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\hyphenation{Mass-achusetts Central Missouri Wis-con-sin Muthuvel}
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\centerline{\bf 2005 Missouri Collegiate Mathematics Competition}
\vskip 6pt
\centerline{\bf Session I}
\bigskip
1. Find the point in the first quadrant on the graph of $y = 7 - x^2$ such that the distance between the $x$- and $y$- intercepts of the tangent line at the point is minimum.
\bigskip
Solution.
\medskip
Let $(a, 7-a^2)$, $a>0$, be a point in the first quadrant on the graph of the function $y = 7 - x^2$.  The slope of the tangent line at this point is $-2a$, and the equation of the tangent line is $y - (7-a^2) = -2a (x-a)$.  The $x$-intercept of this line is
$((a^2+7)/(2a),0)$, and the $y$-intercept is $(0, a^2+7)$.  The distance squared between these two points is
$$s^2 = {(a^2 + 7)^2 \over 4a^2} + (a^2 + 7)^2 = {(a^2 + 7)^2 (4a^2 + 1) \over 4a^2} .$$
The derivative of $s^2$ with respect to $a$ is
$${(a^2 + 7) (8a^4 + a^2 - 7) \over 2a^3} ,$$
and this factors as
$${(a^2 + 7) (8a^2 - 7) (a^2 + 1) \over 2a^3} .$$
The critical point in the first quadrant occurs at
$$a = \sqrt {7 \over 8} = {\sqrt {14} \over 4} .$$
Checking the first or second derivative will show that this critical value produces a minimum value for $s^2$, so the point that makes the distance between the $x$- and $y$- intercepts of the tangent line minimum is
$$\biggl( {\sqrt {14} \over 4} , {49 \over 8} \biggr) .$$
The function $y = 9 - x^2$ would also work, and then the minimum occurs at $(1,8)$.
\bigskip
2.  Let $M(n,k)$ be the number of mappings from a set $X$ of $n$ distinct objects onto a set $Y$ of $k$ distinct objects, and let $P(n,k)$ denote the number of partitions of a set of $n$ distinct objects into $k$ nonempty subsets.  Determine the relationship between $M(n,k)$ and $P(n,k)$ and use it to show that $M(n,k)$ is a multiple of $24$ whenever $k>3$.
\bigskip
Solution.  
\medskip
Consider any particular partition of the elements of $X$ into $k$ nonempty subsets.  Number these subsets from $1$ to $k$.  Similarly, number the elements in $Y$ from $1$ to $k$, but keep this numbering fixed.  Any numbering of the sets of $X$ then defines one surjection from $X$ to $Y$.  There are $k!$ possible numberings of $X$, so this leads to $k!$ surjections for a given partition.  But there are $P(n,k)$ possible partitions, so we have
$$M(n,k) = k! P(n,k) .$$
For any $k>3$ one has that $4!$ divides $k!$, and thus $24$ divides $M(n,k)$.
\medskip
3.  Suppose that $f$ is a polynomial of positive degree $n$ with integer coefficients.  Prove that there are infinitely many integers $x$ for which $f(x)$ is composite.  (Here, composite means those integers, positive or negative, whose absolute value is not $1$ or a prime; thus, $-4$ and $6$ are composite, while $1$ and $-2$ are not.)
\bigskip
Solution I.
\medskip
Let $b = f(0)$ be the constant term of the polynomial.  If $b=0$, then $f(x)$ is even for all even integers $x$.  The only even integers which are not composite are $-2$, $0$, and $2$.  But, since $f$ has degree $n$, it can only take on any of those values at most $n$ times, and therefore, there must be infinitely many even $x$ for which $f(x)$ is even and composite.  Similarly, if $b$ is divisible by any prime $p$, then $f(x)$ will be divisible by $p$ for any $x$ which is a multiple of $p$.  And since $f$ can only take on the values $-p$, $0$, or $p$ at most $n$ times each, there will be infinitely many $x$ for which $f(x)$ is a composite multiple of $p$.  This leaves the cases of $b = \pm 1$.  Note that, for any fixed integer $t$, if we think of $f(x+t)$ as a polynomial in $x$, then as $x$ runs through the integers, it will take on all the same values as $f(x)$ does.  So $f(x+t)$ satisfies the desired property if and only if $f(x)$ does.  To complete the proof, we need only show that there exists some $t$ for which $f(x+t)$ has a constant term other than $\pm 1$ (allowing us to use one of the above arguments to conclude that $f(x+t)$, and hence $f(x)$, has the desired property).  However, the constant term in the polynomial $f(x+t)$ (again thought of as a polynomial in $x$) is $f(t)$.  And since $f$ has positive degree $n$, $f(t)$ can only take on the values $1$ or $-1$ at most $n$ times each, so there must be a positive integer $t$ for which $f(t)$ is not equal to $\pm 1$.
\bigskip
Solution II.
\medskip
Suppose, to the contrary, that there are only finitely many $x$ ($x_1, x_2, \ldots , x_M$), such that $f(x)$ is composite.  Then for $\underline{all}$ $x > x_M$, $f(x)$ is prime:
$$f(x) = c_0 + c_1 x + c_2 x^2 + \cdots + c_N x^N = p,\ \ c_k \in \N \ \ \text{for all  } k .$$
For $x$ sufficiently large, the polynomial is dominated by the leading term, so in this region the polynomial becomes monotonic.  Let $X$ be in this region and exceed $x_M$, and set $F(X) = P$ (a prime).  Also, $(P+1)X$ is in this region; set $f((P+1)X) = Q$.  By earlier remarks $\vert Q \vert > \vert P \vert$.  Then
$$\vert f((P+1)X) - f(X) \vert = \bigg\vert \sum_{k=1}^N c_k ((P+1)^k - 1) X^k \bigg\vert = \vert Q - P \vert .$$
But $P$ divides each term in the summation, so $P$ must divide $Q$ also, a contradiction, since $Q$ is prime.  Hence, the initial supposition is false.
\bigskip
4.  Determine the value of the integral
$$I( \theta ) = \int_{-1}^1 {\sin \theta \, dx \over 1 - 2x \cos \theta + x^2} ,$$
and locate those points $0 \le \theta \le 2\pi$, where $I( \theta )$ is discontinuous.
\vfill\eject
Solution.
\medskip
Replace $1 = \cos ^2 \theta + \sin ^2 \theta$ and let $u = x - \cos \theta$, $du = dx$.  Then the integral becomes 
$$\eqalign{
I( \theta ) &= \int_{-1-\cos \theta}^{1-\cos \theta} {\sin \theta \, du \over u^2 + \sin ^2 \theta} = \text{Tan} ^{-1} \biggl( {u \over \sin \theta} \biggr) \bigg\vert _{-1-\cos \theta}^{1 - \cos \theta} \cr
&= \text{Tan} ^{-1} \biggl( {1 - \cos \theta \over \sin \theta} \biggr) - \text{Tan} ^{-1} \biggl( {-1 - \cos \theta \over \sin \theta} \biggr) . \ \ \ \ \ (\sin \theta \ne 0) \cr
&= \text{Tan} ^{-1} \biggl( {2 \sin ^2 (\theta / 2) \over 2 \sin (\theta / 2) \cos (\theta / 2)} \biggr) - \text{Tan} ^{-1} \biggl( { -2 \cos ^2 (\theta / 2) \over 2 \sin (\theta / 2) \cos (\theta / 2) } \biggr) \cr
&= \text{Tan} ^{-1} ( \tan (\theta / 2) ) - \text{Tan} ^{-1} (-\cot (\theta / 2) ) \cr
&= \text{Tan} ^{-1} (\tan (\theta / 2) ) - \text{Tan} ^ {-1} (\tan (\pi / 2 + \theta / 2) ) .\cr}$$

Here, we recall that the range of $\text{Tan} ^{-1} \theta$ is $(-\pi / 2 , \pi / 2)$.  As an example, $\text{Tan} ^{-1} (\tan (20^\circ)) = 20^\circ$, $\text{Tan} ^{-1} (\tan (100^\circ )) = -80^\circ$, $\text{Tan} ^{-1} ( \tan (220^\circ ) ) = 40^\circ$, and $\text{Tan} ^{-1} (\tan (290^\circ ) ) = -70^\circ $.

Thus, if $0 < \theta < \pi$, then $I(\theta ) = \theta / 2 - ( -(\pi / 2 - \theta / 2) ) = \pi / 2$.

But, if $\pi < \theta < 2\pi$, then $I(\theta ) = -(\pi - \theta / 2) - (\theta / 2 - \pi / 2 ) = -\pi / 2$.

\comment
But
$$\biggl( {1 - \cos \theta \over \sin \theta } \biggr) ^{-1} = {\sin \theta \over 1 - \cos \theta} \cdot {1 + \cos \theta \over 1 + \cos \theta} = - {-1 - \cos \theta \over \sin \theta} ,\ \ \text{when } \sin \theta \ne 0 ,$$
and $\text{Tan} ^{-1} y = - \text{Tan} ^{-1} (-y)$ for any real $y$.  Therefore,
$$I( \theta ) = \text{Tan} ^{-1} \biggl( {1 - \cos \theta \over \sin \theta} \biggr) + \text{Tan} ^{-1} \biggl( {\sin \theta \over 1 - \cos \theta} \biggr) ,$$
provided $\theta \ne \pi$ (which is where $\sin \theta = 0$).  When $\theta = \pi$, inspection of the original integral shows that $I( \theta ) = 0$.

For other values of $\theta$, the value of $I( \theta )$ depends entirely on the sign of $\sin \theta$.
\medskip
$\underline{\text{Case 1}}$. $0 < \theta < \pi$.  Then $\sin \theta > 0$ and
$$I( \theta ) = \text{Tan} ^{-1} x + \text{Tan} ^{-1} \biggl( {1 \over x} \biggr) = {\pi \over 2} . \ \ \ \ \ (x > 0)$$
\medskip
$\underline{\text{Case 2}}$.  $\pi < \theta < 2 \pi$.  Then $\sin \theta < 0$ and
$$I( \theta ) = \text{Tan} ^{-1} x + \text{Tan} ^{-1} \biggl( {1 \over x} \biggr) = -{\pi \over 2} . \ \ \ \ \ (x < 0) $$
\endcomment
Therefore, the function $I( \theta )$ is
$$I( \theta ) = \cases
0,&\text{for $\theta = 0$}\\
{\pi \over 2},&\text{for $0 < \theta < \pi$}\\
0,&\text{for $\theta = \pi$}\\
- {\pi \over 2},&\text{for $\pi < \theta < 2\pi$}\\
0,&\text{for $\theta = 2\pi$}.
\endcases$$
The points of discontinuity occur at $0$, $\pi$, and $2\pi$.
\vfill\eject
5.  Prove that in the MacLaurin series for $\tan \theta$, $-\pi / 2 < \theta < \pi /2$, every coefficient is non-negative.
\bigskip
Solution.
\medskip
Let $D_n$ denote the $n$th derivative of $\tan \theta$.  We shall prove that
$$\cases
D_{2k} = \tan \theta P_k ( \tan ^2 \theta ) &(1a)\\
D_{2k+1} = \sec ^2 \theta Q_k ( \tan ^2 \theta ) ,&(1b)
\endcases $$
where $P_k ( \tan ^2 \theta )$, $Q_k ( \tan ^2 \theta )$ are polynomials of degree $k$ in $\tan ^2 \theta$ with no terms missing and all coefficients positive.  Equations (1) are true for $k=1$; assume they are true for $k=K$.  Then
$$D_{2K+2} = {d \over d \theta} D_{2K+1} = \sec ^4 \theta Q_K ' ( \tan ^2 \theta ) + 2 \sec ^2 \theta \tan \theta Q_K ( \tan ^2 \theta ) .$$
Polynomial $Q_K ' ( \tan ^2 \theta )$ is of degree $2K - 1$ in $\tan \theta$, all terms are odd powers, no odd powers are missing, and all coefficients are positive (by hypothesis).  Write $Q_K ' ( \tan ^2 \theta ) = \tan \theta f_{K-1} ( \tan ^2 \theta )$, and therefore
$$\eqalignno{
&D_{2K+2} = \tan \theta \biggl( \sec ^4 \theta f_{K-1} ( \tan ^2 \theta ) + 2 \sec ^2 \theta Q_K ( \tan ^2 \theta ) \biggr) \cr
&= \tan \theta \biggl( (1 + \tan ^2 \theta ) ^2 f_{K-1} ( \tan ^2 \theta ) + 2 (1 + \tan ^2 \theta) Q_K ( \tan ^2 \theta ) \biggr) \cr
&= \tan \theta P_{K+1} ( \tan ^2 \theta ) .&(2)\cr}$$
Polynomial $P_{K+1} ( \tan ^2 \theta )$ has properties analogous to those of $P_K ( \tan ^2 \theta )$ because only additions and multiplications were involved in constructing it.

Similarly, we obtain
$$\eqalignno{
&D_{2K+3} = {d \over d \theta} D_{2K+2} = \tan \theta \sec ^2 \theta P_{K+1} ' ( \tan ^2 \theta ) + \sec ^2 \theta P_{K+1} ( \tan ^2 \theta ) \cr
&= \sec ^2 \theta \biggl( \tan \theta P_{K+1} ' ( \tan ^2 \theta ) + P_{K+1} ( \tan ^2 \theta ) \biggr) \cr
&= \sec ^2 \theta Q_{K+1} ( \tan ^2 \theta ) .&(3) \cr}$$
Therefore, by the Principle of Mathematical Induction, equations (1) are true for all positive integers.

Equation (1a) gives $D_{2k} = 0$ for $\theta = 0$ and all $k$, while equation (1b) gives $D_{2k+1} > 0$ for $\theta = 0$ and all $k$.  Since the MacLaurin series for $\tan \theta$ is
$$\tan \theta = \sum_{k=0}^\infty {1 \over k!} D_k (0) \theta ^k , \ \ \ \ (-\pi / 2 < \theta < \pi / 2)$$
then this reduces to
$$\tan \theta = \sum_{k=0}^\infty {1 \over (2k+1)!} D_{2k+1} (0) \theta ^{2k+1} ,$$
where $c_k > 0$ for all $k$.  The first few terms are computed to be
$$\tan \theta = \theta + {\theta ^3 \over 3} + {2 \theta ^5 \over 15} + {17 \theta ^7 \over 315} + {62 \theta ^9 \over 2835} + \cdots .$$
\vfill\eject
\centerline{\bf 2005 Missouri Collegiate Mathematics Competition}
\vskip 6pt
\centerline{\bf Session II}
\bigskip
1. If $a < b < c$, $f' (x)$ is strictly increasing on $(a,c)$, and $f(x)$ is continuous on $[a,c]$, then show that
$$(b-a) f(c) + (c-b) f(a) > (c-a) f(b) .$$
\bigskip
Solution.
\medskip
Under the hypotheses, the Mean Value Theorem for derivatives is valid, so there are numbers $\alpha$ and $\beta$ such that
$${f(c) - f(b) \over c-b} = f' ( \beta ), \ \ \ \ \ b < \beta < c$$
and
$${f(b) - f(a) \over b-a} = f' ( \alpha ) , \ \ \ \ \ a < \alpha < b .$$
But $f'$ is strictly increasing, so $f' ( \beta ) > f' ( \alpha )$ and, hence,
$${f(c) - f(b) \over c-b} > {f(b) - f(a) \over b-a} .$$
An elementary rearrangement gives the desired result.
\bigskip
2.  Find all integer solutions $(x,y)$ to the equation $xy = 5x + 11y$.
\bigskip
Solution.
\medskip
First, if we re-write the quation $xy = 5x+11y$ as $xy - 5x - 11y = 0$, we can add $55$ to both sides to get $xy - 5x - 11y + 55 = 55$.  So we have
$$(x-11)(y-5) = 55 .$$
Now, $x$ and $y$ are supposed to be integers, so $x-11$ and $y-5$ should also be integers.  Thus, the question is, how many different pairs of integers can we find whose product is $55$?  There are eight pairs:
$$(55,1),\ (-55,-1),\ (1,55),\ (-1,-55),\ (5,11),\ (-5,-11),\ (11,5),\ (-11,-5) .$$
If $x-5$ is to represent the first number and $y-11$ is to represent the second number, then solving for $x$ and $y$ in each pair yields the eight possible solutions
$$\eqalign{
&x=66,\ y=6\ \ \ \ \ x=-44,\ y=4\ \ \ \ \ x=12,\ y=60\ \ \ \ \ x=10,\ y=-50 \cr
&x=16,\ y=16\ \ \ \ \ x=6,\ y=-6\ \ \ \ \ x=22,\ y=10\ \ \ \ \ x=0,\ y=0 .\cr}$$
\vfill\eject
From Robert Roe, University of Missouri-Rolla.
\medskip
\comment
3.  Define a circle of radius $r$ and center $P$ on a sphere to be the locus of points on the surface of the sphere that are a distance $r$ from $P$, where distance is the usual Euclidean distance in $\R ^3$, assuming that $r$ is less than the diameter of the sphere (otherwise, the locus is empty).  Thus, a circle divides the sphere into two spherical segments.  The area of the circle is the surface area of the segment containing the center point, $P$.  Show that this area is $\pi r^2$. 
\endcomment
3. Define a circle of radius $r$ and center $P$ on a sphere to be the locus of points on the surface of the sphere that are a distance $r$ from $P$, where distance is the usual Euclidean distance in $\R ^3$.  When $r$ is less than the diameter of the sphere, this circle divides the sphere into two spherical segments, or ``caps''.  Show that the area of the cap containing the center point $P$ is $\pi r^2$.
\bigskip
Solution.
\medskip
\comment
Let the sphere be given by $x^2 + y^2 + z^2 = R^2$ and, without loss of generality, let $P$ be the point $(0,0,R)$, the `North Pole''.  The surface area of a spherical segment is $A = 2 \pi R h$, where $h$ is the height of the segment.  This can be derived with an integral for the area of a surface of revolution, or a double integral for surface area, or maybe it is a known formula (but I doubt it).
$$\eqalign{
A &= \int_0^s 2\pi x \sqrt {1 + {x^2 \over R^2 - x^2}} \, dx = \pi R \int_0^s 2x (R^2 - x^2) ^{-1/2} \, dx \cr
&= 2\pi R(R - \sqrt {R^2 - s^2} ) = 2\pi Rh .\cr}$$
Then, from the figure, $s^2 = r^2 - h^2 = R^2 - (R-h)^2$, and we get $r^2 = 2Rh$, making $A = \pi r^2$.
$$\epsfxsize=15pc \epsfbox{SphereCircle.eps}$$
\endcomment
$$\epsfxsize=15pc \epsfbox{circle.eps}$$
Let $O$ and $R$ denote the center and the radius of the sphere.  Any section of the (hollow) sphere by a plane perpendicular to $OP$ is a circle say $\Gamma$.  Let $T$ be a point on the circle $\Gamma$, for which $PT = r$.  When arc $PT$ is revolved about the axis $OP$, it will generate the spherical cap whose area is to be computed.  Let $A(R,\theta )$ and $B(R, \theta + \delta \theta )$ be nearby points on the arc $PT$.  The length of the arc $AB$ is $R \cdot \delta \theta$.  When we rotate arc $AB$ about the line $OP$, the area of the surface generated is
$$2\pi (R \sin \theta ) (R \cdot \delta \theta ) .$$
If $\angle POT = \alpha$, then $r = 2R \sin \alpha / 2$.  The area of the cap is
$$\int_0^\alpha 2\pi R \sin \theta R \, d\theta = 2 \pi R^2 \int_0^\alpha \sin \theta R \, d\theta = 2 \pi R^2 (1 - \cos \alpha ) = 4\pi R^2 \sin ^2 \bigl( {\alpha \over 2} \bigr) = \pi r^2 .$$
\bigskip
4.  Let $p > 2$ be a prime.  Prove or disprove that all prime divisors of $2^p - 1$ have the form $2kp + 1$.
\bigskip
Solution.
\medskip
The fact that all prime divisors of $2^p-1$, where $p>2$ is prime, have the form $2kp+1$ was known to Fermat.  Let $q$ be a prime divisor of $2^p - 1$.  Then $2^p \equiv 1 \pmod q$.  Since $p$ is prime and $2^1 \not\equiv 1 \pmod q$, it follows that the order of 2 mod $q$ is $p$.  By Fermat's little theorem, i.e., $2^{q-1} \equiv 1 \pmod q$, we get $p \ \vert \ q-1$.  Thus, there exists $j$ such that $jp = q-1$.  Since $p$ is odd and $q-1$ is even, $j = 2k$ for some integer $k$.
\vfill\eject
5.  Suppose that $f \colon [0, \infty ) \to [0, \infty )$ is a differentiable function with the property that the area under the curve $y=f(x)$ from $x=a$ to $x=b$ is equal to the arclength of the curve $y=f(x)$ from $x=a$ to $x=b$.  Given that $f(0) = 5/4$, and that $f(x)$ has a minimum value on the interval $(0, \infty )$, find that minimum value.
\bigskip
Solution.
\medskip
The area under the curve $y=f(x)$ from $x=a$ to $x=b$ is
$$\int_a^b f(t) \, dt ,$$
and the arclength of the curve $y=f(x)$ from $x=a$ to $x=b$ is
$$\int_a^b \sqrt {1 + (f' (t))^2 } \, dt .$$
Therefore,
$$\int_a^b f(t) \, dt = \int_a^b \sqrt {1 + (f' (t))^2 } \, dt$$
for all nonnegative $a$ and $b$.  In particular, we can write
$$\int_0^x f(t) \, dt = \int_0^x \sqrt {1 + (f' (t))^2 } \, dt$$
for all nonnegative $x$.  Both sides of the above equation define a function of $x$, and since they are equal, their derivatives are equal; their derivatives are given by the Second Fundamental Theorem of Calculus:
$${d \over dx} \biggl( \int_0^x f(t) \, dt \biggr) = {d \over dx} \biggl( \int_0^x \sqrt {1 + (f' (t))^2 } \, dt \biggr) ,$$
i.e.,
$$f(x) = \sqrt {1 + (f' (x))^2 } .$$
So, we are looking for a function $y$ which satisifies the differential equation
$$y = \sqrt {1 + (y' )^2 } .$$
This equation is separable:
$$\eqalign{
y = \sqrt {1 + (y' )^2 } &\Rightarrow y^2 = 1 + (y' )^2 \cr
&\Rightarrow (y' )^2 = y^2 - 1 \cr
&\Rightarrow y' = \sqrt {y^2 - 1} \cr
&\Rightarrow {dy \over \sqrt {y^2 - 1}} = dx .\cr}$$
Integrating both sides yields
$$\int {dy \over \sqrt {y^2 - 1}} = \int dx \Rightarrow \ln \bigg\vert y + \sqrt {y^2 - 1} \bigg\vert = x + C$$
(where the first integral is evaluated using the trig substitution $y=\sec \theta$ and the two arbitrary constants of integration are combined into one constant on the right hand side).  Next, since $f(0) = 5/4$ is positive, we can drop the absolute value, and solve for $y$:
$$\eqalign{
\ln \bigl( y + \sqrt {y^2 - 1} \bigr) = x+C &\Rightarrow y+ \sqrt {y^2 - 1} = e^{x+C} = A e^x \ \ (\text{where $A = e^C$}) \cr
&\Rightarrow \sqrt {y^2 - 1} = Ae^x - y \cr
&\Rightarrow y^2 - 1 = (Ae^x - y)^2 = A^2 e^{2x} - 2Aye^x + y^2 \cr
&\Rightarrow -1 = A^2 e^{2x} - 2A ye^x \cr
&\Rightarrow 2Aye^x = A^2 e^{2x} + 1 \cr
&\Rightarrow y = {A^2 e^{2x} + 1 \over 2Ae^x} = {A \over 2} e^x + {1 \over 2A} e^{-x} .\cr}$$
Using $f(0) = 5/4$, we find
$${5 \over 4} = {A \over 2} + {1 \over 2A} \Rightarrow A = {1 \over 2} \ \ \text{or}\ \ 2 .$$
This gives two possible functions:
$$y = {1 \over 4} e^x + e^{-x} \ \ \text{or}\ \ y = e^x + {1 \over 4} e^{-x} .$$
This latter has a minimum at $x = -\ln 2$, which is not positive, so we reject that function.  The former has a minimum at $x = \ln 2$, and the $y$ value is $1$.
\medskip
Note: One could also deduce from the differential equation $y' = \sqrt {y^2 - 1}$ that at the minimum value, since $y' = 0$, the $y$-value must be $1$.
\bye 
