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\begin{center}
2014 Missouri Collegiate Mathematics Competition \\ Session I
\end{center}


\noindent 1. Let $0 \le a \le 1$ be given.  Determine all nonnegative continuous functions
$f$ on $\lbrack 0,1 \rbrack$ (or prove there are none) which satisfy the following three
conditions.
\begin{align*}
\int_0^1 f(x) \, dx &= 1 \\
\int_0^1 xf(x) \, dx &= a \\
\int_0^1 x^2 f(x) \, dx &= a^2 .
\end{align*}
\ \\ \ \\
Solution.
\ \\ \ \\
\noindent There is no such function.  Multiply the three equations by $a^2$, $-2a$, and 1,
respectively, and add, getting
$$\int_0^1 f(x) (a-x)^2 \, dx = 0 .$$
But the only nonnegative continuous function satisfying this condition is $f(x) \equiv 0$,
and so no continuous nonnegative function can satisfy the three conditions.
\ \\ \ \\
\noindent 2. Let $T_0$ be an isosceles triangle with base $b$ and base angles $\alpha$.  Define
a sequence $\{ T_n \} _{n=0}^\infty$ of triangles, recursively, as follows.  The base of each
triangle $T_n$ is $b$, and the base angles of triangle $T_{n+1}$ have half the measure of the base
angles of triangle $T_n$.  Evaluate
$$\lim_{n \to \infty} \frac{2^n \cdot \textrm{area} (T_n)}{\textrm{area} (T_0)} .$$
\ \\ \ \\
Solution.
\ \\ \ \\
\noindent The base angles of triangle $T_n$ have measure $\frac{\alpha}{2^n}$.  Thus, the altitude,
$h_n$ of triangle $T_n$ is given by $\tan \left( \frac{\alpha}{2^n} \right) = \frac{h_n}{b/2}$,
so that $h_n = \frac{1}{2} b \tan \left( \frac{\alpha}{2^n} \right)$, and $\textrm{area}(T_n) =
\frac{1}{4} b^2 \tan \left( \frac{\alpha}{2^n} \right)$.  Therefore,
$$\frac{2^n \cdot \frac{1}{4} b^2 \tan \left( \frac{\alpha}{2^n} \right)}
{\frac{1}{4} b^2 \tan \alpha} = \frac{2^n \cdot \tan \left( \frac{\alpha}{2^n} \right)}
{\tan \alpha} = \frac{\alpha \cdot \left( \tan \left( \frac{\alpha}{2^n} \right) / \left(
\frac{\alpha}{2^n} \right) \right)}{\tan \alpha} ,$$
so that
$$\lim_{n \to \infty} \frac{2^n \cdot \textrm{area} (T_n)} {\textrm{area} (T_0)} =
\frac{\alpha}{\tan \alpha} .$$
\ \\ \ \\
\noindent 3. Find the sum of the series $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6}
+ \frac{1}{8} + \frac{1}{9} + \frac{1}{12} + \cdots$, where the terms are the reciprocals of
the positive integers whose only prime factors are twos and threes.
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\begin{align*}
&1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{2} \frac{1}{2} + \frac{1}{2} \frac{1}{3} + \frac{1}{2}
\frac{1}{2} \frac{1}{2} + \frac{1}{3} \frac{1}{3} + \frac{1}{2} \frac{1}{2} \frac{1}{3} + \cdots \\
&= \left( \frac{1}{2} \right)^0 \left( \frac{1}{2} \right)^0 +
\left( \frac{1}{2} \right)^1 \left( \frac{1}{3} \right)^0 + \left( \frac{1}{2} \right)^0
\left( \frac{1}{3} \right)^1 + \left( \frac{1}{2} \right)^1 \left( \frac{1}{2} \right)^1
+ \left( \frac{1}{2} \right)^1 \left( \frac{1}{3} \right)^1 \\
&\ \ \ \ \ + \left( \frac{1}{2} \right)^1
\left( \frac{1}{2} \right)^1 \left( \frac{1}{2} \right)^1 + \cdots \\
&= \left( 1 + \frac{1}{2} + \left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^3 +
\cdots \right) \left( 1 + \frac{1}{3} + \left( \frac{1}{3} \right)^2 + \left( \frac{1}{3} \right)^3
+ \cdots \right) \\
&= \sum_{m=0}^\infty \sum_{n=0}^\infty \left( \frac{1}{2} \right)^m \left( \frac{1}{3} \right)^n
= \sum_{m=0}^\infty \left( \frac{1}{2} \right)^m \sum_{n=0}^\infty \left( \frac{1}{3} \right)^n \\
&= \left( \frac{1}{1 - \frac{1}{2}} \right) \left( \frac{1}{1 - \frac{1}{3}} \right) = 3.
\end{align*}
\ \\ \ \\
\noindent 4. Let $Q$ be an arbitrary convex quadrilateral with vertices $A_0$, $A_1$, $A_2$, and
$A_3$ ordered counterclockwise.  Let $M_i$ be the midpoint of side $A_i A_{i+1}$ interpreting the
subscripts mod 4.  Then draw the medians $A_i M_{i+2}$ (again interpreting the subscripts mod 4).
Prove that the quadrilaterals $A_0 M_0 A_2 M_2$ and $A_1 M_1 A_3 M_3$ have equal areas.
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\begin{center}
\includegraphics[width=4in]{leon2fig.eps}
\end{center}
\ \\ \ \\
\noindent Draw diagonal $A_0 A_2$.  Then because $M_0$ is the midpoint of $A_0 A_1$ the triangles
$A_0 M_0 A_2$ and $M_0 A_1 A_2$ have equal bases and the same altitude, and so have equal areas.
Similarly, triangles $A_2 M_2 A_0$ and $M_2 A_3 A_0$ have equal areas.  Therefore, denoting the
area of the figure by $\vert \ast \vert$, $\vert A_0 M_0 A_2 M_2 \vert = \vert M_0 A_1 A_2 \vert
+ \vert M_2 A_3 A_0 \vert$ making $\vert A_0 M_0 A_2 M_2 \vert = \frac{1}{2} \vert Q \vert$.
Draw diagonal $A_1 A_3$ and use the same reasoning to show that $\vert A_1 M_1 A_3 M_3 \vert
= \frac{1}{2} \vert Q \vert$, making $\vert A_0 M_0 A_2 M_2 \vert = \vert A_1 M_1 A_3 M_3 \vert$.
\ \\ \ \\
\noindent From Rick Mabry, \textit{Crosscut convex quadrilaterals}, Math.~Mag.,
\textbf{84} (2011), 16--25.
\ \\ \ \\
\noindent 5. For which positive integers $a$ does there exist a right triangle with integer
sides, at least one of which is $a$?
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent  If there were such a triangle for $a=1$, then by the Pythagorean Theorem, there would be positive integers $b, c$ such that either $b^2 + c^2 = 1$ or $1 = c^2 - b^2 = (c+b)(c-b)$, neither of which is possible.  Likewise, for $a=2$, neither of the equations $b^2 + c^2 = 4$ nor $4 = c^2 - b^2$ has positive integer solutions.

For all positive integers $a \ge 3$, however, there does exist such a triangle with $a$ as a leg.  To see this, we observe first that if the positive integer $a_0$ is a leg of a right triangle and $k$ is any positive integer, then $k a_0$ is a leg of a right triangle since $a_0^2 + b_0^2 = c_0^2$ implies that $(ka_0)^2 + (kb_0)^2 = (kc_0)^2$.  Since every positive integer greater than or equal to three is a multiple either of an odd number or of $4$, it suffices to consider the cases $a=4$ and $a \ge 3$ odd.  The Pythagorean triple $(3,4,5)$ dispenses with the case $a=4$.  Suppose now that $a \ge 3$ is odd.  We write the Pythagorean Theorem in the form $a^2 = (c+b)(c-b)$.  To find positive integers $b$ and $c$ for which this last equation holds, we solve the system
\begin{align*}
&c+b = a^2 \\
&c-b = 1,
\end{align*}
obtaining
$$b = \frac{a^2 - 1}{2},\ \ c = \frac{a^2 + 1}{2} .$$
Since $a$ is odd, these are integers; since $a \ge 3$, they are positive.
\vfill\eject
\begin{center}
2014 Missouri Collegiate Mathematics Competition \\ Session II
\end{center}

\noindent 1. 101 marbles are numbered from 1 to 101.  The marbles are divided between two baskets A and B.
The marble numbered 40 is in basket A.  This marble is removed from basket A and put in basket B.  The
average of all the
numbers on the marbles in A increases by $\frac{1}{4}$.  The average of all the numbers on the marbles
in B increases by $\frac{1}{4}$ too.  How many marbles were there originally in basket A?
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent Let $x_0, x_1, \ldots , x_n$ (respectively $x_{n+1}, \ldots , x_{100}$)
denote the numbers on the marbles originally in basket A (respectively in basket B) with $x_0 = 40$.
The original averages are
$$m_A = \frac{40+x_1 + \cdots + x_n}{n+1} \ \ \textrm{and}\ \ m_B = \frac{x_{n+1} + \cdots + x_{100}}
{100-n} .$$
After the marble numbered 40 is removed from basket A and put in basket B, the averages become
$$m_A' = \frac{x_1 + \cdots + x_n}{n}\ \ \textrm{and}\ \ m_B' = \frac{40+x_{n+1} + \cdots + x_{100}}
{101-n} .$$
Expressing $m_A' = m_A + \frac{1}{4}$ and $m_B' = m_B + \frac{1}{4}$, we easily get:
$$\frac{x_1 + \cdots + x_n}{n(n+1)} = \frac{40}{n+1} + \frac{1}{4}$$
and
$$\frac{x_{n+1} + \cdots + x_{100}}{(100-n)(101-n)} = \frac{40}{101-n} - \frac{1}{4}$$
which yield
\begin{equation}\label{eq1}
x_1 + \cdots + x_n = \frac{n^2 + 161n}{4}
\end{equation}
and
\begin{equation}\label{eq2}
x_{n+1} + \cdots + x_{100} = \frac{5900 + 41n - n^2}{4} .
\end{equation}
Since $x_1 + \cdots + x_n + x_{n+1} + \cdots + x_{100} = (1 + 2 + \cdots + 100 + 101) - 40 =
\frac{101 \times 102}{2} - 40$, we obtain by addition of (\ref{eq1}) and (\ref{eq2}):
$$5900 + 202n = 204 \times 101 - 160.$$
Hence, $n=72$ and there were initially 73 marbles in basket A.
\ \\ \ \\
\noindent This is Problem 3 from the Dutch Mathematical Olympiad - 1999.  The solution is by Michel
Bataille.
\ \\ \ \\
\noindent 2. If $f(x) = \sin ^6 \left( \frac{x}{4} \right) + \cos ^6 \left( \frac{x}{4} \right)$,
find $f^{(2014)} (0)$.  (Note that $f^{(n)} (x)$ refers to the $n$th derivative of $f$ evaluated
at $x$.)
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent We have
\begin{align*}
&\sin^6 \left( \frac{x}{4} \right) + \cos^6 \left( \frac{x}{4} \right) \\
&= \left( \sin^2 \left( \frac{x}{4} \right) + \cos^2 \left( \frac{x}{4} \right) \right)
\left( \sin^4 \left( \frac{x}{4} \right) - \sin^2 \left( \frac{x}{4} \right) \cos^2 \left( \frac{x}{4} \right) + \cos^4 \left( \frac{x}{4} \right) \right) \\
&= \sin^4 \left( \frac{x}{4} \right) + 2 \sin^2 \left( \frac{x}{4} \right)
\cos ^2 \left( \frac{x}{4} \right) + \cos^4 \left( \frac{x}{4} \right) - 3 \sin^2 \left( \frac{x}{4}
\right) \cos^2 \left( \frac{x}{4} \right) \\
&= 1 - \frac{3}{4} \sin^2 \left( \frac{x}{2} \right) = 1 - \frac{3}{4} \left( \frac{1 - \cos x}{2}
\right) \\
&= \frac{5}{8} + \frac{3}{8} \cos x .
\end{align*}
Thus, $f^{(2014)} (x) = - \frac{3}{8} \cos x$ and $f^{(2014} (0) = - \frac{3}{8}$.
\ \\ \ \\
\noindent Similar to a problem from the 2008 Harvard-MIT Mathematics Tournament.
\ \\ \ \\
\noindent 3. A polynomial $P(x)$ is known to be of the form $P(x) = x^{15} - 9x^{14} + \cdots - 7$,
where the ellipsis ($\cdots$) represent unknown intermediate terms.  It is also known that all
roots of $P(x)$ are integers.  Find the roots (including multiplicities) of $P(x)$.
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent Let $r_1, r_2, \ldots , r_{15}$ be the roots of $P(x)$.  Since $P(x) = \prod_{i=1}^{15}
(x-r_i)$, we can see that the sum of the roots is 9 and the product is 7.  Since all of the
roots are assumed to be integers, the product tells us that one of the roots is $\pm 7$ and all of
the others are $\pm 1$.  Note that $-7$ cannot be a root since the sum could not then be any larger
than 7.  Therefore, one of the roots is 7.  In order for the sum to be 9, eight of the other
roots are 1 and the remaining six are -1.
\ \\ \ \\
\noindent From a U and I contest.
\ \\ \ \\
\noindent 4. Find all real solutions to the equation $4x^2 - 40 \lfloor x \rfloor + 51 = 0$.  (Note
that $\lfloor x \rfloor$ denotes the floor function, the greatest integer less than or equal to
$x$.)
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent Let $n = \lfloor x \rfloor$ and $\alpha = x - n$; then $0 \le \alpha < 1$.  Substituting
for $x$, we obtain $4(n+\alpha)^2 - 40n + 51 = 0$, from which the quadratic formula gives $\alpha
= -n \pm \sqrt {10n - \frac{51}{4}}$.  A necessary condition for real solutions is $n \ge 2$.
From the restrictions on $\alpha$, we see that there is no need to consider the negative square
root, and we also get two inequalities: firstly, $-n + \sqrt {10n - \frac{51}{4}} \ge 0$
(equivalently, $n^2 - 10n + \frac{51}{4} \le 0$), which gives $1.5 \le n \le 8.5$; secondly, $-n
+ \sqrt {10n - \frac{51}{4}} < 1$ (equivalently, $n^2 - 8n + \frac{55}{4} > 0$), which gives
$n < 2.5$ or $n > 5.5$.  Therefore, the only values of $n$ that need to be considered are $n = 2,
6, 7$, or 8.

Substituting these four values into $x = n + \alpha = \sqrt {10n - \frac{51}{4}} = \frac
{\sqrt {40n - 51}}{2}$, we find that the solutions are
$$x = \frac{\sqrt {29}}{2},\ \ \frac{3 \sqrt {21}}{2},\ \ \frac{\sqrt {229}}{2},\ \ \textrm{and}
\ \ \frac{\sqrt {269}}{2} .$$
\ \\ \ \\
\noindent From the 1999 Canada National Olympiad.
\ \\ \ \\
\noindent 5. Given the fact that $\int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt {\pi}$,
evaluate the integral
$$I = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + (y-x)^2 + y^2)} \, dx \, dy .$$
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent Make the change of variables $u = x+y$, $v = -x+y$ and note that $u^2 + v^2 =
2x^2 + 2y^2$.  Also, $x = \frac{1}{2} (u-v)$ and $y = \frac{1}{2} (u+v)$, making the Jacobian
of the transformation: $\frac{\partial (x,y)}{\partial (u,v)} = \frac{1}{2}$.  Thus,
\begin{align*}
I &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{2} e^{-(\frac{1}{2} u^2 + \frac{3}{2}
v^2)} \, du \, dv \\
&= \frac{1}{2} \left( \int_{-\infty}^{\infty} e^{-\frac{1}{2} u^2} \, du \right) \left(
\int_{-\infty}^{\infty} e^{-\frac{3}{2} v^2} \, dv \right) \\
&= \frac{1}{2} \left( \sqrt 2 \int_{-\infty}^{\infty} e^{-w^2} \, dw \right) \left(
\frac{\sqrt 2}{\sqrt 3} \int_{-\infty}^{\infty} e^{-z^2} \, dz \right) \\
&= \frac{\pi}{\sqrt 3} .
\end{align*}

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