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\hyphenation{Mass-achusetts Central Missouri Wis-con-sin Muthuvel}
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\centerline{\bf 1998 Missouri MAA 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 length of segment $PQ$ 
is a minimum.
\bigskip
Solution.
\smallskip
Let $P = (p, p^2)$ 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
$$y - p^2 = - {1 \over 2p} (x - p) .$$
Therefore, the other intersection point of the normal line and the 
parabola, $Q$ has the $x$-coordinate 
$$q = -p - {1 \over 2p} .$$
Let
$$d = \biggl( 2p + {1 \over 2p} \biggr) ^2 + \biggl( p^2 - \biggl( 
p^2 + 1 + {1 \over 4p^2} \biggr) \biggr) ^2$$
represent the square of the distance between $P$ and $Q$.  
Differentiating $d$ with respect to $p$ and setting this expression to 
zero results in the equation
$$8p - {3 \over 2} p^{-3} - {1 \over 4} p^{-5} = 0 .$$
Simplifying this equation, we obtain
$${1 \over 4} p^{-5} ( 32p^6 - 6p^2 - 1 ) = 0 .$$
But, 
$$32p^6 - 6p^2 - 1 = (2p^2 - 1) ( 4p^2 + 1 )^2$$
so the real solutions of the equation are $p = \pm \sqrt {1/2}$.
Hence, the coordinates of $P$ so that the length of segment $PQ$ is a 
minimum is
$$P = \biggl( \pm \sqrt {1 \over 2} , {1 \over 2} \biggl) .$$
\bigskip
2.  Let $0 < a_1 < a_2 < \cdots < a_n$ and let $e_i = \pm 1$.  Prove 
that $\sum_{i=1}^n e_i a_i$ assumes at least ${n+1 \choose 2}$ 
distinct values as the $e_i$'s range over the $2^n$ possible 
combinations of signs.
\bigskip
Solution.
\smallskip
The result is clear if $n=1$.  Assume the result is correct for some $n$ 
and suppose $0 < a_1 < a_2 < \cdots < a_n < a_{n+1}$.  By the induction 
assumption,
$$\sum_{i=1}^n e_i a_i + a_{n+1}$$
assumes at least ${n+1 \choose 2}$ distinct values, the smallest of which 
is
$$- \sum_{i=1}^n a_i + a_{n+1} .$$
Let
$$s_p = - \sum_{i=1}^{p-1} a_i + a_p - \sum_{i=p+1}^{n+1} a_i$$          
for $p = 0, 1, 2, \ldots , n, n+1$.  Note that
$$s_{n+1} = - \sum_{i=1}^n a_i + a_{n+1} .$$
Also, note that $s_p$ is strictly increasing.  Therefore, the $n+1$ 
additional values $s_p$, $p=0, 1, 2, \ldots , n$ are all smaller than 
$s_{n+1}$.  It follows that
$$\sum_{i=1}^{n+1} e_i a_i$$
assumes at least ${n+1 \choose 2} + (n+1) = {n+2 \choose 2}$ distinct 
values.
\bigskip
3.  If $m$ and $n$ are positive integers and $a < b$, find a formula for
$$\int_a^b {(b-x)^m \over m!} {(x-a)^n \over n!} dx$$
and use your formula to evaluate
$$\int_0^1 (1-x^2)^n dx .$$
\bigskip
Solution.
\smallskip
Integration by parts once yields
$$\int_a^b {(b-x)^m \over m!} {(x-a)^n \over n!} dx = \int_a^b 
{(b-x)^{m-1} \over (m-1)!} {(x-a)^{n+1} \over (n+1)!} dx $$
so continuing, we get
$$\int_a^b {(b-x)^m \over m!} {(x-a)^n \over n!} dx = \int_a^b 
{(x-a)^{n+m} \over (n+m)!} dx = {(b-a)^{n+m+1} \over (n+m+1)!} .$$
Now,
$$\int_0^1 (1-x^2)^n dx = {1 \over 2} \int_{-1}^1 (1-x)^n (x-(-1))^n dx = 
{1 \over 2} (n!)^2 {2^{2n+1} \over (2n+1)!} = {2 \cdot 4 \cdot 6 \cdots 
2n \over 1 \cdot 3 \cdot 5 \cdots (2n+1)} .$$
\bigskip
4.  Describe how to fold a rectangular sheet of paper so that the lower 
right corner touches the left edge and the length of the crease is a 
minimum.  Discuss how the dimensions of the rectangle affect the result. 
\bigskip
Solution.
\smallskip
Let the width and the height of the page be given by $w$ and $h$ and the 
length of the crease by $c$.  Assume $h>w$, because otherwise the minimum 
is clearly $h$ when the paper is folded vertically in half.  There are 
three remaining cases: (i) when the crease extends from the bottom edge 
to the top edge, (ii) when the crease extends from the bottom edge to the 
right edge, and (iii) when the crease extends from the left edge to the 
right edge.  In case (i) the minimum is clearly when the crease is 
vertical with $c=h$.  Also clear is case (iii), in which the minimum 
occurs when the lower right corner is placed on the upper left corner.  
Here
$$c = {w \over h} \sqrt {w^2 + h^2} .$$
In case (ii), let $a$ and $b$ be the short leg and the long leg of the 
right triangle whose hypotenuse is $c$ and whose right angle is the 
corner which was folded over to the left edge.  Let $x$ be the distance 
from this corner (after the fold) to the lower left corner of the page.  
From the Pythagorean Theorem,
$$x^2 + (w-a)^2 = a^2 \text{  and  } w^2 + (b-x)^2 = b^2 .$$
Solving these equations for $a$ and $b$ yields
$$a = {x^2 + w^2 \over 2w} \text{  and  } b = {x^2 + w^2 \over 2x} .$$
Now, since $c^2 = a^2 + b^2$, we have $c^2$ in terms of $x$.  
Differentiation yields a relative minimum at $x = w / \sqrt 2$, where $c 
= {3 \sqrt 3 \over 4} w$.  The minimum is therefore
$$\min \bigg\{ h, {3 \sqrt 3 \over 4} w, {w \over h} \sqrt {w^2 + h^2} 
\bigg\} ,$$
but we can eliminate the case (ii) value because that leads to the 
inequality
$${3 \sqrt 3 \over 4} < {h \over w} < {4 \sqrt {11} \over 11}$$
which is impossible because ${3 \sqrt 3 \over 4} > {4 \sqrt {11} \over 
11}$.  The minimum is $h$ (case (i)) whenever
$$0 < {w \over h} < \sqrt {{\sqrt 5 - 1 \over 2}}$$
and
$${w \over h} \sqrt {w^2 + h^2}$$
(case (iii)) otherwise. 
\bigskip
Modified from the 1967 All-Soviet Mathematics Competition - Problem 88
\smallskip
5.  Let $n$ be a positive integer.  Prove that there exists a number 
divisible by $5^n$ that does not contain a single zero in its decimal 
notation.
\bigskip
Solution.
\smallskip
Perform the following algorithm.
\smallskip
$\underline{\text{Algorithm}}$.
{\obeylines \sfcode`;=3000
$m := 5^n$;
{\bf for $i := 0$ to $n$ }
\quad {\bf if} the digit in the $10^i\text{th}$ place of $m$ is $0$
\qquad {\bf then} $m := m + 10^i \cdot 5^n$; \par}
\medskip
The resulting $m$ has no zero from its $10^0\text{th}$ to
$10^n\text{th}$ places and is a multiple of $5^n$.
\smallskip
Next, change each zero in $m$ to any nonzero digit and call this 
number $p$.  Note that any zero in $m$ must be in its 
$10^{n+1}\text{th}$ place or higher.
\smallskip
The number $p$ is a multiple of $5^n$ and does not contain a single 
zero in its decimal notation.
\vfill\eject
\centerline{\bf 1998 Missouri MAA Collegiate Mathematics Competition}
\vskip 6pt
\centerline{\bf Session II}
\bigskip
1.  Let $I$ be the $n \times n$ identity matrix.  Prove that $AB - BA \ne 
I$ for any $n \times n$ matrices $A$ and $B$. 
\bigskip
Solution.
\smallskip
Let $A = (a_{ij})_{n \times n}$ and $B = (b_{ij})_{n \times n}$.  Then 
the sum of the diagonal (i.e., the trace) of $AB - BA$ is
$$\sum_{i=1}^n \Biggl( \sum_{j=1}^n a_{ij} b_{ji} - \sum_{j=1}^n b_{ij} 
a_{ji} \Biggr) = 0 .$$
Therefore, $AB - BA \ne I$.
\bigskip
2.  Let $a_1, \ldots , a_n$ be positive real numbers and let $s$ denote 
their sum.  Show that
$$(1+a_1) (1+a_2) \cdots (1+a_n) \le 1 + {s \over 1!} + {s^2 \over 2!} + 
\cdots + {s^n \over n!} .$$
\bigskip
Solution.
\smallskip
By the Arithmetic Mean-Geometric Mean inequality,
$$\Biggl( \prod_{j=1}^n (1 + a_j) \Biggr) ^{1 \over n} \le {1 \over n} 
\sum_{j=1}^n (1 + a_j) = {1 \over n} (n+s) = 1 + {s \over n} ,$$
so
$$\prod_{j=1}^n (1 + a_j) \le \biggl( 1 + {s \over n} \biggr) ^n = 
\sum_{j=0}^n {n \choose j} {s^j \over n^j} = \sum_{j=0}^n {n! \over 
(n-j)! \cdot n^j} \cdot {s^j \over j!} \le \sum_{j=0}^n {s^j \over j!} .$$
\vfill\eject
From {\it Crux Mathematicorum}, vol. 8 (1982), p. 99 and pp. 271--272.
\smallskip  
Proposed by Jack Brennen, student, Poolesville, Maryland.
\smallskip
3.  Sum the series
$$\sum_{i=1}^\infty {36 i^2 + 1 \over (36 i^2 - 1)^2 } .$$   
\bigskip
Solution.
\smallskip
If $S$ is the required sum, then we have
$$2S = \sum_{i=1}^\infty \biggl( {1 \over (6i-1)^2} + {1 \over (6i+1)^2} 
\biggr) = {1 \over 5^2} + {1 \over 7^2} + {1 \over 11^2} + {1 \over 13^2} 
+ \cdots .$$
Now, using the result
$$1 + {1 \over 2^2} + {1 \over 3^2} + \cdots = {\pi ^2 \over 6} ,$$
we obtain successively
$$\eqalign{
&{1 \over 2^2} + {1 \over 4^2} + {1 \over 6^2} + \cdots = {\pi ^2 \over 
24} ,\cr
&1 + {1 \over 3^2} + {1 \over 5^2} + \cdots = {\pi ^2 \over 6} - {\pi ^2 
\over 24} = {\pi ^2 \over 8} ,\cr
&{1 \over 3^2} + {1 \over 9^2} + {1 \over 15^2} + \cdots = {\pi ^2 \over 
72} ,\cr}$$
and
$$2S + 1 = 1 + {1 \over 5^2} + {1 \over 7^2} + {1 \over 11^2} + {1 \over 
13^2} + \cdots = {\pi ^2 \over 8} - {\pi ^2 \over 72} = {\pi ^2 \over 9} 
,$$
from which
$$S = {\pi ^2 - 9 \over 18} .$$
\bigskip
4.  Circle $O$ has a diameter of $3$; let $AOD$ be a diameter.  A second 
circle, of radius 1, is inscribed in circle $O$ so that its center lies 
along $AOD$ and such that this circle is tangent to circle $O$ at point 
$D$.  A third circle, of radius $1/2$, is next inscribed in circle $O$ so 
that its center also lies along $AOD$ and such that it is tangent to 
circle $O$ at point $A$.  Determine the radius of a fourth circle that 
could be constructed inside circle $O$ and which would be simultaneously 
tangent to all three circles.
\bigskip
Solution.
\smallskip
\input circle
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\smallskip
We have the following information from the diagram.  $OD = OF = {3 
\over 2}$, $O'D = O'E = 1$, $BG = {1 \over 2}, \text{ and } GC = EC = 
r$.  In $\triangle BCO'$, $BO' = {3 \over 2}$, $BC = {1 \over 2} + r$, 
$O'C = 1 + r$, and the semi-perimeter of $\triangle BCO'$ is ${3 \over 
2} + r$.  In $\triangle BCO$, $BO = 1$, $BC = {1 \over 2} + r$, $OC = 
{3 \over 2} - r$, and the semi-perimeter of $\triangle BCO$ is ${3 
\over 2}$.  In $\triangle OCO'$, $OO' = {1 \over 2}$, $OC = {3 \over 2} 
- r$, $O'C = 1 + r$ and the semi-perimeter of $\triangle OCO'$ is ${3 
\over 2}$.
\vfill\eject
Now, 
$$\text{area } \triangle BCO' = \text{ area } \triangle BCO + 
\text{ area } \triangle OCO' .$$
Hence, by Heron's formula applied to each triangle, we obtain
$$\sqrt {\biggl( {3 \over 2} +r \biggr) (r) (1) \biggl( {1 \over 2} 
\biggr) } = \sqrt {\biggl( {3 \over 2} \biggr) \biggl( {1 \over 2} \biggr) 
(1-r)(r)} + \sqrt { \biggl( {3 \over 2} \biggr) (1)(r) \biggl( {1 \over 2} 
-r \biggr) }$$
or
$$\sqrt {3+2r} = \sqrt {3(1-r)} + \sqrt {3(1-2r)} .$$
Squaring both sides leads to
$$11r - 3 = 6 \sqrt {(1-r)(1-2r)}$$
and squaring again gives the quadratic equation
$$49r^2 + 42r - 27 = (7r+9)(7r-3) = 0 ,$$
from which $r = {3 \over 7}$.
\bigskip
5.  A sequence of polynomials $\{ P_i (x) \} _{i=0}^\infty$ is defined by 
the generating function
$${2 e^{tx} \over e^t + 1} = \sum_{i=0}^\infty P_i (x) {t^i \over i!} .$$
Show that 1 is a zero of $P_i (x)$ for all even $i>0$, and $1/2$ 
is a zero of $P_i (x)$ for all odd $i>0$.
\bigskip
Solution.
\smallskip
$\underline{\text{Part 1}}$.  In the generating function, replace $t$ 
by $-t$ and set $x = {1 \over 2}$.
$$\sum_{i=0}^\infty P_i \biggl( {1 \over 2} \biggr) {(-t)^i \over i!} = 
{2 e^{-t/2} \over e^{-t} + 1} \cdot {e^t \over e^t} = {2 e^{t/2} \over 
1 + e^t} = \sum_{i=0}^\infty P_i \biggl( {1 \over 2} \biggr) {t^i \over 
i!} .$$
Thus, the series represents an even function of $t$ and each $P_i ( 
{1 \over 2} ) = 0$ for odd $i$.
\vfill\eject
$\underline{\text{Part 2}}$.  In the generating function, replace $x$ 
by $x+1$ to give
$${2 e^{xt + t} \over e^t + 1} = \sum_{i=0}^\infty P_i (x+1) {t^i \over 
i!} ,$$
so
$$\sum_{i=0}^\infty ( P_i (x+1) + P_i (x) ) {t^i \over i!} = {2e^{xt} 
\over e^t + 1} (e^t + 1) = 2e^{xt} = 2 \sum_{i=0}^\infty x^i {t^i \over 
i!} .$$  As $t$ is arbitrary, then for each $i$
$$2x^i = P_i (x+1) + P_i (x) .$$
In this, set $x=0$ to obtain
$$0 = P_i (1) + P_i (0) .\eqno(*)$$
Again, in the generating function, replace $t$ by $-t$ and set $x=0$.  
Subtraction of the two series gives, for even $i$,
$$P_i (1) - P_i (0) = 0 .$$
In combination with ($*$), we obtain
$$P_i (1) = P_i (0) = 0$$
for all even $i>0$.
\bye
