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\begin{center}
2012 Missouri Collegiate Mathematics Competition \\ Session I
\end{center}


\noindent 1. Two circles in the first quadrant are tangent to each other and both are tangent
to the $x$-axis and the line $y=mx$, where $m$ is a positive constant.  Find the ratio
of the radius of the larger circle to the radius of the smaller circle as a function of $\theta$,
the angle between the $x$-axis and the line $y=mx$.
\ \\ \ \\
Solution.
\ \\ \ \\
\noindent Solution (see attached diagram):  Let $O$ be the origin, $A$ the center of the
smaller circle, and $B$ the center of the larger circle.  The bisector of angle $O$ goes
through $A$ and $B$.  Without loss of generality, let the radius of the smaller circle be
equal to $1$, and let $r$ be the radius of the larger circle, $x$ the length of segment
$\overline{OA}$, and $y$ the length of segment $\overline{OB}$.

By similar triangles, we have $x/1=y/r$, so $y=rx$.  By considering segment $\overline{OB}$,
we also find that $x+1+r=y$.  Therefore, we have $x+1+r=rx$ or $r=(x+1)/(x-1)$.  Since
$y=\csc(\theta/2)$, we find that $\displaystyle r=\frac{\csc(\theta/2)+1}{\csc(\theta/2)-1}$,
or equivalently  $\displaystyle r=\frac{1+\sin(\theta/2)}{1-\sin(\theta/2)}$.
$$
\includegraphics[width=3.0in]{Morgan1b.eps}
$$
\ \\ \ \\
\noindent (Similar to a C of C problem.)
\ \\ \ \\
\noindent 2. Prove that $\dps AB - BA \not = I_n$ for any $n \times n$ matrices
$A$ and $B$ over the real numbers, where $I_n$ denotes the $n \times n$ identity matrix.
\ \\ \ \\
Solution.
\ \\ \ \\
\noindent We consider the sum of the elements of main diagonals.  Let $A=(a_{ij})$ and $B=(b_{ij})$.
Then the sum of the elements of the main diagonal of $AB-BA$ is
$\dps \sum_{i=1}^n \sum_{j=1}^n a_{ij}b_{ji} - \sum_{i=1}^n \sum_{j=1}^n b_{ij}a_{ji} = 0$
while the sum of the main diagonal of $I_n$ is $n$. Therefore, $\dps AB - BA \not = I_n$.
\ \\ \ \\
\noindent 3. Find
$$\lim_{n \to \infty} \left( \frac{1}{\sqrt n \sqrt {n+1}} + \frac{1}{\sqrt n \sqrt {n+2}} + \cdots
+ \frac{1}{\sqrt n \sqrt {2n}} \right) .$$
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\begin{align*}
&\lim_{n \to \infty} \left( \frac{1}{\sqrt n \sqrt {n+1}} + \frac{1}{\sqrt n \sqrt {n+2}} + \cdots
+ \frac{1}{\sqrt n \sqrt {2n}} \right) \\
&= \lim_{n \to \infty} \frac{1}{n} \left( \sqrt {\frac{n}{n+1}} + \sqrt {\frac{n}{n+2}} + \cdots
+ \sqrt {\frac{n}{n+n}} \right) \\
&= \lim_{n \to \infty} \frac{1}{n} \left( \frac{1}{\sqrt {1 + 1/n}} + \frac{1}{\sqrt {1 + 2/n}} + \cdots
+ \frac{1}{\sqrt {1 + 1}} \right) \\
&= \int_0^1 \frac{1}{\sqrt {1+x}} \,dx = 2 \sqrt {1+x} \bigg\vert _0^1 = 2 ( \sqrt 2 - 1 ) .
\end{align*}
\ \\ \ \\
\noindent 4.
\ \\ \ \\
\noindent Part 1. Let $n$ be a positive integer and let $P_{n-1} (x)$ be the polynomial equal to
$P_{n-1} (x) = \frac{x^n - 1}{x-1}$.  Find $P_{n-1} (1)$.
\ \\ \ \\
\noindent Part 2. On the circle of radius 1, let $V_1$, $V_2$, $\ldots$, $V_n$ be the vertices of a
regular $n$-gon
inscribed in the circle.  Let $\lambda _k = dist (V_1, V_k)$, $k=1, 2, \ldots , n$.  Show that
$$\prod_{k=2}^n \lambda _k = n.$$
\ \\ \ \\
\noindent Part 3. For the regular $n$-gon in Part 2, find the product of the lengths of all the line
segments joining the vertices.
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent Part 1. $P_{n-1} (x) = n$, either by using L'Hopital's Rule to evaluate $\lim_{x \to 1} P_{n-1} (x)$,
or by factoring the numerator as $(x-1) (x^{n-1} + x^{n-2} + \cdots + x + 1)$ to see that $P_{n-1} (x)$ is the
second factor.
\ \\ \ \\
\noindent Part 2. Consider the vertices as the $n$th roots of $1$ and let $V_1 = 1$.  Then the $V_k$'s are
complex numbers and
$$\prod_{k=1}^n (z-V_k) = z^n - 1 $$
because the product is simply the factored form of $z^n - 1$.  Thus,
$$\prod_{k=2}^n (z-V_k) = P_{n-1} (z)$$
and $\lambda _k = \vert V_1 - V_k \vert = \vert 1 - V_k \vert $.  Therefore, from Part 1,
$$\prod_{k=2}^n \lambda _k = \prod_{k=2}^n \vert 1 - V_k \vert = \vert P_{n-1} (1) \vert = n .$$
\ \\ \ \\
\noindent Part 3. From Part 2, for each vertex, the product of the lengths of the segments to the other
vertices is $n$, so the product for all vertices is $n^n$.  But because each segment joins two vertices,
every segment's length appears twice as a factor in $n^n$.  Thus, if the required product is $P$, we have
$P^2 = n^n$ giving $P = \sqrt {n^n}$.
\ \\ \ \\
\noindent Parts 1 and 2 are from de Souza, P. N. and Silva, J-N, \textit{Berkeley Problems in Mathematics},
Springer, 1998.
\ \\ \ \\
\noindent 5.  Let $S$ be a set of twelve distinct positive integers such that for distinct
$a$, $b$, $c$, and $d$ in $S$,
$a+b \ne c+d$.  Prove that the largest element in $S$ is greater than 56.
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent Consider the ${12 \choose 2} = 66$ possible differences $b-a$, where $b>a$ are elements
of $S$.
We observe that if $a, b, c, d \in S$ are all distinct and $a-c = d-b$, then $a+b = c+d$.  And if
$a,b,c \in S$ are all distinct and $c-a = b-c$, then $c = (a+b)/2$ so $c$ is not the largest or
smallest
element of $S$.  And if $c-a_1 = b_1-c$ and $c-a_2 = b_2-c$, then $a_1+b_1 = 2c = a_2+b_2$.  So for
each $c$, not the largest or smallest element of $S$, at most one pair of $a, b$ can be found such
that $c-a=b-c$.  Hence, we can have at most 10 duplicates for each $c$ not the largest or
smallest element of $S$.  Therefore, there are at least
$66 - 10 = 56$ distinct differences.  Hence, the largest element of $S$ is greater than 56.

\ \\ \ \\

\noindent In fact, one such set of 12 numbers is
$$\{ 1, 2, 3, 8, 13, 23, 38, 41, 55, 64, 68, 72 \}.$$

\ \\ \ \\

\noindent This problem is from G. Berzsenyi, \textit{Math Investigations: Distinct Sums of Twosomes},
Quantum, \textbf{5.4} (March/April 1995), 39.
\vfill\eject
\begin{center}
2012 Missouri Collegiate Mathematics Competition \\ Session II
\end{center}

\noindent 1. Express 2012 as a sum of (two or more) consecutive integers.
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent We need to find positive integers $k$ and $m$, $m\ge 2$, such
that
$$(k+1)+(k+2)+\dots+(k+m)=2012.$$
Note that the left side of this equation is the difference between the sum of the first
$k+m$ positive integers and the sum of the first $k$ positive integers, so we have
$$\aligned2012&=\frac{(k+m)(k+m+1)}2-\frac{k(k+1)}2\\
&=\frac12(k^2+2km+m^2+k+1-k^2-k)\\
&=\frac12m(m+2k+1).\endaligned$$
Thus,
$$m(m+2k+1)=4024=8\cdot503.$$
Note that $m$ and $m+2k+1$ have opposite parity and $m+2k+1>m>1$.  Therefore,
$m$ must be 8 and $k$ must be 247.  The desired sum is then $2012=248+249+250+251
+252+253+254+255$.
\ \\ \ \\
\noindent (Similar to an Iowa contest problem by Jacek Fabrykowski.)
\ \\ \ \\
\noindent 2. Let $f$ be a function that has a continuous derivative over the interval $[a, b]$, and
let $f(a)=f(b)=0$.  Prove that
\[\max_{a\le x \le b} |f'(x)| \ge \frac{4}{(b-a)^2} \int_a^b |f(x)| dx. \]
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent By the mean value theorem, for any $x \in (a, b)$,
\[\dps f(x) = f(x)-f(a) = f'(\xi_1)(x-a) = f(x)-f(b) = f'(\xi_2)(x-b),\]
where $\xi_1 \in (a,x)$ and $\xi_2 \in (x,b)$.  Let $\dps M=\max_{a\le x \le b} |f'(x)|$.
Then, $|f(x)| \le M(x-a)$ and $|f(x)| \le M(b-x)$.  Therefore,
\begin{eqnarray}
&&\frac{4}{(b-a)^2} \int_a^b |f(x)| dx \nonumber \\
&\le& \frac{4}{(b-a)^2} \left ( \int_a^{(a+b)/2} M(x-a) dx + \int_{(a+b)/2}^b M(b-x) dx \right )
\nonumber \\
&=& \frac{4}{(b-a)^2} \left ( \frac{(b-a)^2}{8}M + \frac{(b-a)^2}{8}M \right ) = M. \nonumber
\end{eqnarray}
\ \\ \ \\
\noindent 3. For real $a>0$ define the sequence $\{x_n\}$ by
$$x_{n+1}=a(x_n^2 + 4), \qquad x_0 = 0 .$$
Determine necessary and sufficient conditions on $a$ for $\ds{\lim_{n\to \infty}x_n}$ to exist
and be finite.
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent Assume $\ds{\lim_{n\to \infty}x_n} = x$, with $x$ finite.  Then $x=a(x^2+4)$, making
$$x=\frac{1}{2a}\left(1\pm\sqrt{1-16a^2}\right)$$
To make $x$ real and finite, then, we need $1-16a^2\ge0$, and since $a>0$ was given,
we get $0<a\le \frac{1}{4}$.
\ \\
\noindent Now assume $0<a\le \frac{1}{4}$.  By induction, we can show that $\{x_n\}$ is a
nondecreasing sequence.
$$x_1=4a>x_0=0 .$$
If we assume $x_n \ge x_{n-1}$, then
$$x_{n+1}-x_{n}=a(x_n^2-x_{n-1}^2)$$
and so $x_{n+1}\ge x_n$.
\ \\
\noindent Note that $x_1=4a < 2$, so another induction will show that $\{x_n\}$ is bounded above.
Let $x_n<2$.  Then
$$x_{n+1}=a(x_n^2+4)<\frac{1}{4}(4+4) = 2 .$$
This means $\{x_n\}$ is a nondecreasing sequence that is bounded above, and thus $\{x_n\}$ converges
to a real finite limit.
\ \\
\noindent The required necessary and sufficient condition is $0<a \le\frac{1}{4}$.
\ \\ \ \\
\noindent 4. Let $P(x)$ be a polynomial in $x$ of degree $n>2$.  Suppose that $a \ne b$,
$P(a) = n_1$, and
$P(b) = n_2$.  What is the remainder $R(x)$ when $P(x)$ is divided by $(x-a)(x-b)$?
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent By the Factor Theorem for polynomials over a field,
\begin{equation}
\frac{P(x)}{x-a} = Q_1 (x) + \frac{n_1}{x-a},\ \ \ \frac{P(x)}{x-b} = Q_2 (x) + \frac{n_2}{x-b} ,
\end{equation}
where $Q_1 (x)$, $Q_2 (x)$ are unique polynomials of degree $n-1$.  The Division Algorithm for polynomials
over a field leads to the conclusion that if $P(x)$ is divided by $(x-a)(x-b)$, then there is a unique quotient
$Q(x)$ and a unique remainder $cx+d$, $c,d \in \mathbb{R}$ such that
\begin{equation}
\frac{P(x)}{(x-a)(x-b)} = Q(x) + \frac{cx+d}{(x-a)(x-b)} .
\end{equation}
Multiplications of eq. (2) by either $(x-b)$ or $(x-a)$ then lead to the equations
\begin{align*}
\frac{P(x)}{x-a} &= (x-b) Q(x) + \frac{cx+d}{x-a} = \left( (x-b) Q(x) + c \right) + \frac{d+ac}{x-a} , \\
\frac{P(x)}{x-b} &= (x-a) Q(x) + \frac{cx+d}{x-b} = \left( (x-a) Q(x) + c \right) + \frac{d+bc}{x-b} .
\end{align*}
Comparison of these results with the equations in (1) then imply (since $a \ne b$)
if $d+ac = n_1$, $d+bc = n_2$.  Thus, $c = \frac{n_2 - n_1}{b-a}$ and $d = \frac{n_1 b - n_2 a}{b-a}$,
and the desired remainder is
$$R(x) = \frac{1}{b-a} \left( (n_2 - n_1) x + (n_1 b - n_2 a) \right) .$$
\ \\ \ \\
\noindent 5. Let $S$ be the set of points in $\mathbb{R}^2$ that constitute the graph of $y=x^2$, $-2 \le x \le
2$, and let $d(p_1, p_2)$ denote the Euclidean distance between $p_1, p_2 \in S$.  Determine $p_1$
and $p_2$ that maximizes $d(p_1,p_2)$.
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent It is geometrically apparent that one of the points of a maximally separated pair $\{ p_1, p_2 \}$
must be either $(2,4)$ or $(-2,4)$.  Consider the point $(-2,4)$ as $p_2$ and $(x,x^2)$ as $p_1$.   Let
$g(x) = d^2 (p_1, p_2) = (x+2)^2 + (x^2 - 4)^2$.  Then
$$g'(x) = 2(x+2) + 2(x^2 - 4) \cdot 2x = 2(x+2)(2x^2 - 4x +1).$$
Setting this equal to 0 and eliminating trivial conditions we have
$$x^{\pm} = 1 \pm \frac{\sqrt 2}{2}.$$
Calculating the second derivative we have
$$g''(x) = 2 \left( 2x^2 - 4x + 1 + (x+2)(4x-4) \right) = 2 ( 6x^2 - 7) .$$
Using the second derivative test, we find that $g''(x^{-}) < 0$ and $g''(x^{+}) > 0$, so the distance
is maximized at $x^{-}$.  Therefore, the points which maximize the distance are
$$(-2,4) \ \ \ \textrm{and}\ \ \ \left( 1 - \frac{\sqrt 2}{2}, \left(1 - \frac{\sqrt 2}{2} \right)^2 \right)$$
and by symmetry
$$(2,4) \ \ \ \textrm{and}\ \ \ \left(-1 + \frac{\sqrt 2}{2}, \left(-1 + \frac{\sqrt 2}{2}\right)^2 \right).$$

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