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\begin{center}
2016 Missouri Collegiate Mathematics Competition \\ Session I
\end{center}

\noindent 1.  Let $\mathcal{R}$ be the region in the first quadrant bounded by the $x$-axis and the graph of $y=\sin x$ from $x=0$ to $x=\pi $. The graph of $y=\cos x$ divides $\mathcal{R}$ into two regions. Let $\mathcal{R}_{1}$ be
the subregion of $\mathcal{R}$ bounded on the left by $y=\cos x$, on the right by $y=\sin x$, and below by the $x$-axis. Find a positive number $b$ such that
the line $y=b$ divides $\mathcal{R}_{1}$ into two regions of equal area.
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\begin{center}
\includegraphics[height=4.5in]{FigureI-1eps}
\end{center}
\ \\ \ \\
\noindent In the figure, the region $\mathcal R_1$ is $KDFHJGK$.  Note that the regions $GEJ$ and $IBD$ have equal areas, as do $ABC$ and $FED$.

Let $\mathcal A_1$ be the area of region $GFH$, and let $\mathcal A_2$ be the area of the region $KFG$.  Then the area of $\mathcal R_1$ is $\mathcal A_1 +\mathcal A_2$.

Let $\mathcal B_1$ be the area of region $GEJ$ ($=$ area of region $IBD$), and let $\mathcal B_2$ be the area of the region $FED$ ($=$ area of region $ABC$).

We thus want to find $b$ such that $\mathcal B_1 +(\mathcal A_2- \mathcal B_2) = \frac{1}{2}(\mathcal A_1 + \mathcal A_2)$.

$\ds{\mathcal A_1 = \int_{\frac{\pi}{2}}^{\pi}\sin{x}\,dx = \int_0^{\frac{\pi}{2}}\cos{x}\,dx}=1$

$\ds{\mathcal A_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin{x}-\cos{x})\,dx = \sqrt{2}-1}$

The coordinates of $D$ are $(\cos^{-1}{b}, b)$, so that

$\ds{\mathcal B_1 = \int_0^{\cos^{-1}{b}}(\cos{x}-b)\,dx = \sin{(\cos^{-1}{b})} - b\,\cos^{-1}{b}}$

$\ds{\mathcal B_2 = \int_{\cos^{-1}{b}}^{\frac{\pi}{2}}(b-\cos{x})\,dx = b\frac{\pi}{2}-b\cos^{-1}{b} -1 + \sin{(\cos^{-1}{b})} }$

Thus, the condition $\mathcal B_1 +(\mathcal A_2- \mathcal B_2) = \frac{1}{2}(\mathcal A_1 + \mathcal A_2)$ becomes

$$\sin{(\cos^{-1}{b})}- b\,\cos^{-1}{b}+\sqrt{2}-1 - \frac{b\pi}{2}+b\cos^{-1}{b} +1 -\sin{(\cos^{-1}{b})}= \frac{1}{2}(1+\sqrt{2}-1).$$

Several things cancel nicely, giving $\ds{b=\frac{\sqrt{2}}{\pi}}$.
%\noindent
%Let $x_{1}$ and $x_{2}$, with $x_{1}<x_{2}$, be the $x-$coordinates of the
%points in $R_{1}$ at which the line $y=b$ intersects the graphs of $y=\cos x$
%and $y=\sin x$, respectively. Note that
%\[
%x_{1}=\cos ^{-1}b\text{ and }x_{2}=\pi -\sin ^{-1}b.
%\]%
%The area of $R_{1}$ is
%\begin{eqnarray*}
%&&\int_{\pi /4}^{\pi }\sin x\,dx-\int_{\pi /4}^{\pi /2}\cos x\,dx \\
%&=&-\left. \cos x\right\vert _{\pi /4}^{\pi }-\left. \sin x\right\vert _{\pi
%/4}^{\pi /2} \\
%&=&1+\dfrac{\sqrt{2}}{2}-1+\dfrac{\sqrt{2}}{2} \\
%&=&\sqrt{2}.
%\end{eqnarray*}
%The area of the region below the line $y=b$ is
%\begin{eqnarray*}
%&&\int_{x_{1}}^{\pi /2}\left( b-\cos x\right) \,dx+\int_{\pi
%/2}^{x_{2}}b\,dx+\int_{x_{2}}^{\pi }\sin x\,dx \\
%&=&\left. \left( bx-\sin x\right) \right\vert _{x_{1}}^{\pi /2}+b\left(
%x_{2}-\dfrac{\pi }{2}\right) -\left. \cos x\right\vert _{x_{2}}^{\pi } \\
%&=&\dfrac{\pi b}{2}-1-bx_{1}+\sin x_{1}+bx_{2}-\dfrac{b\pi }{2}+1+\cos x_{2}
%\\
%&=&b\left( x_{2}-x_{1}\right) +\sin x_{1}+\cos x_{2} \\
%&=&b\left( \pi -\sin ^{-1}b-\cos ^{-1}b\right) +\sin \left( \cos
%^{-1}b\right) -\cos \left( \sin ^{-1}b\right) \\
%&=&\dfrac{\pi b}{2}+\sqrt{1-b^{2}}-\sqrt{1-b^{2}} \\
%&=&\dfrac{\pi b}{2}.
%\end{eqnarray*}
%Thus,
%\[
%\dfrac{\pi b}{2}=\dfrac{\sqrt{2}}{2},
%\]
%so that
%\[
%b=\dfrac{\sqrt{2}}{\pi }.
%\]
\ \\ \ \\
\noindent 2. Find, with proof, the sum of the series
$$\sum_{n=1}^\infty\sin\frac1{2^{n+1}}\cos\frac3{2^{n+1}}.$$
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent Using the identity $\sin\alpha\cos\beta=\frac12[\sin(\alpha+\beta)+
\sin(\alpha-\beta)]$, the $n$th term of the series is $\displaystyle\frac12\left[\sin\left(\frac1{2^{n-1}}\right)
+\sin\left(\frac{-1}{2^n}\right)\right]=\frac12\left[\sin\left(\frac1{2^{n-1}}\right)-\sin\left(\frac{1}{2^n}\right)\right]$
This yields a telescoping sum whose $n$th partial sum equals $\frac12(\sin1-\sin\frac1{2^n})$.
As $n$ approaches infinity, the second term approaches 0, so the sum of the series is $\frac12\sin 1$.
\ \\ \ \\
\noindent From a Michigan MATH Challenge.
\ \\ \ \\
\noindent 3. Find, with proof, all positive integers $x$ satisfying $\displaystyle{3^{(2^{x!})}=2^{(3^{x!})}+1}$.
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent Clearly, $x=1$ is a solution. If $x>1$, then $x!$ is even and so $3^{x!} \equiv 1 \mod 4$. If $x>1$ is a solution, then this implies that the last digit of $\displaystyle{2^{(3^{x!})}+1}$ is $3$. But this is impossible since the last digit of an even power of $3$ is either $1$ or $9$. Thus, $x=1$ is the unique solution.
\ \\ \ \\
\noindent This problem appeared in the 1985 Romanian Mathematical Olympiad.
\ \\ \ \\
\noindent 4. Find a polynomial with integer coefficients that has $\displaystyle{2^{\frac{1}{5}} + 2^{-\frac{1}{5}}}$ as a root.
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent Let $\displaystyle{x = 2^{\frac{1}{5}} + 2^{-\frac{1}{5}}}$ so that
$$x^3 = 2^{\frac{3}{5}} + 2^{-\frac{3}{5}} + 3(2^{\frac{1}{5}} + 2^{-\frac{1}{5}}) = 2^{\frac{3}{5}} + 2^{-\frac{3}{5}} + 3x ,$$
or
$$2^{\frac{3}{5}} + 2^{-\frac{3}{5}} = x^3 - 3x .$$
Also,
\begin{align*}
x^5 &= 2 + 2^{-1} + 5(2^{\frac{3}{5}} + 2^{-\frac{3}{5}}) + 10(2^{\frac{1}{5}} + 2^{-\frac{1}{5}}) \\
&= 5/2 + 5(x^3 - 3x) + 10x .
\end{align*}
Hence, $\displaystyle{2^{\frac{1}{5}} + 2^{-\frac{1}{5}}}$ is a root of
$$P(x) = 2x^5 - 10x^3 + 10x - 5.$$
It is easy to adapt this argument to find a polynomial with integer coefficients that has $\displaystyle{q^{\frac{1}{n}} + q^{-\frac{1}{n}}}$ as a root, where $q$ is a nonzero rational and $n$ is an odd positive integer.
\ \\ \ \\
\noindent Problem 1187 from Crux Mathematicorum, Vol. 14, No. 1 (January 1988), pp. 30--31.  Proposed by Stanley Rabinowitz, Digital Equipment Corp., Nashua, New Hampshire and solution by Bruce Shawyer, Memorial University of Newfoundland, St. John's.
\ \\ \ \\
\noindent 5. For each positive integer $n$, let
\[
\mathcal{R}_{n}=\left\{ \left( x,y\right) :0\leq x\leq n\text{ and }0\leq
y\leq \sqrt{x}\right\} .
\]
Let $N\left( n\right) $ denote the number of points in $\mathcal{R}_{n}$
whose coordinates are both integers. Compute
\[
\lim\limits_{n\rightarrow \infty }\dfrac{N\left( n\right) }{n^{3/2}}.
\]
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent  For each positive integer $n$, the number of integers $y$ with $0 \leq y \leq \sqrt{n}$ is $\lfloor{\sqrt{n}}\rfloor + 1$, so that

\[
N(n) = \sum_{k=0}^{n} \left ( \lfloor{\sqrt{k}}\rfloor + 1\right ) = (n+1) + \sum_{k=1}^{n}  \lfloor{\sqrt{k}}\rfloor.
\]
Since $\sqrt{k} - 1 < \lfloor{\sqrt{k}}\rfloor \leq \sqrt{k}$, and
\[
\int_{0}^{n} \sqrt{x}\, dx \leq \sum_{k=1}^{n} \sqrt{k} \leq \int_{0}^{n+1} \sqrt{x}\, dx,
\]
we have
\[
1 + \int_{0}^{n}\sqrt{x}\, dx\leq N(n) \leq (n+1) + \int_{0}^{n+1}\sqrt{x}\, dx,
\]
and so
\[
1+\frac{2}{3}n^{3/2} \leq N(n) \leq (n+1) + \frac{2}{3}(n+1)^{3/2}.
\]
Dividing by $n^{3/2}$ and applying the squeeze theorem gives
\[
\lim_{n\to\infty}\frac{N(n)}{n^{3/2}} = \frac{2}{3}.
\]

\vfill\eject
\begin{center}
2016 Missouri Collegiate Mathematics Competition \\ Session II
\end{center}

\noindent 1. Evaluate
$$\sum_{k=1}^n k 2^{k-1}$$
for $n=1, 2, 3, \ldots $.
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent Let
$$f(x) = \sum_{k=0}^n x^k ,$$
then
$$f' (x) = \sum_{k=1}^n k x^{k-1}$$
and
$$f'(2) = \sum_{k=1}^n k 2^{k-1} .$$
Also,
$$f'(x) = \frac{d}{dx} \left( \frac{x^{n+1} - 1}{x-1} \right) = \frac{(x-1)((n+1)x^n) - (x^{n+1} - 1)(1)}{(x-1)^2} $$
and
$$f'(2) = 2^n (n+1) - 2^{n+1} + 1 = 2^n (n-1) + 1.$$
\ \\ \ \\
\noindent 2. Evaluate
\[
\int_{1}^{\infty }\dfrac{\left\lfloor x\right\rfloor }{x^{3}}\,dx,
\]
where $\left\lfloor x\right\rfloor $ denotes the greatest integer function.
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent Denoting the integral by $I$, we have
\[
I=\sum\limits_{k=1}^{\infty }\int_{k}^{k+1}\dfrac{\left\lfloor
x\right\rfloor }{x^{3}}\,dx.
\]
On the interval $[k,k+1)$, we make the substitution $x=k+\alpha $, with $0\leq \alpha <1$. Then
\begin{eqnarray*}
I &=&\sum\limits_{k=1}^{\infty }\int_{\alpha=0}^{\alpha=1}\dfrac{k}{\left( k+\alpha
\right) ^{3}}\,d\alpha \\
&=&-\dfrac{1}{2}\sum\limits_{k=1}^{\infty }\left. \dfrac{k}{\left( k+\alpha
\right) ^{2}}\right\vert _{\alpha =0}^{\alpha =1} \\
&=&-\dfrac{1}{2}\sum\limits_{k=1}^{\infty }\left[ \dfrac{k}{\left(
k+1\right) ^{2}}-\dfrac{1}{k}\right] \\
&=&-\dfrac{1}{2}\sum\limits_{k=1}^{\infty }\left[ \dfrac{k+1}{\left(
k+1\right) ^{2}}-\dfrac{1}{k}-\dfrac{1}{\left( k+1\right) ^{2}}\right] \\
&=&\dfrac{1}{2}\sum\limits_{k=1}^{\infty }\left( \dfrac{1}{k}-\dfrac{1}{k+1}
\right) +\dfrac{1}{2}\sum\limits_{k=1}^{\infty }\dfrac{1}{\left( k+1\right)
^{2}} \\
&=&\dfrac{1}{2}\sum\limits_{k=1}^{\infty }\left( \dfrac{1}{k}-\dfrac{1}{k+1}
\right) +\dfrac{1}{2}\sum\limits_{k=2}^{\infty }\dfrac{1}{k^{2}} \\
&=&\dfrac{1}{2}+\dfrac{1}{2}\left( \dfrac{\pi ^{2}}{6}-1\right) \\
&=&\dfrac{\pi ^{2}}{12}.
\end{eqnarray*}
\ \\ \ \\
\noindent 3. Find the smallest positive angle $\theta$ in degrees satisfying
$$\sin^2(2015\theta) + \cos^2 (2016\theta)=1.$$
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent Using the identity $1-\sin^2(2015\theta)=\cos^2(2015\theta)$, the given equation
can be rewritten as $\cos^2(2016\theta)=\cos^2(2015\theta)$ which is equivalent
to $\cos(2016\theta)=\pm\cos(2015\theta)$.  As $\cos x$ is decreasing for
$x\in[0,\pi]$, we have $\cos(2016\theta)\ne+\cos(2015\theta)$ for $\theta\in
(0,\pi/2016]$.  We find the $\theta\in(0,\pi/2016]$ for which $\cos(2016\theta)=-\cos
(2015\theta)$.  Equality occurs if and only if $2016\theta=\pi/2+t$ and $2015\theta
=\pi/2-t$ for some positive $t$ in the appropriate range.  Solving for $\theta$, we
obtain $\theta=\pi/4031$.  Hence, the answer in degrees is $(180/4031)^\circ$.
\ \\ \ \\
\noindent Similar to a Univ. of South Carolina practice problem for the American Regions
Mathematics Competition.
\ \\ \ \\
\noindent 4. If $a$ and $b$ are the radii of two spheres, tangent to each other and to a
plane (on the same side of the plane), the radius $x$ of the largest sphere
which can pass between them is given by what formula?
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent This is really a two-dimensional problem, so we replace the spheres by circles
and the plane by a line $l$. Let $\mathcal{C}_{1}$ be the circle of radius
$a,\mathcal{C}_{2}$ the circle of radius $b$, and $\mathcal{C}$ the circle of
radius $x$. Let $A$ be the point of tangency of $\mathcal{C}_{1}$ and $l$, $B$
the point of tangency of $\mathcal{C}_{2}$ and $l$, and $D$ the point of
tangency of $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$. Let $Q_{1}$ be the center
of $\mathcal{C}_{1}$, $Q_{2}$ be the center of $\mathcal{C}_{2}$, and $O$ the
center of $\mathcal{C}$. Let $E$ be the point of tangency of $\mathcal{C}_{1}$
and $\mathcal{C}$ and $F$ the point of tangency of $\mathcal{C}_{2}$ and
$\mathcal{C}$. Let $G$ be the point of tangency of $\mathcal{C}$ and $l$.

Since $D$ is on $\overline{Q_{1}Q_{2}}$, $Q_{1}Q_{2}=a+b$. Since $E$ and $F$
are on $\overline{OQ_{1}}$ and $\overline{OQ_{2}}$, respectively, $OQ_{1}=a+x$
and $OQ_{2}=b+x$. Set $u=AG$ and $v=BG$.

\ \\ \ \\
\begin{center}
\includegraphics[height=2.5in]{Chip17eps}
\end{center}
\ \\ \ \\

We have the following system of equations.
\[
\left\{
\begin{array}
[c]{c}%
\left(  a-x\right)  ^{2}+u^{2}=\left(  a+x\right)  ^{2}\\
\left(  b-x\right)  ^{2}+v^{2}=\left(  b+x\right)  ^{2}\\
\left(  b-a\right)  ^{2}+\left(  u+v\right)  ^{2}=\left(  a+b\right)  ^{2}
\end{array}
\right.  .
\]
Expanding and simplifying gives
\[
\left\{
\begin{array}
[c]{c}
u^{2}=4ax\\
v^{2}=4bx\\
\left(  u+v\right)  ^{2}=4ab
\end{array}
\right.  .
\]
Substituting $u$ and $v$ from the first two of these into the third gives
\begin{align*}
\left(  \sqrt{4ax}+\sqrt{4bx}\right)  ^{2}  & =4ab\\
\sqrt{ax}+\sqrt{bx}  & =\sqrt{ab}\\
\sqrt{x}\left(  \sqrt{a}+\sqrt{b}\right)    & =\sqrt{ab}\\
\sqrt{x}  & =\dfrac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\\
x  & =\dfrac{ab}{(\sqrt{a}+\sqrt{b})^2}.
\end{align*}
\ \\ \ \\
\noindent American Mathematical Monthly, E432, ODMPB p. 56.
\ \\ \ \\
\noindent 5. In a regular octagon each side is colored blue or yellow.  From such a coloring, another
coloring will be obtained ``in one step'' as follows: if the two neighbors of a side have different
colors, the ``new'' color of the side will be blue, otherwise the color will be yellow.  (Note that
the colors
are modified simultaneously.)  Show that after a finite number, say $N$, of moves, all sides will be
colored yellow.  What is the least value of $N$ that works for all possible colorings?
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent Consider a coloring of the sides of a regular octagon where each side
is colored blue or yellow.  Label the sides of the octagon $S_1$, $\ldots$, $S_8$ and code the colors
of the sides of the regular octagon by the column $\overline{x} = (x_1, \ldots , x_8)^T$, where
$$x_i = \begin{cases}
1 &\quad \textrm{if $S_i$ is colored blue},\\
0 &\quad \textrm{if $S_i$ is colored yellow},
\end{cases}$$
for $i=1, 2, \ldots , 8$.  Let
$$A = \begin{pmatrix}
0 &1 &0 &0 &0 &0 &0 &1 \\
1 &0 &1 &0 &0 &0 &0 &0 \\
0 &1 &0 &1 &0 &0 &0 &0 \\
0 &0 &1 &0 &1 &0 &0 &0 \\
0 &0 &0 &1 &0 &1 &0 &0 \\
0 &0 &0 &0 &1 &0 &1 &0 \\
0 &0 &0 &0 &0 &1 &0 &1 \\
1 &0 &0 &0 &0 &0 &1 &0
\end{pmatrix} .$$
Then $A^n \overline{x}$ is the coded coloring of the sides of the regular octagon after $n$ steps, where
the addition and multiplication in the matrix operations are performed mod 2.  Since $A^4 = 0$ it
follows that after 4 steps, all sides will be colored yellow.  This is the least possible value of $N$
that works for all possible colorings, since for $\overline{x} = (1, 0, 0, 0, 0, 0, 0, 0)^T$,
$$A\overline{x} = (0,1,0,0,0,0,0,1)^T,\ \ A^2 \overline{x} = (0,0,1,0,0,0,1,0)^T,\ \ A^3\overline{x}
= (0,1,0,1,0,1,0,1)^T .$$
\ \\ \ \\
\noindent Problem 7 from the 11th Austrian-Polish Mathematics Competition - December 1989.

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