\input amstex
%\input epsfx
\input graphicx
\hyphenation{Mass-achusetts Central Missouri Wis-con-sin Muthuvel}
\raggedbottom
\loadmsam
\loadmsbm
\loadeufm
\def\R{\Bbb R}
\def\Z{\Bbb Z}
\def\Q{\Bbb Q}
\def\C{\Bbb C}
\def\N{\Bbb N}
\tolerance=1600
\comment
\hsize=33pc
\vsize=42pc
\endcomment
\hsize=6.5in
\vsize=9in
\baselineskip=12pt
\lineskip=12pt
\centerline{\bf 2010 Missouri Collegiate Mathematics Competition}
\vskip 6pt
\centerline{\bf Session I}
\bigskip
1.  For the parabola having equation $y = -x^2$ let $a<0$ and $b>0$ with points $P : (a,-a^2)$
and $Q : (b,-b^2)$.  Let $M$ be the midpoint of $PQ$ and let $R$ be the intersection of
the vertical line through $M$ with the parabola.  Show that the area of the region
bounded by the parabola and the line segment $PQ$ is ${4 \over 3}$ of the area of triangle
$PQR$. (Archimedes, 3rd century B.C.)
\bigskip
Solution. \medskip The equation of the line containing $P$ and $Q$ is
$$y = -(b+a) x + ab .$$
Thus, the area of the segment of the parabola is
$$\text{Area} (\text{seg} PQR) = \int_a^b (-x^2 + (b-a) x - ab) dx = {(b-a)^2 \over 6} .$$
Triangle $PQR$ is divided into two triangles by line segment $MR$ and
$$| MR | = - \left( {a+b \over 2} \right)^2 + {a^2 + b^2 \over 2} = \left( {b-a \over 2} \right) ^2 .$$
For each of these triangles, the altitude to $MR$ is $(b-a)/2$, so they have equal areas, giving
$$\text{Area}( \triangle PQR ) = \left( {b-a \over 2} \right) \left( {b-a \over 2} \right) ^2 =
{(b-a)^3 \over 8} .$$
\bigskip
2. Suppose that the length of the larger base of an isosceles trapezoid equals the length of a
diagonal and
the length of the smaller base equals the altitude.  Find the ratio of the length of the
larger base to the length of the smaller base.
\bigskip
Solution.
\medskip
Without loss of generality, consider the figure below where $r =
{\overline{CD} \over \overline{AB}} >1$ is the ratio of the length of the larger base to the length
of the smaller base and $\overline{AB} = 1$ unit.
$$
\includegraphics[width=2.25in]{im2.eps}
$$
Then $\overline{OD} = {1 \over 2} (r-1)$, and
in $\triangle BOC$ we have $(r - {1 \over 2} (r-1))^2 + 1 = r^2$, or $3r^2 - 2r
- 5 = (3r-5)(r+1) = 0$.  Hence $r = {5 \over 3}$.
\comment
Let $h$ denote the length of the smaller base and $x$ the larger base. If we consider a right
triangle formed by a diagonal, an altitude dropped from the endpoint of the
diagonal at the shorter base, and the appropriate part of the longer base, we
obtain the equation $x^2 = h^2 + ((x+h)/2)^2$. Rewriting, we find that $3x^2 -
2hx - 5h^2 = 0$. Considering $x$ as the variable and $h$ as the constant, we
obtain $x = (2h + \sqrt {4h^2 + 60h^2})/6$ which yields $x = (5/3)h$.
\endcomment
\medskip
$\underline{\text{Remark}}$.  This problem comes from the 1953 AHSME.
\bigskip
3. Consider the Diophantine equation $x(2x^2 + 3x + 3) = y^3 -1$.  Prove that
it does not have a solution $(x,y)$ in positive integers, or find such a
solution if it does.
\medskip
Solution.
\medskip
The equation is equivalent to
$x^3 + (x+1)^3 = y^3$.  By Fermat's Last Theorem, the equation $x^n + y^n =
z^n$ has no solutions in positive integers $x$, $y$, $z$ if the integer $n$ is larger than $2$.
Hence, the
given Diophantine equation has no solution.
\medskip
4. The Fibonacci and Lucas
numbers are sequences defined by the following initial conditions and second
order recurrence relations.  Let $F_0 = 0$, $F_1 = 1$, and for $n \ge 2$ define
$F_n = F_{n-1} + F_{n-2}$; let $L_0 = 2$, $L_1 = 1$, and for $n \ge 2$ define
$L_n = L_{n-1} + L_{n-2}$.

Let $\alpha = (1 + \sqrt 5)/2$ and $\beta = (1-\sqrt 5)/2$.  The Binet form of the
Fibonacci and Lucas numbers are given by
$$F_n = {\alpha^n - \beta^n \over \sqrt 5} \ \ \text{and}\ \ L_n = \alpha^n + \beta^n ,$$
where $n$ is any nonnegative integer.

Let $k$ be a fixed positive integer and define
$$U_n = {F_{kn} \over L_k} ,$$
where $n$ is any nonnegative integer.  Find $U_0$ and $U_1$ and find a second
order recurrence relation involving $L_k$ satisfied by the sequence $\{ U_n \}_{n=0}^\infty$.
\medskip
Solution.
$$U_0 = {F_0 \over L_k} = 0 \ \ \text{and}\ \ U_1 = {F_k \over L_k} .$$
We shall express $U_{n+1}$ in terms of one or more $U_n$'s of lower
order.  We have the identity
$$\eqalign{
\alpha ^{k(n+1)} - \beta ^{k(n+1)} &= (\alpha ^k + \beta ^k) (\alpha^{kn} -
\beta^{kn}) - (\alpha^{kn} \beta^k - \alpha^k \beta^{kn}) \cr &= (\alpha^k +
\beta^k) (\alpha^{kn} - \beta^{kn}) - (\alpha \beta)^k (\alpha^{k(n-1)} -
\beta^{k(n-1)}) \cr}$$
Multiplication of both sides by ${1 \over \sqrt 5 (\alpha^k
+ \beta^k)}$ and use of the definition of the $U_n$'s, and of the fact that
$\alpha \beta = -1$, gives
$$U_{n+1} = L_k U_n - (-1)^k U_{n-1} .$$
\comment
Using the Binet forms for the Fibonacci and Lucas numbers and doing
some algebra, we have for $n \ge 1$
$$\eqalign{
U_{n+1} &= \frac{F_{k(n+1)}}{L_k} = \frac{\alpha^{k(n+1)} - \beta^{k(n+1)}}{\sqrt 5 (\alpha^k +
\beta^k)} \cr
&= \frac{1}{\sqrt 5} \left( \frac{\alpha^{kn} \alpha^k - \beta^{kn} \beta^k}{\alpha^k + \beta^k}
\right) \cr
&= \frac{1}{\sqrt 5} \left( \frac{\alpha^{kn} \alpha^k - \beta^{kn} \alpha^k + \beta^{kn}
\alpha^k - \beta^{kn} \beta^k}{\alpha^k + \beta^k} \right) \cr
&= \frac{1}{\sqrt 5} \left( \frac{\alpha^{kn} \alpha^k - \beta^{kn} \alpha^k + \alpha^{kn}
\beta^k - \beta^{kn} \beta^k - \alpha^{kn} \beta^k + \beta^{kn}
\alpha^k}{\alpha^k + \beta^k} \right) \cr
&= \alpha^k \cdot U_n + \beta^k \cdot U_n + \frac{1}{\sqrt 5} \left( \frac{\alpha^k
\beta^{kn} - \beta^k \alpha^{kn}}{\alpha^k + \beta^k} \right) \cr
&= (\alpha^k + \beta^k) U_n - (\alpha \beta)^k \cdot \frac{1}{\sqrt 5} \cdot
\left( \frac{\alpha^{(k-1)n} - \beta^{(k-1)n}}{\alpha^k + \beta^k} \right) \cr
&= L_k U_n - (-1)^k U_{n-1} .\cr}$$
\endcomment
\bigskip
5. If your calculator is set to radian mode, and you enter any number and then repeatedly push the
``cosine'' button, the displayed value will converge to $0.73908 \ldots$.
Call this number $d$.  If $f(x) = x - \cos x$, then $f(d) = 0$.  The number $d$ can be
expressed as a series in odd powers of $\pi$:
$$d = \sum_{n=0}^\infty a_n \pi^{2n+1} .$$
Find $a_0$ and $a_1$.
\vfill\eject
Solution.
\medskip
The function $f$ is
increasing on the interval $(-{\pi \over 2}, {3\pi \over 2})$, so has an
inverse on this interval.  Call the inverse $g$.  Then $g(0) = d$ and $g({\pi
\over 2}) = {\pi \over 2}$. We will develop the Taylor series expansion of
$g(x)$ around ${\pi \over 2}$.
$$g'(x) = {1 \over f'(g(x))} = {1 \over 1 + \sin (g(x)) } .$$
So, $g'({\pi \over 2}) = {1 \over 2}$.
$$g''(x) = -(1 + \sin (g(x)))^{-2} \cos (g(x)) \left( {1 \over 1+\sin(g(x))} \right) = {-\cos (g(x))
\over (1+\sin(g(x)))^3} .$$
So, $g''({\pi \over 2}) = 0$.

\noindent One more differentiation yields
$$g'''(x) = {\sin (g(x)) \over (1+\sin(g(x)))^4} + \cos (g(x)) (\text{bounded terms}) .$$
So, $g'''({\pi \over 2}) = {1 \over 16}$, and we have
$$g(x) = {\pi \over 2} + {1 \over 2} \left( x - {\pi \over 2} \right) + {1 \over 16}
{(x - {\pi \over 2})^3
\over 3!} + \cdots ,$$ making
$$d = g(0) = {\pi \over 2} + {1 \over 2} \left( 0 - {\pi \over 2} \right) + {1 \over 16}
{(0 - {\pi \over 2})^3 \over 3!} + \cdots = {\pi \over 4} - {\pi^3 \over 768} + \cdots ,$$
so $a_0 = {1 \over 4}$ and $a_1 = -{1 \over 768}$.
\bigskip
$\underline{\text{Remark}}$.  See ``The Dottie Number,'' {\it Math. Mag.}, 80 (2007), 73--74.
\vfill\eject
\centerline{\bf 2010 Missouri Collegiate Mathematics Competition}
\vskip 6pt
\centerline{\bf Session II}
\bigskip
1. Let $S$ be the collection of ordered pairs $(x,y)$ in $[0,1] \times [0,1]$ such that
either $x$ or $y$ is irrational. Prove or disprove that for any two distinct ordered pairs in
$S$, we can find a path in $S$ connecting the two points.
\bigskip
Solution.
\medskip
We can always find a path in $S$ between any two distinct ordered pairs in $S$.  To prove
this, let $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$ be two distinct points in $S$.  We need to
consider 4 cases.

$\underline{\text{Case 1}}$. If $x_1$ and $y_2$ are irrational, then draw the vertical
line $x=x_1$ and the horizontal line $y=y_2$.  These lines intersect at the point
$P_3 = (x_1, y_2)$.  Start at $P_1$ and follow the vertical line to $P_3$ and then follow
the horizontal line to $P_2$.  This is a path in $S$ that connects $P_1$ and $P_2$.

$\underline{\text{Case 2}}$.  If $x_2$ and $y_1$ are irrational, then draw the horizontal line
$y=y_1$ and the vertical line $x=x_2$. These lines intersect at the point $P_3 = (x_2, y_1)$.
Start at $P_1$ and follow the horizontal line to $P_3$ and then follow the vertical line
to $P_2$. This is a path in $S$ that connects $P_1$ and $P_2$.

$\underline{\text{Case 3}}$. If $x_1$ and $x_2$ are irrational, then draw the two vertical
lines $x=x_1$ and and $x=x_2$. Also, draw the horizontal line $y=\sqrt 2/2$.
Let $P_3 = (x_1,\sqrt 2/2)$ and $P_4 = (x_2,\sqrt 2/2)$. Start at $P_1$ and go to $P_3$
along the vertical line $x=x_1$.  Then go from $P_3$ to $P_4$ along the horizontal line.
Finally, go from $P_4$ to $P_2$ along the vertical line $x=x_2$.  This is a path in $S$ that
connects $P_1$ and $P_2$.

$\underline{\text{Case 4}}$.  If $y_1$ and $y_2$ are irrational, then draw the two horizontal
lines $y=y_1$ and and $y=y_2$. Also, draw the vertical line $x=\sqrt 2/2$.  Let $P_3 =
(\sqrt 2/2,y_1)$ and $P_4 = (\sqrt 2/2,y_2)$. Start at $P_1$ and go to $P_3$ along the
horizontal line $y=y_1$.  Then go from $P_3$ to $P_4$ along the vertical line.  Finally,
go from $P_4$ to $P_2$ along the horizontal line $y=y_2$.  This is a path in $S$ that
connects $P_1$ and $P_2$.

This completes the proof.
\bigskip
2. If $a, b, c$ are positive real numbers, find the value of $x$ that minimizes the function
$$f(x) = \sqrt {a^2 + x^2} + \sqrt {(b-x)^2 + c^2} .$$
(Hint: Think geometrically.)
\bigskip
Solution. \medskip The simplest solution is to use geometry.  Consider the
figure below where $AB = a$, $BC = b$, $CD = c$, and $BP = x$.
$$
\includegraphics[width=3.0in]{ma1.eps}
$$
We note that $f(x) = AP + PD$, which is a minimum when $P$ lies at
the intersection of lines $BC$ and $AD$.  Then
$${BP \over PC} = {x \over b-x} = {a \over c} .$$
Hence,
$$x = {ab \over a+c} .$$
\bigskip
3. A sequence of $2 \times 2$ matrices, $\{ M_n \}_{n=1}^\infty$, is defined as follows:
$$M_n = \pmatrix
m_{11} = {1 \over (2n+1)!} & m_{12} = {1 \over (2n+2)!} \\
m_{21} = \sum_{k=0}^n {(2n+2)! \over (2k+2)!} & m_{22} = \sum_{k=0}^n {(2n+1)! \over (2k+1)!}
\endpmatrix .$$
For each $n$, let $\det M_n$ denote the determinant of $M_n$.  Determine the value of
$$\lim_{n \to \infty} \det M_n .$$
\bigskip Solution.
\medskip
$$\eqalign{
&\det M_n = m_{11} m_{22} - m_{12} m_{21} = \sum_{k=0}^n {1 \over (2k+1)!} -
\sum_{k=0}^n {1 \over (2k+2)!} \cr
&= \sum_{k=1}^{2n+2} (-1)^{k+1} {1 \over k!} = \sum_{k=0}^{2n+2} (-1)^{k+1}
{1 \over k!} - (-1) = 1 - \sum_{k=0}^{2n+2} (-1)^k {1 \over k!} .\cr}$$
Hence,
$$\lim_{n \to \infty} \det M_n = 1 - \lim_{n \to \infty} \sum_{k=0}^{2n+2} (-1)^k {1
\over k!} = 1 - e^{-1} .$$
\bigskip
4. Evaluate the integral
$$I = \int_{1 \over 2}^2 {\ln x \over 1+x^2} dx .$$
\vfill\eject Solution. \medskip
$$\eqalign{
I = \int_{1 \over 2}^1 {\ln x \over 1+x^2} dx + \int_1^2 {\ln x \over 1+x^2} dx
&= \int_{1 \over 2}^1 {\ln x \over 1+x^2} dx + \int_1^{1 \over 2} {-\ln u \over
1+u^{-2}} \left( {-1 \over u^2} du \right) \cr &= \int_{1 \over 2}^1 {\ln x
\over 1+x^2} dx - \int_{1 \over 2}^1 {\ln u \over 1+u^2} du = 0.\cr}$$ \comment
The key to this problem is to discover that the integral has hidden symmetry;
observe that the limits are mutual reciprocals. Let $x=u^{-1}$, $dx = -u^{-2}
du$. Then
$$I = \int_2^{1 \over 2} {(-\ln u) \over 1+u^{-2}} (-u^{-2} du) = -\int_{1 \over 2}^2
{\ln u \over 1+u^2} du = -I .$$
Hence, $I=0$.
\endcomment
\bigskip
5. A function $f$ has the following properties:
\medskip
\item{(a)} For all $x \ge 1$, $f(x)$ is a positive, differentiable, decreasing function;
\item{(b)} Whenever $x$ equals a natural number $k$, we set $f(k) = f_k$, an element of a numerical
sequence;
\item{(c)} The series $\sum_{k=1}^\infty f_k$ diverges to $\infty$;
\item{(d)} $F$ is an arbitrary antiderivative of $f$, but with a fixed constant of integration,
and is defined for all $x \ge 1$;
\item{(e)} $\lim_{x \to \infty} F(x) = \infty$.
\medskip
Prove that
$$\lim_{n \to \infty} \left( \sum_{k=1}^n f_k - F(n) \right)$$
is finite.
\bigskip
Solution.
\medskip
By the Mean Value Theorem, for each natural number $k$ there is a
constant $M_k$ such
that $k < M_k < k+1$ and
$$F(k+1) - F(k) = ((k+1)-k) F' (M_k) = f(M_k) .$$
The MVT applies here because $f$ differentiable on $[ 1, \infty )$ implies that
$f$ is integrable on $[ 1, x ]$ for all real $x>1$, which implies that $F$ is
continuous on $[ 1,x ]$ by the Fundamental Theorem of Calculus.
%$$\epsfxsize=30pc \epsfbox{integraltest.eps}$$
$$
\includegraphics[width=5.0in]{integraltest.eps}
$$
Since $f$ decreases on $[ k,k+1 ]$, then $f_k \ge F(k+1) - F(k) \ge f_{k+1}$.
Then
$$\sum_{k=1}^{n+1} f_k - F(n+1) - \left( \sum_{k=1}^n f_k - F(n) \right) = f_{n+1} -
( F(n+1) - F(n) ) \le 0$$
from the preceding line.  Since $n$ is arbitrary, this shows that the sequence
$\{ c_n \}_{n=1}^\infty$, where
$$c_n = \sum_{k=1}^n f_k - F(n) $$
is a decreasing sequence.

Finally, since $f$ is Riemann integrable over any interval $[1, n]$ and since f is positive on this
interval, then from the proof of the Integral Test, one has (by definition of $c_n$)
$$\int_1^n f(x) dx = \sum_{k=1}^{n-1} f_k = c_n + F(n) - f_n ,$$
or
$$F(n) - F(1) \le c_n + F(n) - f_n .$$
Hence,
$$c_n \ge f_n - F(1) > - F(1) ,$$
and the terms of $\{ c_n \}_{n=1}^\infty$ are bounded below by
$-F(1)$.  It follows from the Bounded Monotone Sequence Theorem that
$\lim_{n \to \infty} c_n$ is finite.
\bigskip
 \bye
