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\hyphenation{Mass-achusetts Central Missouri Wis-con-sin Muthuvel}
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\def\N{\Bbb N}
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\centerline{\bf 2008 Missouri Collegiate Mathematics Competition}
\vskip 6pt
\centerline{\bf Session I}
\bigskip
1. A straight line segment of arbitrary positive slope $m$ is drawn from the origin $O$ to the point of
intersection $A$ in quadrant $I$ with the ellipse $x^2 + 4y^2 = 4$. A line segment is then drawn parallel
to the $x$-axis from $A$ over to the $y$-axis, which the segment meets at point $B$. Points $O$, $A$, $B$
are thus the vertices of a right triangle. Deduce the value of $m$ that maximizes the area $R$ of triangle
$OAB$, and prove that this really is the maximum area.
\bigskip
Solution.
\vglue -5pc
$$\epsfxsize=15pc \epsfbox{Prob2fig.eps}$$
\bigskip
Combination of $y=mx$ and $x^2 + 4y^2 = 4$ yields the following coordinates for point $A$:
$$\biggl( {2 \over \sqrt {1 + 4m^2}} , {2m \over \sqrt {1 + 4m^2}} \biggr) .$$
Hence, the area of the triangle is
$$R = {1 \over 2} ( \overline{AB} ) ( \overline{OB} ) = {1 \over 2} {2 \over \sqrt {1 + 4m^2}} {2m \over \sqrt
{1 + 4m^2}} = {2m \over 1+4m^2} .$$
Then
$${dR \over dm} = {2 - 8m^2 \over (1+4m^2)^2} = 0 ,$$
the only solution of which that is consistent with $m>0$, is $m=1/2$. To this corresponds
$$R = {2 (1/2) \over 1 + 4(1/2)^2} = {1 \over 2} .$$
That this is the maximum area we can see from the computation
$$\eqalign{
{d^2 R \over dm^2} &= {(1+4m^2)^2 (-16m) - (2-8m^2) (2) (1+4m^2) (8m) \over (1+4m^2)^4} \cr
&= {16m \over (1+4m^2)^3} (4m^2 - 3) ,\cr}$$
the value of which is $-2$ at $m=1/2$.
\bigskip
2. Let $a$ be a positive real number. The Lemniscate of Bernoulli is defined by
$$(x^2 + y^2)^2 = a^2 (x^2 - y^2) .$$
Find the area bounded by the Lemniscate of Bernoulli.
\bigskip
Solution.
\medskip
In polar coordinates, the Lemniscate of Bernoulli is defined by
$$r^2 = a^2 \cos 2\theta .$$
Therefore, the area bounded by the Lemniscate of Bernoulli is
$$2 \int {1 \over 2} r^2 \, d\theta = a^2 \int_{-{\pi \over 4}}^{\pi \over 4} \cos 2\theta \, d\theta = a^2 .$$
\bigskip
3. The sequence of Catalan numbers, $\{ C_n \}_{n=1}^\infty$, is defined by
$$C_n = {1 \over n+1} {2n \choose n} .$$
Does there exist a member of the sequence that is not a natural
number? Find one, or prove that there is none.
\bigskip
Solution.
\medskip
The idea is to rewrite each $C_n$ as a linear combination of binomial coefficients. We have
$$C_n = {1 \over n+1} {2n \choose n} = {(2n)! \over (n+1)! n!} = {1 \over n} {(2n)! \over (n+1)! (n-1)!} $$
so $$\cases (n+1) C_n = {2n \choose n} &\\
n C_n = {2n \choose n-1} .
\endcases$$
Subtraction gives
$$C_n = {2n \choose n} - {2n \choose n-1} .$$
But the two binomial coefficients are natural numbers because they are numbers of combinations of countable
objects. Additionally,
$${2n \choose n} > {2n \choose n-1}$$
if and only if
$${1 \over n} > {1 \over n+1} ,$$
which is true for all $n \ge 1$. Hence, $C_n$ is the positive difference of two natural numbers for all
$n$, and so is itself always a natural number.
\bigskip
4. Suppose a belt is stretched tightly over two circular pulleys with radii $r_1$ and $r_2$, whose centers
are $d$ units apart with $d > r_1 + r_2$. If $r_1 > r_2$, find a formula for the total length of the belt
in terms of $r_1$, $r_2$, and $d$.
\bigskip
Solution.
\medskip
The total length is composed of two equal straight parts and two circular arcs. View the pulleys as
side-by-side horizontally with the larger one on the left. Denote the centers of the pulleys by $P_1$
(larger pulley) and $P_2$ and the top points of tangency by $T_1$ (on larger pulley) and $T_2$. Let $\theta$
be the angle $T_1 P_1 P_2$. Then we get:
$$L = 2 \big\lbrack \sqrt {d^2 - (r_1 - r_2)^2} + r_1 (\pi - \theta ) + r_2 \theta \big\rbrack ,$$
where
$$\theta = \cos ^{-1} \biggl( {r_1 - r_2 \over d} \biggr) $$
is in radians.
\bigskip
5. Evaluate
$$\sum_{k=1}^\infty {1 \over {k+n \choose k}}$$
for $n \ge 2$. What is this series when $n=1$?
\vfill\eject
Solution.
\medskip
$$\eqalign{
&\sum_{k=1}^\infty {1 \over {k+n \choose k}} = \sum_{k=1}^\infty {1 \over {k+n \choose n}} = \sum_{k=1}^\infty
{1 \over {(k+1) (k+2) \cdots (k+n) \over n!}} \cr
&= n! \sum_{k=1}^\infty {1 \over (k+1)(k+2) \cdots (k+n)} \cr
&= {n! \over n-1} \sum_{k=1}^\infty \biggl( {1 \over (k+1)(k+2) \cdots (k+n-1)} - {1 \over (k+2) (k+3) \cdots
(k+n)} \biggr) \cr
&= {n! \over n-1} \biggl( {1 \over 2 \cdot 3 \cdot \cdots \cdot n} - {1 \over 3 \cdot 4 \cdot \cdots \cdot
(n+1)} + {1 \over 3 \cdot 4 \cdot \cdots \cdot (n+1)} - {1 \over 4 \cdot 5 \cdot \cdots \cdot (n+2)} \cr
&\ \ \ \ \ + {1 \over 4 \cdot 5 \cdot \cdots \cdot (n+2)} - {1 \over 5 \cdot 6 \cdot \cdots \cdot (n+3)} +
\cdots \biggr) \cr
&= {n! \over n-1} \cdot {1 \over n!} = {1 \over n-1} .\cr}$$
For $n=1$, the series is a harmonic series
$${1 \over 2} + {1 \over 3} + {1 \over 4} + {1 \over 5} + \cdots$$
which is divergent, and the formula $1/(n-1)$ would indicate that the series should be divergent.
\vfill\eject
\centerline{\bf 2008 Missouri Collegiate Mathematics Competition}
\vskip 6pt
\centerline{\bf Session II}
\bigskip
1.
\vskip 1pt
\item{(a)} Let $f(x) = x^3 + x$. Let $g(x)$ be the inverse function of $f(x)$. Find $g' (10)$.
\smallskip
\item{(b)} For $x>0$, define $h(x) = 1/f(x)$. Prove that the function $f(x) + h(x)$ has its absolute minimum
when $x=g(1)$.
\bigskip
Solution.
\medskip
\item{(a)} Since $f$ and $g$ are inverse functions, $f(g(x)) = x$. (The equation $g(f(x)) = x$ will work as well.)
On differentiating we get
$$f' (g(x)) \cdot g'(x) = 1 .$$
Since $f(2) = 10$, we have $g(10) = 2$. Substituting $x=10$ in the previous equation,
$$f' (g(10)) \cdot g' (10) = 1,\ \ \text{or}\ \ f'(2) \cdot g' (10) = 1 .$$
But $f' (x) = 3x^2 + 1$, so $f'(2) = 13$. Hence, $g' (10) = 1/13$.
\smallskip
\item{(b)}
\comment
$$f(x) + h(x) = f(x) + 1/f(x) .$$
We know (using the AM-GM argument) that the minimum value of $a +
1/a$ is 2, and it happens when $a=1$. Thus, $f(x) + h(x)$ has a
minimum value when $f(x) = 1$, or $x=g(1)$.
\endcomment
Let $F(x) = f(x) + h(x) = f(x) + 1/f(x)$. Then
$$F' (x) = f'(x) - {f' (x) \over \lbrack f(x) \rbrack ^2} = 0, \ \ \text{so}\ \
f'(x) \{ \lbrack f(x) \rbrack ^2 - 1 \} = 0.$$ As $f'(x) > 1$, then
division by $f'(x)$ is permitted, and $f(x_0)=1$ (since $f(x)>0$) at
the critical point $x_0$. Thus, $g(f(x_0)) = g(1)$. To determine
the nature of the critical point, we compute
$$\eqalign{
F''(x) &= f''(x) - {\lbrack f(x) \rbrack ^2 f''(x) - 2 f(x) \lbrack
f'(x) \rbrack ^2 \over \lbrack f(x) \rbrack ^2} \cr &= 6g(1) - {1^2
\lbrack 6g(1) \rbrack - 2 (1) \lbrack 3(g(1))^2 + 1 \rbrack ^2 \over
1^2} \cr &= 2 \lbrack 3(g(1))^2 + 1 \rbrack ^2 \ \ \text{at $x=x_0$}
\cr &>0, \ \ \text{necessarily}. \cr}$$
Hence, $x=x_0$ is the location of a relative minimum. It is the
location of the absolute minimum because at the endpoints
$$\lim_{x \to 0} F(x) = \lim_{x \to \infty} F(x) = \infty .$$
Note: We do not need to compute $x_0$ explicitly; for completeness,
it is roughly $x_0 = 0.682328$.
\bigskip
2. Consider the lattice $L = \{ (x,y) \ :\ x,y \in \Z \}$. Color the lattice using an arbitrary coloring
scheme with 2008 available colors. Prove or disprove: In every coloring scheme, there must be a rectangle
whose four vertices all lie in $L$ and are colored with the same color.
\bigskip
Solution.
\medskip
Such a rectangle must exist. First, note that, by the Pigeonhole Principle, among any set of 2009 points from $L$,
at least two must have the same color. In particular, for a fixed integer $x$, at least two of the points $(x,0)$,
$(x,1)$, $(x,2)$ $\ldots$, $(x, 2008)$, must have the same color. For these 2009 points, there are $2008^{2009}$
different possible colorings. Therefore, by the Pigeonhole Principle (again), as $x$ runs from 0 to $2008^{2009}$,
there must be at least two values of $x$ (say $x_1$ and $x_2$) for which the coloring for
$$(x_1,0),\ (x_1, 1),\ (x_1,2),\ \ldots ,\ (x_1,2008)\ \ \text{and}\ \ (x_2,0),\ (x_2,1),\ (x_2,2),\ \ldots ,\
(x_2, 2008)$$
are the same. (By the same, we mean that the corresponding points $(x_1,y)$ and $(x_2,y)$ are the same color for
each $y=0, 1, 2, \ldots , 2008$.) And, as noted before, among the points
$$(x_1,0),\ (x_1,1),\ (x_1,2),\ \ldots ,\ (x_1,2008) ,$$
there must be two $y$-values (say $y_1$ and $y_2$) for which $(x_1,y_1)$ and $(x_1,y_2)$ have the same color.
Thus, the points $(x_1,y_1)$, $(x_1,y_2)$, $(x_2,y_1)$, and $(x_2,y_2)$ have the same color, and are the vertices
for some rectangle.
\medskip
Exercise 1.6 (p. 17) from {\it Ramsey Theory on the Integers,} by Bruce M. Landman and Aaron Robertson,
AMS publication.
\bigskip
3. Find $b>1$ such that the graphs of $y = \log _b (x)$ and $y=b^x$ intersect in exactly one point, i.e., are
tangent to one another.
\bigskip
Solution.
\medskip
Because $y = \log _b (x)$ and $y=b^x$ are inverse functions, any intersections must occur on the line $y=x$,
and for the case when the graphs are tangent, the common tangent line must be $y=x$ as well. At the point
of tangency, say $x_0$, then, the derivative of each function must be one, giving:
$$\eqalign{
(b^x)' _{x=x_0} &= b^{x_0} \ln b = 1 \cr (\log _b x)' _{x=x_0} &= {1
\over x_0 \ln b} = 1 .\cr}$$ From the second equation we get $x_0 =
1/(\ln b)$, and substituting this into the first equation gives
$${1 \over \ln b} = \log _b \biggl( {1 \over \ln b} \biggr) = {\ln \bigl( {1 \over \ln b} \bigr) \over \ln b}$$
which leads to
$$\ln \biggl( {1 \over \ln b} \biggr) = 1 ,$$
which then gives
$$b = e^{1 \over e} = \exp \biggl( {1 \over e} \biggr) .$$
(Substituting back to get $x_0$ yields $x_0 = e$.)
\bigskip
4. Suppose that
$${2x+3 \over x^2 - 2x + 2}$$
has the Taylor series
$$\sum_{k=0}^\infty a_k x^k .$$
Find the sum of the odd numbered coefficients, i.e., find
$$\sum_{k=0}^\infty a_{2k+1} = a_1 + a_3 + a_5 + \cdots .$$
\bigskip
Solution.
\medskip
Let
$$F(x) = {2x+3 \over x^2 - 2x + 2} .$$
If the Taylor series for $F$ is to represent $F$, then the series
must converge inside the circle in the complex plane that is
centered at the origin and extends out to the nearest point at which
the denominator of $F$ vanishes. This occurs when $x = 1 \pm i$, so
the radius $R$ of the circle of convergence is
$$R = \{ (1 \pm i)(1 \mp i) \}^{1/2} = \sqrt 2.$$
Therefore,
$${2x+3 \over x^2 - 2x + 2} = \sum_{k=0}^\infty a_k x^k $$
for all which satisfy $\vert x \vert < \sqrt 2$. This gives us
$$\sum_{k=0}^\infty a_k = \sum_{k=0}^\infty a_k (1)^k = F(1)\ \ \text{and}\ \ \sum_{k=0}^\infty (-1)^k a_k
= \sum_{k=0}^\infty a_k (-1)^k = F(-1) .$$
Thus,
$$\eqalign{
&\sum_{k=0}^\infty a_{2k+1} = {1 \over 2} \biggl( (a_0 + a_1 + a_2 + a_3 + \cdots ) - (a_0 - a_1 + a_2 - a_3
+ \cdots ) \biggr) \cr
&= {1 \over 2} \biggl( \sum_{k=0}^\infty a_k - \sum_{k=0}^\infty (-1)^k a_k \biggr) = {1 \over 2} \biggl(
F(1) - F(-1) \biggr) \cr
&= {5 - {1 \over 5} \over 2} = {12 \over 5} .\cr}$$
\bigskip
5. For each integer $n$, let $a_n = 8n^2 + 3n + 10$ and $b_n = 3n^2 + n + 3$. Since $a_1 = 21$ and
$b_1 = 7$, we can write $\gcd (a_1, b_1) = 7$, where $\gcd$ denotes the greatest common divisor.
Find $\max _{n \in \Z} \gcd (a_n, b_n)$.
\bigskip
Solution.
\medskip
Suppose that $d$ is a common divisor of $a_n$ and $b_n$. Then $d$ must also divide
$$3a_n - 8b_n = n+6 .$$
Dividing $a_n$ and $b_n$ by $n+6$, we get
$$8n^2 + 3n + 10 = (n+6)(8n-45) + 280\ \ \text{and}\ \ 3n^2 + n + 3 = (n+6)(3n-17) + 105 .$$
Therefore, $d$ must divide both 280 and 105. Since the greatest common divisor of 280 and 105 is 35,
this is our (potential) answer. However, we need to check that this $\gcd$ is actually attained.
To do this, we simply use $n=-6$:
$$a_{-6} = 8(-6)^2 + 3(-6) + 10 = 280\ \ \text{and}\ \ b_{-6} = 3(-6)^2 + (-6) + 3 = 105 .$$
\bye