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\hyphenation{Mass-achusetts Central Missouri Wis-con-sin Muthuvel}
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\centerline{\bf 2003 Missouri Collegiate Mathematics Competition}
\vskip 6pt
\centerline{\bf Session I}
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1. Let $P \ne (0,0)$ be a point on the parabola $y=x^2$. The normal
line to the parabola at $P$ will intersect the parabola at another point,
say $Q$. Find the coordinates of $P$ so that the sum of the $y$-coordinates (or the sum of the ordinates) of $P$ and $Q$ is a minimum.
\comment
area under the normal line between $P$ and $Q$ is a minimum.
\endcomment
\bigskip
Solution.
\medskip
Let $p>0$, $P = (p,p^2)$ and $Q = (q,q^2)$. The slope of the tangent line to
the parabola at $P$ is $2p$, so the slope of the normal line to the
parabola at $P$ is $-1/2p$. Thus, the equation of the normal line to
the parabola at $P$ is
$$y - p^2 = -{1 \over 2p} (x - p) .$$
Since $Q = (q,q^2)$ lies on the normal line,
$$q^2 - p^2 = - {1 \over 2p} (q-p)\ \ \text{or}\ \ q + p = - {1 \over
2p} .$$
Solving for $q$, we obtain
$$q = -p - {1 \over 2p} .$$
\comment
Now the area under the normal line between $P$ and $Q$ is
$$\eqalign{
A &= \biggl( {p^2 + q^2 \over 2} \biggr) (p-q) \cr
&= \biggl( {2p^2 + 1 + {1 \over 4p^2} \over 2} \biggr) \cdot \biggl( 2p + {1 \over 2p} \biggr) \cr
&= 2p^3 + {3 \over 2} p + {1 \over 2} p^{-1} + {1 \over 16} p^{-3} .\cr}$$
Differentiating this quantity with respect to $p$, we obtain
$${dA \over dp} = 6p^2 + {3 \over 2} - {1 \over 2} p^{-2} - {3 \over 16} p^{-4} = {1 \over 16p^4} ( 96p^6 + 24p^4 - 8p^2 - 3) .$$
Setting this quantity equal to 0 and solving for $p$ (with the aid of a calculator), we have
$$p = 0.564641 \ldots$$
Therefore, the coordinates of $P$ so that the area under the normal line between $P$ and $Q$ is a minimum are $( 0.564641 \ldots , 0.318819 \ldots )$.
\endcomment
Now the sum of the $y$-coordinates of $P$ and $Q$ is
$$\eqalign{
S &= p^2 + q^2 = p^2 + \biggl( -p - {1 \over 2p} \biggr) ^2 \cr
&= p^2 + p^2 + 1 + {1 \over 4p^2} = 2p^2 + 1 + {1 \over 4} p^{-2} .\cr}$$
Differentiating this quantity with respect to $p$, we obtain
$${dS \over dp} = 4p - {1 \over 2} p^{-3} = {1 \over 2} p^{-3} (8p^4 - 1) .$$
Setting this quantity equal to 0 and solving for $p$, we have
$$p = {1 \over {\root 4 \of 8}} .$$
Therefore, the coordinates of $P$ so that the sum of the $y$-coordinates of $P$ and $Q$ is a minimum are
$$\biggl( {1 \over {\root 4 \of 8}} , {1 \over {\root 2 \of 8}} \biggr) .$$
\bigskip
2. There is an $m \times n$ rectangular array of office mailboxes in which mail for $p \le mn$ people is distributed. Initially, the mailboxes are assigned alphabetically beginning at the upper left and proceeding down each column (the ``next'' mailbox to one at the bottom of a column is the one at the top of the next column to the right). A new secretary is hired, and decides that the mailboxes will now be assigned alphabetically beginning at the upper left and proceeding to the right across each row. Discuss, as completely as possible, whose mailboxes will be unchanged.
\bigskip
Solution.
\medskip
The people in mailboxes $(1,1)$ and $(m,n)$ (if $p = mn$) never change mailboxes. Consider mailbox $(x,y)$. By simple counting using the two arrangements, $(x,y)$ will not change if and only if
$$m(y-1) + x = n(x-1) + y $$
or
$$(n-1)(x-1) = (m-1)(y-1) .$$
This is a linear equation in $x$ and $y$ whose graph is the diagonal containing both $(1,1)$ and $(m,n)$ (note the positive $y$-axis points down with the given setup). Any point on this diagonal between $(1,1)$ and $(m,n)$ with integer coordinates is a solution. If $m-1$ and $n-1$ are relatively prime, there are no such points, if $m-1$ and $n-1$ are not relatively prime there is a solution for each common divisor, and if $m=n$ every mailbox on the diagonal is unchanged. If $k > 1$ is the smallest nontrivial common divisor of $m-1$ and $n-1$, then the penultimate solution is
$$\left(\frac{m-1}{k} + 1, \frac{n-1}{k} + 1\right) ,$$
and this solution counts if
$$p \ge \frac{mn + k -1}{k} .$$
In fact, for any common divisor $d$, the corresponding solution is counted whenever
$$p \ge \frac{mn + d -1}{d} .$$
\bigskip
3. For sufficiently small but positive $\theta$, the relation $\tan \theta > \theta$ holds. Prove, in the other direction, that for $0 < \theta < \pi / 4$ one has
$$\tan \theta < {4 \theta \over \pi } .$$
\bigskip
Solution I.
\medskip
Let $f(x) = \tan x$ and let $x = \theta$ be an arbitrary number in $(0, \pi /4)$. Then from the Mean Value Theorem (since $f(x)$ is continuous on $[0, \pi /4]$ and differentiable on $(0, \pi /4)$)
$$\cases \tan \theta = \tan 0 + \theta \sec ^2 \xi &\\
\tan \pi /4 = \tan \theta + ({\pi \over 4} - \theta ) \sec ^2 \xi ' &, \endcases$$
where $0 < \xi < \theta$ and $\theta < \xi ' < \pi /4$. But on the interval $(0, \pi /4)$ $\sec ^2 x$ is an increasing function, so $\sec ^2 \xi ' > \sec ^2 \xi$. This yields
$${1 - \tan \theta \over {\pi \over 4} - \theta} > {\tan \theta \over \theta} ,$$
and algebraic rearrangement results in
$${4 \theta \over \pi} > \tan \theta .$$
\medskip
Solution II.
\medskip
Let
$$f( \theta ) = \tan \theta - {4 \theta \over \pi} \text{ for } \theta \in [0, \pi / 4] .$$
Then,
$$f' ( \theta ) = \sec ^2 \theta - {4 \over \pi} .$$
Therefore, there exists a unique $\xi \in (0, \pi / 4)$ such that $f'(\xi) = 0$. Also, $f' (\theta) < 0$ for $\theta \in (0, \xi)$ and $f'(\theta) > 0$ for $\theta \in (\xi, \pi / 4)$. Therefore, $f$ is decreasing on $(0, \xi)$ and $f$ is increasing on $(\xi, \pi / 4)$. In addition, $f(0) = f(\pi / 4) = 0$. Therefore, $f(\theta ) < 0$ for $(0, \pi / 4)$.
\bigskip
4. Let $ABCD$ be a quadrilateral, with sides $AB = a$, $BC = b$, $CD = c$, where $a$, $b$, and $c$ are fixed positive quantities. Prove that when the quadrilateral $ABCD$ has a maximum area, then $ABCD$ can be inscribed in a semicircle.
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Solution.
\medskip
$$\epsfxsize=15pc \epsfbox{quad.eps}$$
\noindent Let $\theta$ denote $\langle ABC$ and $\phi$ denote $\langle BCD$. Since the lengths $a$, $b$, and $c$ are given, the area $ABCD$ depends upon two independent variables $\theta$ and $\phi$. A fixed value of $\phi$ will keep triangle $BCD$ rigid, and the area $ABCD$ will then depend upon $\triangle ABD$.
\noindent Given any value of $\phi$,
$$\text{the area of triangle } ABD = {1 \over 2} AB \cdot BD \cdot \sin \langle ABD .$$
Since $AB$ and $BD$ are fixed, this area is maximum when $\langle ABD$ is a right angle.
\noindent Similarly, given a value of $\theta$, the area $ABCD$ is maximized when $\langle ACD$ is a right angle. In other words, for the maximum area of $ABCD$, the points $B$ and $C$ must lie on a semicircle with $AD$ as a diameter.
\bigskip
5. Define a sequence $\{ x_n \} _{n=2}^\infty$ by
$$(n + x_n) [ \root n \of 2 - 1 ] = \ln 2 .$$
Find $\lim _{n \to \infty} x_n$.
\bigskip
Solution.
\medskip
Solving for $x_n$, we obtain
$$x_n = {\ln 2 \over \root n \of 2 - 1} - n ,$$
which is of the form $\infty - \infty$. Then, using the substitution $u=1/n$ and l-Hospital's Theorem twice, we obtain
$$\eqalign{
\lim_{n \to \infty} x_n &= \lim_{n \to \infty} {\ln 2 - n \cdot 2^{1/n} + n \over 2^{1/n} - 1} \cr
&= \lim_{u \to 0} {u \ln 2 - 2^u + 1 \over u 2^u - u} \cr
&= \lim_{u \to 0} {1 - 2^u \over u2^u + {2^u - 1 \over \ln 2}} \cr
&= \lim_{u \to 0} {-\ln 2 \over u\ln 2 + 2} \cr
&= -{1 \over 2} \ln 2 .\cr}$$
\vfill\eject
\centerline{\bf 2003 Missouri Collegiate Mathematics Competition}
\vskip 6pt
\centerline{\bf Session II}
\bigskip
From {\it Crux Mathematicorum}, Problem 2532, April, 2000. Proposed by Hojoo Lee, student, Kwangwoon University, Kangwon-Do, South Korea. Solution by Richard B. Eden, Ateneo de Manila University, Manila, The Philippines.
\medskip
1. Suppose that $a$, $b$ and $c$ are positive real numbers satisfying $a^2 + b^2 + c^2 = 1$. Prove that
$${1 \over a^2} + {1 \over b^2} + {1 \over c^2} \ge 3 + {2(a^3 + b^3 + c^3) \over abc} .$$
\bigskip
Solution.
\medskip
$$\eqalign{
{1 \over a^2} + {1 \over b^2} + {1 \over c^2} &- 3 - {2 (a^3 + b^3 + c^3) \over abc} \cr
= &{a^2 + b^2 + c^2 \over a^2} + {a^2 + b^2 + c^2 \over b^2} + {a^2 + b^2 + c^2 \over c^2} \cr
&-3 - 2 \biggl( {a^2 \over bc} + {b^2 \over ca} + {c^2 \over ab} \biggr) \cr
= a^2 &\biggl( {1 \over b^2} + {1 \over c^2} \biggr) + b^2 \biggl( {1 \over c^2} + {1 \over a^2} \biggr) + c^2 \biggl( {1 \over a^2} + {1 \over b^2} \biggr) \cr
&-2 \biggl( {a^2 \over bc} + {b^2 \over ca} + {c^2 \over ab} \biggr) \cr
= a^2 &\biggl( {1 \over b} - {1 \over c} \biggr) ^2 + b^2 \biggl( {1 \over c} - {1 \over a} \biggr) ^2 + c^2 \biggl( {1 \over a} - {1 \over b} \biggr) ^2 \ge 0 ,\cr}$$
with equality if and only if $a=b=c$.
\bigskip
2. Let $x_1 > 1$ be odd and define the sequence $\{ x_n \} _{n=1}^\infty$ recursively by $x_n = x_{n-1}^2 - 2$, $n \ge 2$. Prove that for any pair of integers $j$, $k$ satisfying $1 \le j < k$, the terms $x_j$, $x_k$ are relatively prime.
\bigskip
Solution.
\medskip
We have, immediately, $x_2 = x_1^2 - 2$, $x_3 = x_2^2 - 2 = (x_1^2 - 2)^2 - 2 = x_1^4 - 4x_1^2 + 2$, and in general, $x_k = P(x_j^2) \pm 2$, where $P(x_j^2)$ is a homogeneous polynomial in $x_j^2$, and $+2$ obtains if $k>j+1$ and $-2$ obtains if $k=j+1$. Let $p$ be the largest prime that divides $x_j$, $x_k$. Then $p$ must also divide 2, so either $p=2$ or no prime $p$ exists. But $x_1$ is odd, which implies all $x_n$'s are odd. Hence, $(x_j , x_k) = 1$ is the only possibility.
\bigskip
3. Let $d(n)$ denote the number of divisors of $n$. Call $n$ a round number if $m