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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}
\bigskip
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.
\bigskip
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 $m<n$ implies $d(m) < d(n)$.  Prove that if $n$ is round and
$$n = p_1^{\alpha _1} p_2^{\alpha _2} \cdots p_m^{\alpha _m}$$
is the unique prime factorization of $n$ with $p_1 < p_2 < \cdots < p_m$,
then there is no prime missing between $p_1$ and $p_m$.
\bigskip
Solution.
\medskip
First observe that the number of positive, integral divisors of a power of a prime $p^\alpha$ is just $\alpha + 1$ (the divisors are $1$, $p$, $p^2$, $\ldots$ , $p^\alpha$).  Suppose, now, that
$$n = p_1^{\alpha _1} p_2^{\alpha _2} \cdots p_r^{\alpha _r} $$
is a round number, but that the $j$th natural prime $P_j$ is missing from the chain of primes, $1<j<r$.  That is, $P_j < p_j$ and $P_j$ is not a divisor of $n$.  Form the new number $n'$ by replacing $p_j^{\alpha _j}$ by $P_j^{\alpha _j}$.

Then, clearly, $n' < n$, but the number of positive, integral divisors of $n'$ is, combinatorically,
$$(\alpha _1 + 1)(\alpha _2 + 1) \cdots (\alpha _j + 1) \cdots (\alpha _r + 1) .$$
This, however, is also the number of positive, integral divisors of $n$, so $n$ could not have been a round number, a contradiction.  Thus, no prime $P_j$ is missing from the sequence.
\bigskip
Modification of a problem from the XI Italian Mathematical Olympiad 1995.  The solution is a modification of a solution by Edward T. H. Wang.
\medskip
4.  In a class of 10 students no two of them have the same ordered pair (written and oral examinations) of scores in mathematics.  We say that student $A$ is better than $B$ if his two scores are greater than or equal to the corresponding scores of $B$.  The scores are integers between 1 and 5.
\item{(a)} Show that there exist three students $A$, $B$, and $C$ such that $A$ is better than $B$ and $B$ is better than $C$.
\item{(b)} Would the same be true for a class of 9 students?
\bigskip
Solution.
\medskip
a) Let the ordered pairs be denoted by $(a_i, b_i)$, $i= 1, 2, \ldots , 10$, where $a_i$ and $b_i$ denote, respectively, the scores of the $i$th student.  For convenience of notation, write $(a_i, b_i) < (a_j, b_j)$ if the $j$th student is better than the $i$th student.  Thus $(a_i, b_i) < (a_j, b_j)$ if and only if $a_i \le a_j$ and $b_i \le b_j$.  Since there are 10 $a_i$'s which are all in $\{ 1, 2, \ldots , 5 \}$ we must have either
\smallskip
\item{(1):} For some $m \in \{ 1, 2, \ldots , 5 \}$, $a_i = a_j = a_k = m$ where $i \ne j \ne k \ne i$,
\smallskip
\noindent or
\smallskip
\item{(2):} Every $m \in \{ 1, 2, \ldots , 5 \}$ appears exactly twice as the first component in the ordered pairs.
\smallskip
\noindent
Similarly, either
\smallskip
\item{(3):} For some $n \in \{ 1, 2, \ldots , 5 \}$, $b_i = b_j = b_k = n$ where $i \ne j \ne k \ne i$,
\smallskip
\noindent or
\smallskip
\item{(4):} Every $n \in \{ 1, 2, \ldots , 5 \}$ appears exactly twice as the second component in the ordered pairs.

In case (1), we would have three ordered pairs of the form $(m,b_i)$, $(m,b_j)$, and $(m,b_k)$.  Since $b_i$, $b_j$, and $b_k$ must all be distinct, we may assume, without loss of generality, that $b_i < b_j < b_k$, and then $(m,b_i) < (m,b_j) < (m,b_k)$.  Similarly, in case (3), we would have three ordered pairs of the form $(a_i,n)$, $(a_j,n)$, and $(a_k,n)$.  Assuming $a_I < a_j < a_k$, we then have $(a_i,n) < (a_j,n) < (a_k,n)$.  If neither (1) nor (3) holds, then both (2) and (4) must hold and so every $l \in \{ 1, 2, \ldots , 5 \}$ appears exactly twice as the first component and exactly twice as the second component in the ordered pairs.  In particular, there must be ordered pairs of the form $(1,b_i)$ and $(1,b_j)$ for some $b_i < b_j$.  Since 5 must appear twice as the second component and $b_i \ne b_j$, there must be at least one ordered pair of the form $(a_k,5)$ for some $a_k \ne 1$.  Then we have $(1,b_i) < (1,b_j) < (a_k,5)$ and the conclusion follows.
\smallskip
\item{(b)} No.  For example, if the 9 ordered pairs are $(1,5), (2,4), (3,3), (4,2), (5,1)$ and $(2,5), (3,4), (4,3), (5,2)$, then no two ordered pairs from the first five are comparable and no two ordered pairs from the other four are comparable.  Hence the required ``chain'' of three students does not exist.
\bigskip
From Problem 2.2.11, {\it Berkeley Problems in Mathematics}, Springer, 1998, QA43.S695.
\medskip
5.  For a real $2 \times 2$ matrix
$$X = \pmatrix x&y\\ z&t \endpmatrix$$
let $\Vert X \Vert = x^2 + y^2 + z^2 + t^2$, and define a distance function by $d(X,Y) = \Vert X-Y \Vert$.  Let $\Sigma = \{ X \mid \det (X) = 0 \}$ and let
$$A = \pmatrix 1&0\\ 0&2 \endpmatrix .$$
Find the minimum distance from $A$ to $\Sigma$ and exhibit a specific matrix $S \in \Sigma$ that achieves this minimum.
\bigskip
Solution.
\medskip
The problem reduces to minimizing 
$$F(x,y,z,t) = (x-1)^2 +y^2 + z^2 + (t-2)^2$$
subject to
$$g(x,y,z,t) = xt - yz = 0 .$$
A Lagrange multiplier solution requires solving the five equations
$$\eqalign{
2(x-1) &= \lambda t \cr
2y &= -\lambda z  \cr
2z &= -\lambda y  \cr
2(t-2) &= \lambda x  \cr
xt &= yz .\cr}$$
The second and third equations have only the trivial solution unless $\lambda^2 = 4$, and the first and fourth equations have no solution if $\lambda^2 = 4$, so we must have $y = z = 0$ and either $x=0$ or $t=0$.  The two possible solutions are $(1,0,0,0)$ and $(0,0,0,2)$.  Because $F(1,0,0,0) = 4$ and $F(0,0,0,2) = 1$, the minimum distance is $1$ and occurs when 
$$X = \pmatrix 0&0\\ 0&2 \endpmatrix .$$
\bye 