\documentclass [12pt]{article} \usepackage{amsfonts} \usepackage{geometry} \usepackage{amsmath} \setcounter{MaxMatrixCols}{30} \usepackage{amssymb} \usepackage{graphicx} \setlength{\topmargin}{-0.5in} \setlength{\leftmargin}{0in} \setlength{\oddsidemargin}{0in} \setlength{\evensidemargin}{0in} \setlength{\textwidth}{6in} \setlength{\textheight}{8.84in} \newcommand{\lbin}{\left(\!\begin{array}{c}} \newcommand{\rbin}{\end{array}\!\right)} \newcommand{\ldet}{\left|\begin{array}} \newcommand{\rdet}{\end{array}\right|} \newcommand{\lmat}{\left(\begin{array}} \newcommand{\rmat}{\end{array}\right)} \newcommand{\dspace}{\baselineskip 22pt} \newcommand{\sspace}{\baselineskip 14pt} \newcommand{\dps}{\displaystyle} \newcommand{\qed}{\hfill\rule{2mm}{2mm} \\} \newcommand{\ds}{\displaystyle} \pagestyle{myheadings} \begin{document} \thispagestyle{empty} \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 0x_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 02. 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 isR(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).\$ \end{document}