exam: HMMT

problem: Find the sum of the positive integer divisors of $2^{2007}$.

solution:
$\mathbf{2}^{\mathbf{2 0 0 8}}-\mathbf{1}$. The divisors are the powers of two not exceeding $2^{2007}$. So the sum is
$1+2+2^{2}+\cdots+2^{2007}=-1+2+2+2^{2}+\cdots+2^{2007}=-1+2^{2}+2^{2}+\cdots+2^{2007}=\cdots=-1+2^{2008}$.

solution_match: Answer:


problem:
The four sides of quadrilateral $A B C D$ are equal in length. Determine the perimeter of $A B C D$ given that it has area 120 and $A C=10$.
![](https://cdn.mathpix.com/cropped/2025_01_24_f658dbe1668619ae64f9g-1.jpg?height=302&width=646&top_left_y=928&top_left_x=783)

solution: 52. Let $M$ be the midpoint of $A C$. Then triangles $A M B, B M C, C M D$, and $D M A$ are all right triangles having legs 5 and $h$ for some $h$. The area of $A B C D$ is 120 , but also $4 \cdot\left(\frac{1}{2} \cdot 5 \cdot h\right)=10 h$, so $h=12$. Then $A B=B C=C D=D A=\sqrt{12^{2}+5^{2}}=13$, and the perimeter of $A B C D$ is 52 .


problem: Five people are crowding into a booth against a wall at a noisy restaurant. If at most three can fit on one side, how many seating arrangements accomodate them all?

solution: 240. Three people will sit on one side and two sit on the other, giving a factor of two. Then there are 5 ! ways to permute the people.


problem: Thomas and Michael are just two people in a large pool of well qualified candidates for appointment to a problem writing committee for a prestigious college math contest. It is 40 times more likely that both will serve if the size of the committee is increased from its traditional 3 members to a whopping $n$ members. Determine $n$. (Each person in the pool is equally likely to be chosen.)

solution:
16. Suppose there are $k$ candidates. Then the probability that both serve on a 3 membered committee is $(k-2) /\binom{k}{3}$, and the odds that both serve on an $n$ membered committee are $\binom{k-2}{n-2} /\binom{k}{n}$. The ratio of the latter to the former is

$$
\frac{\binom{k}{3}\binom{k-2}{n-2}}{(k-2)\binom{k}{n}}=\frac{k!(k-2)!1!(k-3)!n!(k-n)!}{k!(k-2)!(n-2)!(k-n)!3!(k-3)!}=\frac{n \cdot(n-1)}{3!}
$$

Solving $n \cdot(n-1)=240$ produces $n=16,-15$, and we discard the latter.


problem: The curves $y=x^{2}(x-3)^{2}$ and $y=\left(x^{2}-1\right)(x-2)$ intersect at a number of points in the real plane. Determine the sum of the $x$-coordinates of these points of intersection.

solution:
7. Because the first curve touches the $x$-axis at $x=0$ and $x=3$ while the second curve crosses the $x$-axis at $x= \pm 1$ and $x=2$, there are four points of intersection. In particular, the points of intersection have $x$-coordinates determined by the difference of the two curves:

$$
0=x^{2}(x-3)^{2}-\left(x^{2}-1\right)(x-2)=\left(x^{4}-6 x^{3}+\cdots\right)-\left(x^{3}+\cdots\right)=x^{4}-7 x^{3}+\cdots
$$

We need only the first two coefficients to determine $x_{1}+x_{2}+x_{3}+x_{4}=-\left(\frac{-7}{1}\right)=7$.


problem: Andrew has a fair six sided die labeled with 1 through 6 as usual. He tosses it repeatedly, and on every third roll writes down the number facing up as long as it is not the 6 . He stops as soon as the last two numbers he has written down are squares or one is a prime and the other is a square. What is the probability that he stops after writing squares consecutively?

solution: $4 / \mathbf{2 5}$. We can safely ignore all of the rolls he doesn't record. The probability that he stops after writing two squares consecutively is the same as the probability that he never rolls a prime. For, as soon as the first prime is written, either it must have been preceded by a square or it will be followed by a nonnegative number of additional primes and then a square. So we want the probability that two numbers chosen uniformly with replacement from $\{1,2,3,4,5\}$ are both squares, which is $(2 / 5)^{2}$.


solution:
18. Add the three equations to obtain

$$
x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x-4 x-4 y-4 z+6=258
$$

which rewrites as $(x+y+z-2)^{2}=256$. Evidently, $x+y+z=2 \pm 16$. Since $x, y$, and $z$ are positive, $x+y+z>0$ so $x+y+z=2+16=18$.


problem: Find the positive real number(s) $x$ such that $\frac{1}{2}\left(3 x^{2}-1\right)=\left(x^{2}-50 x-10\right)\left(x^{2}+25 x+5\right)$.

solution: $\mathbf{2 5}+\mathbf{2} \sqrt{\mathbf{1 5 9}}$. Write $a=x^{2}-50 x-10$ and $b=x^{2}+25 x+5$; the given becomes $\frac{a+2 b-1}{2}=a b$, so $0=2 a b-a-2 b+1=(a-1)(2 b-1)$. Then $a-1=x^{2}-50 x-11=0$ or $2 b-1=2 x^{2}+50 x+9=0$. The former has a positive root, $x=25+2 \sqrt{159}$, while the latter cannot, for obvious reasons.


problem:
Cyclic quadrilateral $A B C D$ has side lengths $A B=1, B C=2, C D=3$, and $A D=4$. Determine $A C / B D$.
![](https://cdn.mathpix.com/cropped/2025_01_24_f658dbe1668619ae64f9g-2.jpg?height=581&width=592&top_left_y=1623&top_left_x=818)

solution:
5/7. Let the diagonals intersect at $P$. Note that triangles $A B P$ and $D C P$ are similar, so that $3 A P=D P$ and $3 B P=C P$. Additionally, triangles $B C P$ and $A D P$ are similar, so that $2 B P=A P$. It follows that

$$
\frac{A C}{B D}=\frac{A P+P C}{B P+P D}=\frac{2 B P+3 B P}{B P+6 B P}=\frac{5}{7}
$$


problem:
A positive real number $x$ is such that

$$
\sqrt[3]{1-x^{3}}+\sqrt[3]{1+x^{3}}=1
$$

Find $x^{2}$.

solution:
$\frac{\sqrt[3]{28}}{\mathbf{3}}$. Cubing the given equation yields

$$
1=\left(1-x^{3}\right)+3 \sqrt[3]{\left(1-x^{3}\right)\left(1+x^{3}\right)}\left(\sqrt[3]{1-x^{3}}+\sqrt[3]{1+x^{3}}\right)+\left(1+x^{3}\right)=2+3 \sqrt[3]{1-x^{6}}
$$

Then $\frac{-1}{3}=\sqrt[3]{1-x^{6}}$, so $\frac{-1}{27}=1-x^{6}$ and $x^{6}=\frac{28}{27}$ and $x^{2}=\frac{\sqrt[3]{28}}{3}$.

## Adult Acorns - Gee, I'm a Tree! [200]

In this section of the team round, your team will derive some basic results concerning tangential quadrilaterals. Tangential quadrilaterals have an incircle, or a circle lying within them that is tangent to all four sides. If a quadrilateral has an incircle, then the center of this circle is the incenter of the quadrilateral. As you shall see, tangential quadrilaterals are related to cyclic quadrilaterals. For reference, a review of cyclic quadrilaterals is given at the end of this section.

Your answers for this section of the team test should be proofs. Note that you may use any standard facts about cyclic quadrilaterals, such as those listed at the end of this test, without proving them. Additionally, you may cite the results of previous problems, even if you were unable to prove them.

For these problems, $A B C D$ is a tangential quadrilateral having incenter $I$. For the first three problems, the point $P$ is constructed such that triangle $P A B$ is similar to triangle $I D C$ and lies outside $A B C D$.


problem: Show that $P A I B$ is cyclic by proving that $\angle I A P$ is supplementary to $\angle P B I$.

solution:
Note that $I$ lies on the angle bisectors of the angles of quadrilateral $A B C D$. So writing $\angle D A B=2 \alpha, \angle A B C=2 \beta, \angle B C D=2 \gamma$, and $\angle C D A=2 \delta$, we have

$$
\begin{aligned}
\angle I A P+\angle P B I & =\angle I A B+\angle B A P+\angle P B A+\angle A B I \\
& =\angle I A B+\angle C D I+\angle I C D+\angle A B I \\
& =\alpha+\beta+\gamma+\delta .
\end{aligned}
$$

We are done because the angles in quadrilateral $A B C D$ add up to $360^{\circ}$.
![](https://cdn.mathpix.com/cropped/2025_01_24_f658dbe1668619ae64f9g-3.jpg?height=385&width=581&top_left_y=1610&top_left_x=794)

solution_match: Solution.


problem:
Show that triangle $P A I$ is similar to triangle $B I C$. Then conclude that

$$
P A=\frac{P I}{B C} \cdot B I
$$

solution:
We have $\angle I B C=\angle A B I$ because $I$ lies on the angle bisector, and $\angle A B I=\angle A P I$ because $P A I B$ is cyclic. Additionally,

$$
\angle B C I=\angle I C D=\angle P B A=\angle P I A
$$

by the angle bisector $C I$, that triangles $P A B$ and $I D C$ are similar, and the fact that $P A I B$ is cyclic, respectively. It follows that triangles $P A I$ and $B I C$ are similar. In particular, it follows that $I P / P A=B C / B I$, as required.


problem:
Deduce from the above that

$$
\frac{B C}{A D} \cdot \frac{A I}{B I} \cdot \frac{D I}{C I}=1
$$

solution:
Exchanging the roles of $A$ and $D$ with $B$ and $C$, respectively, converts the formula from problem 2 into another formula:

$$
P B=\frac{P I}{A D} \cdot A D
$$

Then one the one hand, dividing the two gives $P A / P B=(A D \cdot B I) /(B C \cdot A I)$. On the other hand, $P A / P B=D I / C I$ because triangles $P A B$ and $I D C$ are similar. Clearing the denominators in the equation

$$
\frac{D I}{C I}=\frac{A D \cdot B I}{B C \cdot A I}
$$

yields the desired form.


problem:
Show that $A B+C D=A D+B C$. Use the above to conclude that for some positive number $\alpha$,

$$
\begin{array}{ll}
A B=\alpha \cdot\left(\frac{A I}{C I}+\frac{B I}{D I}\right) & B C=\alpha \cdot\left(\frac{B I}{D I}+\frac{C I}{A I}\right) \\
C D=\alpha \cdot\left(\frac{C I}{A I}+\frac{D I}{B I}\right) & D A=\alpha \cdot\left(\frac{D I}{B I}+\frac{A I}{C I}\right) .
\end{array}
$$

solution:
Draw in the points of tangency $P, Q, R$, and $S$, of the incircle with sides $A B, B C, C D$, and $A D$, as shown. Then we have equal tangents $A P=A S, B P=B Q, C Q=C R$, and $D R=D S$. Then

$$
A B+C D=A P+B P+C R+D R=A S+(B Q+C Q)+D S=B C+A D
$$

Using the result of problem 3, we set $B C=x \cdot B I \cdot C I$ and $A D=x \cdot A I \cdot D I$ for some $x$, and $A B=y \cdot A I \cdot B I$ and $C D=y \cdot C I \cdot D I$ for some $y$. Now because $A B+C D=B C+A D$, we obtain

$$
y(A I \cdot B I+C I \cdot D I)=x(B I \cdot C I+A I \cdot D I)
$$

So it follows that the ratio $A B: B C: C D: D A$ is uniquely determined. One easily checks that the posed ratio satisfies the three required relations.
![](https://cdn.mathpix.com/cropped/2025_01_24_f658dbe1668619ae64f9g-4.jpg?height=473&width=643&top_left_y=1615&top_left_x=779)


problem:
Show that

$$
A B \cdot B C=B I^{2}+\frac{A I \cdot B I \cdot C I}{D I}
$$

solution:
Returning to the original set up, Ptolemy's theorem applied to quadrilateral $P A I B$ gives $A B \cdot P I=P A \cdot B I+P B \cdot A I$. Substituting equation $P A=\frac{P I}{B C} \cdot B I$ from problem 2 and its cousin $P B=\frac{P I}{A D} \cdot A I$ allows us to write

$$
A B \cdot P I=\frac{P I}{B C} \cdot B I^{2}+\frac{P I}{A D} \cdot A I^{2}
$$

or

$$
A B \cdot B C=B I^{2}+\frac{B C}{A D} \cdot A I^{2}
$$

Substituting the formula $B C / A D=\frac{B I \cdot C I}{A I \cdot D I}$ from problem 3 finishes the problem.


problem: Let the incircle of $A B C D$ be tangent to sides $A B, B C, C D$, and $A D$ at points $P, Q, R$, and $S$, respectively. Show that $A B C D$ is cyclic if and only if $P R \perp Q S$.

solution:
Let the diagonals of $P Q R S$ intersect at $T$. Because $\overline{A P}$ and $\overline{A S}$ are tangent to $\omega$ at $P$ and $S$, we may write $\alpha=\angle A S P=\angle S P A=\angle S Q P$ and $\beta=\angle C Q R=\angle Q R C=\angle Q P R$. Then $\angle P T Q=\pi-\alpha-\beta$. On the other hand, $\angle P A S=\pi-2 \alpha$ and $\angle R C Q=\pi-2 \beta$, so that $A B C D$ is cyclic if and only if

$$
\pi=\angle B A D+\angle D C B=2 \pi-2 \alpha-2 \beta
$$

or simply

$$
\pi / 2=\pi-\alpha-\beta=\angle P T Q
$$

as desired.

A brief review of cyclic Quadrilaterals.

The following discussion of cyclic quadrilaterals is included for reference. Any of the results given here may be cited without proof in your writeups.
A cyclic quadrilateral is a quadrilateral whose four vertices lie on a circle called the circumcircle (the circle is unique if it exists.) If a quadrilateral has a circumcircle, then the center of this circumcircle is called the circumcenter of the quadrilateral. For a convex quadrilateral $A B C D$, the following are equivalent:

- Quadrilateral $A B C D$ is cyclic;
- $\angle A B D=\angle A C D$ (or $\angle B C A=\angle B D A$, etc.);
- Angles $\angle A B C$ and $\angle C D A$ are supplementary, that is, $m \angle A B C+m \angle C D A=180^{\circ}$ (or angles $\angle B C D$ and $\angle B A D$ are supplementary);

Cyclic quadrilaterals have a number of interesting properties. A cyclic quadrilateral $A B C D$ satisfies

$$
A C \cdot B D=A B \cdot C D+A D \cdot B C
$$

a result known as Ptolemy's theorem. Another result, typically called Power of a Point, asserts that given a circle $\omega$, a point $P$ anywhere in the plane of $\omega$, and a line $\ell$ through $P$ intersecting $\omega$ at points $A$ and $B$, the value of $A P \cdot B P$ is independent of $\ell$; i.e., if a second line $\ell^{\prime}$ through $P$ intersects $\omega$ at $A^{\prime}$ and $B^{\prime}$, then $A P \cdot B P=A^{\prime} P \cdot B^{\prime} P$. This second theorem is proved via similar triangles. Say $P$ lies outside of $\omega$, that $\ell$ and $\ell^{\prime}$ are as before and that $A$ and $A^{\prime}$ lie on segments $B P$ and $B^{\prime} P$ respectively. Then triangle $A A^{\prime} P$ is similar to triangle $B^{\prime} B P$ because the triangles share an angle at $P$ and we have

$$
m \angle A A^{\prime} P=180^{\circ}-m \angle B^{\prime} A^{\prime} A=m \angle A B B^{\prime}=m \angle P B B^{\prime}
$$

The case where $A=B$ is valid and describes the tangents to $\omega$. A similar proof works for $P$ inside $\omega$.