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\hyphenation{Mass-achusetts Central Missouri Wis-con-sin Muthuvel}
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\centerline{\bf 2001 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 area bounded by the
normal line and the parabola is a minimum.
\bigskip
Solution.
\medskip
Let $P = (p,p^2)$ and $Q = (q,q^2)$. Without loss of generality,
let $p>0$. 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
$$q = -p - {1 \over 2p} .$$
Now the area bounded by the normal line and the parabola (as a function
of $p$) is
$$A = \int_q^p \biggl( -{1 \over 2p} (x-p) + p^2 - x^2 \biggr) dx .$$
Integrating, substituting the $p$ expression for $q$, and simplifying,
we obtain
$$\eqalign{
A &= -{1 \over 4p} (x-p)^2 + p^2 x - {1 \over 3} x^3 \bigg\vert _q^p
= {2 \over 3} p^3 + {1 \over 4p} (q-p)^2 - p^2 q + {1 \over 3} q^3 \cr
&= {2 \over 3} p^3 + {1 \over 4p} \biggl( -2p - {1 \over 2p} \biggr) ^2
- p^2 \biggl( -p - {1 \over 2p} \biggr) + {1 \over 3} \biggl( -p - {1
\over 2p} \biggr) ^3 \cr
&= {4 \over 3} p^3 + p + {1 \over 4p} + {1 \over 48 p^3} .\cr}$$
Differentiating with respect to $p$ yields
$$\eqalign{
{dA \over dp} &= 4p^2 + 1 - {1 \over 4p^2} - {1 \over 16p^4} \cr
&= {1 \over 16p^4} (64p^6 + 16p^4 - 4p^2 - 1) \cr
&= {1 \over 16p^4} (4p^2 - 1) (16p^4 + 8p^2 + 1) .\cr}$$
Setting the derivative equal to 0 results in $p = 1/2$ so the
coordinates of $P$ so that the area bounded by the normal line and the
parabola is a minimum are $(1/2, 1/4)$.
\bigskip
2. Let $\{ x_i \}$ denote any finite sequence with the
following properties:
\medskip
\item{(a)} $x_i \in \{ -2, 1, 2 \}$ for each $x_i$,
\item{(b)} $\sum_i x_i = 29$,
\item{(c)} $\sum_i x_i^2 = 59$.
\medskip
In considering the family of all such sequences, let $M = \max \{
\sum_i x_i^3 \}$ and $m = \min \{ \sum_i x_i^3 \}$. Determine $M/m$.
\bigskip
Solution.
\medskip
In a sequence $\{ x_i \}$, let $a = \text{number of -2's}$, $b = \text{number of 1's}$, and $c = \text{number of 2's}$. Properties (b) and (c) then yield
$$\cases -2a + b + 2c = 29 &\\
4a + b + 4c = 59 ,&
\endcases$$
or upon elimination of $b$: $3a + c = 15$. To obtain $M$ the sum $\sum_i x_i^3$ must be dominated by the 2's. The maximum value of $c$ consistent with $3a + c = 15$ is $c=15$. But this leads to $b=-1$; hence, $c=12$, $a=1$, $b=7$, and $M = 1(-2)^3 + 7(1)^3 + 12(2)^3 = 95$.
To obtain $m$ the sum $\sum_i x^3$ must be dominated by the -2's. The maximum value of $a$ consistent with $3a + c = 15$ is $a=5$, so $b=39$ and $c=0$. Hence,
$m = 5(-2)^3 + 39(1)^3 + 0(2)^3 = -1$.
It follows that $M/m = -95$.
\vfill\eject
3. Let $a$, $b$, and $c$ be the sides of a triangle with perimeter 2.
Prove that
$$3/2 < a^2 + b^2 + c^2 + 2abc < 2.$$
\bigskip
Solution.
\medskip
\comment
The semiperimeter of the triangle, $s$, is $s = (a+b+c)/2 = 2/2 = 1$.
From Heron's formula,
$$A = \sqrt {s(s-a)(s-b)(s-c)} = \sqrt {(1-a)(1-b)(1-c)} .$$
Hence, $A > 0$ implies $(1-a)(1-b)(1-c) > 0$, or
$$1 - (a+b+c) + (ab + ac + bc) - abc > 0.$$
But $a+b+c=2$, so
$$2abc - 2(ab + ac + bc) < 2(1-2) = -2 .$$
Finally,
$$a^2 + b^2 + c^2 + 2abc = (a+b+c)^2 + \bigl( 2abc - 2(ab + ac + bc)
\bigr) < 2^2 + (-2) = 2.$$
\endcomment
Since the perimeter is 2, no side can be greater than 1, so the area,
$$A = {1 \over 2} bc \sin \alpha ,$$
is less than 1/2. Then $A^2 = (1-a)(1-b)(1-c)$ implies
$$\eqalign{
0 &< (1-a)(1-b)(1-c) < 1/4 ,\cr
0 < 1 - (a&+b+c) + (ab+ac+bc) - abc < 1/4 ,\cr
0 < 1 &- 2 + (ab+ac+bc) - abc < 1/4 ,\cr
1 &< (ab+ac+bc) - abc < 5/4 ,\cr}$$
and
$$2 < 2(ab+ac+bc) - 2abc < 5/2 .$$
Finally,
$$a^2 + b^2 + c^2 + 2abc = (a+b+c)^2 + (2abc - 2(ab+ac+bc)) ,$$
so,
$$2^2 - 5/2 < a^2 + b^2 + c^2 + 2abc < 2^2 - 2 ,$$
or,
$$3/2 < a^2 + b^2 + c^2 + 2abc < 2 .$$
\vfill\eject
4. Find the sum
$$S = \sum_{k=1}^\infty {k^2 \over 3^k} .$$
\bigskip
Solution.
\medskip
Define
$$F(x) = \sum_{k=1}^\infty {x^{k+1} \over 3^k} ,\ \ \vert x \vert < 3 .$$
As this series is geometric, we have, explicitly,
$$F(x) = {x^2 \over 3-x} .$$
Differentiation gives
$$F'(x) = \sum_{k=1}^\infty {(k+1) x^k \over 3^k} = {6x - x^2 \over (3-x)^2} ,$$
so
$$\sum_{k=1}^\infty {kx^k \over 3^k} = {6x - x^2 \over (3-x)^2} - {F(x) \over x} = {3x \over (3-x)^2} .$$
A second differentiation gives, from the previous line
$$\sum_{k=1}^\infty {k^2 x^{k-1} \over 3^k} = {3 (x + 3) \over (3-x)^3} ,$$
so
$$S = \sum_{k=1}^\infty {k^2 \over 3^k} = {3(1+3) \over (3-1)^3} = {3 \over 2} .$$
\vfill\eject
5. A set of five cubical dice has the following properties:
\medskip
\item{(a)} On each face of each die is a 3-digit integer. No two
integers on a given face are the same.
\item{(b)} Every integer has a nonzero hundred's digit.
\item{(c)} The sum, when the dice are rolled, of the five integers is
a 4-digit integer.
\item{(d)} Whenever the dice are rolled, their sum $S$ can be found
quickly as follows: the sum of the unit's digits of the five dice is the
last two digits of $S$, and the first two digits of $S$ are 50 minus the
sum of the unit's digits.
\medskip
For example, if the dice come up 189, 256, 275, 845, and 168, the
sum of the unit's digits is $9+6+5+5+8=33$, so the value of $S$ is 1733,
since $50-33=17$.
Explain, justifying your statements, how such a set of dice can be
constructed.
\bigskip
Solution.
\medskip
Since the last two digits of $S$ equals the sum of the unit's digits of
the five dice, the sum of the ten's digits must be a multiple of 10.
This can be accomplished in a number of ways, such as making the ten's
digit on all faces of each die the same: die 1 ten's digit = 4, die 2
ten's digit = 5, die 3 ten's digit = 6, die 4 ten's digit = 7, die 5
ten's digit = 8 will make the sum of the ten's digits always be 30.
The quick-sum property, assuming the sum of the ten's digits is 30,
gives
$$S = 100 \biggl( \sum_{i=1}^5 {h_i} \biggr) + 300 + \sum_{i=1}^5
{u_i} = 100 \biggl( 50 - \sum_{i=1}^5 {u_i} \biggr) + \sum_{i=1}^5
{u_i} .$$
Here, $h_i$ and $u_i$ are the hundred's and unit's digits on die $i$.
This equation reduces to
$$\sum_{i=1}^5 (h_i + u_i) = 47.$$
So, whenever the dice are rolled, the sum of the 5 hundred's digits
plus the sum of the 5 unit's digits must be 47. This can occur if the
hundred's digit plus the unit's digit on all faces of each die is a
constant and the five constants add to 47. One possibility for the
five constants is 13, 8, 9, 7, and 10, which was the case in the
example.
This problem can be solved in more generality - there is nothing sacred
about the 50, for instance. A pertinent result is that the above
solution guarantees that the difference between any two faces of a
given die is a multiple of 99, and this property can be used as a
starting point to solve the problem.
\vfill\eject
\centerline{\bf 2001 Missouri MAA Collegiate Mathematics Competition}
\vskip 6pt
\centerline{\bf Session II}
\bigskip
1. Circle $B$ lies wholly in the interior of circle $A$. Find the
loci of points equidistant from the two circles?
\bigskip
Solution.
\medskip
The locus of points is an ellipse. Let circle $A$ have radius $r_A$
and center $(x_A, y_A)$ and circle $B$ have radius $r_B$ and center
$(x_B, y_B)$. If $(x,y)$ is a point equidistant from the two circles,
then the distance $(x,y)$ is from outer circle $A$,
$$r_A - \sqrt {(x-x_A)^2 + (y-y_A)^2}$$
is equal to the distance $(x,y)$ is from inner circle $B$,
$$\sqrt {(x-x_B)^2 + (y-y_B)^2} - r_B .$$
That is,
$$r_A - \sqrt {(x-x_A)^2 + (y-y_A)^2} = \sqrt {(x-x_B)^2 + (y-y_B)^2} -
r_B .$$
Therefore,
$$r_A + r_B = \sqrt {(x-x_A)^2 + (y-y_A)^2} + \sqrt {(x-x_B)^2 +
(y-y_B)^2} .$$
But this is exactly the equation of an ellipse.
\bigskip
2. Show that if $x$, $y$, and $z$ are positive reals such that
$x+y+z=1$, then
$$\biggl( {1 \over x} - 1 \biggr) \biggl( {1 \over y} - 1 \biggr)
\biggl( {1 \over z} - 1 \biggr) \ge 8.$$
\bigskip
Solution.
$$\biggl( {1 \over x} - 1 \biggr) \biggl( {1 \over y} - 1 \biggr)
\biggl( {1 \over z} - 1 \biggr) = {1-x \over x} \cdot {1-y \over
y} \cdot {1-z \over z} = {(y+z) (z+x) (x+y) \over xyz} .$$
The problem then reduces to showing that
$$(x+y) (y+z) (z+x) \ge 8 xyz .$$
But, by the A.M.--G.M. inequality,
$$\eqalign{
&x+y \ge 2 \sqrt {xy} \cr
&y+z \ge 2 \sqrt {yz} \cr
&z+x \ge 2 \sqrt {zx} .\cr}$$
Multiplying the left and right sides of the inequalities, we obtain
$$(x+y) (y+z) (z+x) \ge 8 xyz$$
which establishes the result.
\bigskip
3. A convex decagon and all of its diagonals are drawn. How many {\it
interior} points of intersection of the diagonals are there, if it is
assumed that no 3 diagonals share a common {\it interior} point?
\bigskip
Solution.
\medskip
Let $N(n)$ be the number of such interior intersection points in a convex
$n$-gon. Label the vertices $1$, $2$, $3$, $\ldots$, $n$. Any four
labels selected from this list (e.g., $(1, 3, n, 6)$) define a
quadrilateral, which is convex because the $n$-gon is convex. The
quadruple of integers therefore also defines uniquely an interior
intersection point of the diagonals of the quadrilateral, and hence also
of the $n$-gon.
The total number of interior intersection points of diagonals of the
$n$-gon (assuming no 3 of the diagonals meet at a common point) is thus
the number of ways that four integers can be selected (without regard to
order) from a list of $n$ distinct integers. Therefore,
$$N(n) = {n \choose 4} = {n(n-1)(n-2)(n-3) \over 24} ,$$
which for the decagon gives $N(10) = 210$.
\vfill\eject
From {\it From Erd\H{o}s to Kiev} by Ross Honsberger, Dolciani
Mathematical Exposition, 1995, pp. 8--9.
\medskip
4. No matter what $n$ real numbers $x_1$, $x_2$, $\ldots$, $x_n$ may be
selected in the closed unit interval $[0,1]$, prove that there always
exists a real number $x$ in this interval such that the average unsigned
distance from $x$ to the $x_i$'s is exactly $1/2$.
\bigskip
Solution.
\medskip
Adopting a perfectly straightforward approach, look at the function
$$f(x) = {1 \over n} \sum_{i=1}^n \vert x - x_i \vert .$$
At $x=0$, we have (since $x_i \ge 0$)
$$f(0) = {1 \over n} \sum_{i=1}^n \vert - x_i \vert = {1 \over n}
\sum_{i=1}^n x_i .$$
For $x=1$, we get (since $x_i \in [0,1]$)
$$\eqalign{
f(1) &= {1 \over n} \sum_{i=1}^n \vert 1 - x_i \vert \cr
&= {1 \over n} \sum_{i=1}^n (1-x_i) \cr
&= {1 \over n} \biggl( n - \sum_{i=1}^n x_i \biggr) \cr
&= 1 - f(0) ,\cr}$$
and we have the crucial relation
$$f(0) + f(1) = 1.$$
In view of this, the two values $f(0)$ and $f(1)$ are either each equal
to $1/2$, providing two solutions to the problem, or their values {\it
straddle} $1/2$, in which case the continuity of the function implies
$f(x) = 1/2$ for some $x$ between 0 and 1.
\bigskip
From {\it From Erd\H{o}s to Kiev} by Ross Honsberger, Dolciani
Mathematical Exposition, 1995, pp. 6--8.
\medskip
5. Consider the polynomial
$$f(x) = x^n + a_1 x^{n-1} + a_2 x^{n-2} + \cdots + a_{n-1} x + 1,$$
where $a_i \ge 0$. If the equation $f(x) = 0$ happens to have $n$ real roots, is it not
remarkable that the value of $f(2)$ must then be at least $3^n$? Prove
this unlikely consequence: $f(2) \ge 3^n$.
\bigskip
Solution.
\medskip
Since all the coefficients $a_i$ are $\ge 0$, the substitution of any
nonnegative number for $x$ would make $f(x)$ at least 1, implying that
all the roots of $f(x) = 0$ must be negative numbers, say $-r_1$,
$-r_2$, $\ldots$, $-r_n$. Using these roots to factor $f(x)$, we obtain
$$\eqalign{
f(x) &= x^n + a_1 x^{n-1} + a_2 x^{n-2} + \cdots + a_{n-1} x + 1 \cr
&= (x+r_1) (x+r_2) (x+r_3) \cdots (x+r_n) \cr
&= x^n + (r_1 + r_2 + \cdots + r_n) x^{n-1} + (r_1 r_2 + r_1 r_3 +
\cdots ) x^{n-2} + \cdots + r_1 r_2 \cdots r_n .\cr}$$
Thus each coefficient $a_k$ is given by
$$\eqalign{
a_k &= \text{ the sum of the } {n \choose k} \text{ products of the
roots taken } k \text{ at a time}\cr
&= \sum r_{i_1} r_{i_2} \cdots r_{i_k} .\cr}$$
Also, the absolute term $r_1 r_2 \cdots r_n = 1$.
Now, applying the arithmetic mean - geometric mean inequality to the
${n \choose k}$ terms that make up $a_k$, we obtain
$${a_k \over {n \choose k}} = {\sum r_{i_1} r_{i_2} \cdots r_{i_k}
\over {n \choose k}} \ge \biggl( \prod r_{i_1} r_{i_2} \cdots r_{i_k}
\biggr) ^{1 / {n \choose k}} ,$$
from which we get
$$a_k \ge {n \choose k} \biggl( \prod r_{i_1} r_{i_2} \cdots r_{i_k}
\biggr) ^{1/{n \choose k}} .$$
Admittedly, this doesn't seem to be very promising. However, this
awkward product melts away completely with the brilliant observation
that, since there is no preference for one $r_i$ over another, it
follows that each $r_i$ occurs in the product the same number of times
altogether, with the result that, for some positive integer $t$,
$$\prod r_{i_1} r_{i_2} \cdots r_{i_k} = (r_1 r_2 \cdots r_n)^t = 1^t
= 1.$$
Therefore,
$$a_k \ge {n \choose k} .$$
Now, $k$ only runs from 1 to $n-1$, and so if we set $a_0 = a_n = 1$,
then $a_0$ and $a_n$ would respectively equal ${n \choose 0}$ and ${n
\choose n}$, making
$$a_k \ge {n \choose k}$$
for all $k = 0, 1, 2, \ldots , n$. Thus we can write
$$f(x) = \sum_{k=0}^n a_k x^{n-k} ,$$
and the value of $f(2)$ is given by
$$\sum_{k=0}^n a_k 2^{n-k} .$$
Since $a_k \ge {n \choose k}$, then (by the binomial theorem)
$$f(2) \ge \sum_{k=0}^n {n \choose k} 2^{n-k} = \sum_{k=0}^n {n \choose
k} 1^k 2^{n-k} = (1 + 2)^n = 3^n .$$
\bye