_1, \dots, x_n$ as an arrangement of the numbers $1, 2, \dots,
n$ on a circle.
We prove that the optimal arrangement is
\[
\dots, n-4, n-2, n, n-1, n-3, \dots
\]
To show this, note that if
$a, b$ is a pair of adjacent numbers and $c,d$ is another pair (read
in the same order around the circle) with $a < d$ and $b > c$, then
the segment from $b$ to $c$ can be reversed, increasing the sum by
\[
ac + bd - ab - cd = (d-a)(b-c) > 0.
\]
Now relabel the numbers so they appear in order as follows:
\[
\dots, a_{n-4}, a_{n-2}, a_n = n, a_{n-1}, a_{n-3}, \dots
\]
where without loss of generality we assume $a_{n-1} > a_{n-2}$. By
considering the pairs $a_{n-2}, a_n$ and $a_{n-1}, a_{n-3}$ and using
the trivial fact $a_n > a_{n-1}$, we deduce $a_{n-2} > a_{n-3}$. We
then compare the pairs $a_{n-4}, a_{n-2}$ and $a_{n-1}, a_{n-3}$, and
using that $a_{n-1} > a_{n-2}$, we deduce $a_{n-3} > a_{n-4}$. 
Continuing in this
fashion, we prove that $a_n > a_{n-1} > \dots > a_1$ and
so $a_k = k$ for $k = 1, 2, \dots, n$, i.e.\ that the optimal
arrangement is as claimed. In particular, the maximum value of the sum
is
\begin{align*}
1 &\cdot 2 + (n-1)\cdot n + 1 \cdot 3 + 2 \cdot 4 + \cdots + (n-2)\cdot n \\
&= 2 + n^2 - n + (1^2 - 1) + \cdots + [(n-1)^2 - 1] \\
&= n^2 - n + 2 - (n-1) + \frac{(n-1)n(2n-1)}{6} \\
&= \frac{2n^3 + 3n^2 - 11n + 18}{6}.
\end{align*}

Alternate solution: We prove by induction that the value given above
is an upper bound; it is clearly a lower bound because of the
arrangement given above. Assume this is the case for $n-1$. The optimal
arrangement for $n$ is obtained from some arrangement for $n-1$ by
inserting $n$ between some pair $x, y$ of adjacent terms. This
operation increases the sum by $nx + ny - xy = n^2 - (n-x)(n-y)$,
which is an increasing function of both $x$ and $y$. In particular,
this difference is maximal when $x$ and $y$ equal $n-1$ and $n-2$.
Fortunately, this yields precisely the difference between the claimed
upper bound for $n$ and the assumed upper bound for $n-1$, completing
the induction.

\item[B-4]
Suppose such a matrix $A$ exists.  If the eigenvalues of $A$ (over
the complex numbers) are distinct, then there exists a complex 
matrix $C$ such that $B=CAC^{-1}$ is diagonal.  Consequently,
$\sin B$ is diagonal.  But then $\sin A=C^{-1}(\sin B)C$ must
be diagonalizable, a contradiction.  Hence the eigenvalues of $A$
are the same, and $A$ has a conjugate $B=CAC^{-1}$ over
the complex numbers of the form
\[
\left(
\begin{array}{cc}
x & y\\
0 & x
\end{array}
\right).
\]
A direct computation shows that
\[
\sin B = \left(
\begin{array}{cc}
\sin x & y\cdot \cos x\\
0 & \sin x
\end{array}
\right).
\]
Since $\sin A$ and $\sin B$ are conjugate, their eigenvalues
must be the same, and so we must have $\sin x=1$.  This implies
$\cos x=0$, so that $\sin B$ is the identity matrix, as must be $\sin 
A$, a contradiction. 
Thus $A$ cannot exist.

Alternate solution (due to Craig Helfgott and Alex Popa):
Define both $\sin A$ and $\cos A$ by the usual power series.
Since $A$ commutes with itself, the power series identity
\[
\sin^2 A+\cos^2 A = I
\]
holds.  But if $\sin A$ is the given matrix, then by the
above identity, $\cos^2 A$ must equal 
$\left(
\begin{array}{cc}
0 & -2\cdot 1996\\
0 & 0
\end{array}
\right)$
which is a nilpotent matrix.  Thus $\cos A$ is also nilpotent.
However, the square of any $2\times 2$ nilpotent matrix must be zero
(e.g., by the Cayley-Hamilton theorem).  This is a contradiction.

\item[B-5]
Consider a $1 \times n$ checkerboard, in which we write an $n$-letter
string, one letter per square. If the string is balanced, we can cover
each pair of adjacent squares containing the same letter with a $1
\times 2$ domino, and these will not overlap (because no three in a
row can be the same). Moreover, any domino is separated from the next
by an even number of squares, since they must cover opposite letters,
and the sequence must alternate in between.

Conversely, any arrangement of dominoes where adjacent dominoes are
separated by an even number of squares corresponds to a unique
balanced string, once we choose whether the string starts with $X$ or
$O$. In other words, the number of balanced strings is twice the
number of acceptable domino arrangements.

We count these arrangements by numbering the squares $0,1,\dots,n-1$
and distinguishing whether the dominoes start on even or odd numbers.
Once this is decided, one simply chooses whether or not to put a 
domino in each eligible position. Thus
we have $2^{\lfloor n/2 \rfloor}$ arrangements in the first case and $2^{\lfloor
(n-1)/2 \rfloor}$ in the second, but note that the case of no dominoes has
been counted twice. Hence the number of balanced strings is
\[
2^{\lfloor (n+2)/2 \rfloor} + 2^{\lfloor (n+1)/2 \rfloor} - 2.
\]

\item[B-6]
We will prove the claim assuming only that the convex hull of the 
points $(a_{i}, b_{i})$ contains the origin in its interior. (Thanks 
to Marshall Buck for pointing out that the last three words are 
necessary in the previous sentence!) Let $u = \log x, v = \log
y$ so that the left-hand side of the given equation is
\begin{multline}
(a_1, b_1) \exp(a_1 u + b_1 v) + (a_2, b_2) \exp(a_2 u + b_2 v) + \\
\cdots + (a_n, b_n) \exp(a_n u + b_n v).
\end{multline}
Now note that (1) is the gradient of the function
\begin{gather*}
f(u,v) = exp(a_1 u + b_1 v) + 
exp(a_2 u + b_2 v) + \\ 
\cdots + exp(a_n u + b_n v),
\end{gather*}
and so it suffices to show $f$ has a critical point. We will in fact
show $f$ has a global minimum.

Clearly we have
\[
f(u,v) \geq \exp\left( \max_i (a_i u + b_i v) \right).
\]
Note that this maximum is positive for $(u,v) \neq (0,0)$:  if we had
$a_i u + b_i v < 0$ for all $i$, then the subset $ur + vs < 0$ of the
$rs$-plane would be a half-plane containing all of the points $(a_i, 
b_i)$, whose convex hull would then not contain the origin, a
contradiction. 

The function $\max_{i} (a_{i}u + b_{i}v)$ is clearly 
continuous on the unit circle $u^{2} + v^{2} = 1$, which is compact. 
Hence it has a global minimum $M > 0$, and so for all $u,v$,
\[
\max_{i} (a_{i} u + b_{i} v) \geq M \sqrt{u^{2} + v^{2}}.
\]
In particular, $f \geq n+1$ on the disk of radius $\sqrt{(n+1)/M}$.
Since $f(0,0) = n$, the infimum of $f$ is the same over the entire
$uv$-plane as over this disk, which again is compact.
Hence $f$ attains its infimal value at some point in the disk,
which is the desired global minimum.

Noam Elkies has suggested an alternate solution as follows: for $r >
0$, draw the loop traced by (1) as $(u,v)$ travels
counterclockwise around the circle $u^2 + v^2 = r^2$. For $r=0$, this
of course has winding number 0 about any point, but for $r$ large, one
can show this loop has winding number 1 about the origin, so somewhere in
between the loop must pass through the origin. (Proving this latter
fact is a little tricky.)

\end{itemize}
\end{document}