# $11^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament 

## Saturday 23 February 2008

## Individual Round: Geometry Test

1. [3] How many different values can $\angle A B C$ take, where $A, B, C$ are distinct vertices of a cube?

Answer: 5 . In a unit cube, there are 3 types of triangles, with side lengths $(1,1, \sqrt{2}),(1, \sqrt{2}, \sqrt{3})$ and $(\sqrt{2}, \sqrt{2}, \sqrt{2})$. Together they generate 5 different angle values.
2. [3] Let $A B C$ be an equilateral triangle. Let $\Omega$ be its incircle (circle inscribed in the triangle) and let $\omega$ be a circle tangent externally to $\Omega$ as well as to sides $A B$ and $A C$. Determine the ratio of the radius of $\Omega$ to the radius of $\omega$.
Answer: $\quad 3$ Label the diagram as shown below, where $\Omega$ and $\omega$ also denote the center of the corresponding circles. Note that $A M$ is a median and $\Omega$ is the centroid of the equilateral triangle. So $A M=3 M \Omega$. Since $M \Omega=N \Omega$, it follows that $A M / A N=3$, and triangle $A B C$ is the image of triangle $A B^{\prime} C^{\prime}$ after a scaling by a factor of 3 , and so the two incircles must also be related by a scale factor of 3 .
![](https://cdn.mathpix.com/cropped/2025_01_24_684c7884f8823da5eea3g-1.jpg?height=289&width=329&top_left_y=1048&top_left_x=936)
3. [4] Let $A B C$ be a triangle with $\angle B A C=90^{\circ}$. A circle is tangent to the sides $A B$ and $A C$ at $X$ and $Y$ respectively, such that the points on the circle diametrically opposite $X$ and $Y$ both lie on the side $B C$. Given that $A B=6$, find the area of the portion of the circle that lies outside the triangle.
![](https://cdn.mathpix.com/cropped/2025_01_24_684c7884f8823da5eea3g-1.jpg?height=253&width=253&top_left_y=1543&top_left_x=974)

Answer: $\pi-2$ Let $O$ be the center of the circle, and $r$ its radius, and let $X^{\prime}$ and $Y^{\prime}$ be the points diametrically opposite $X$ and $Y$, respectively. We have $O X^{\prime}=O Y^{\prime}=r$, and $\angle X^{\prime} O Y^{\prime}=90^{\circ}$. Since triangles $X^{\prime} O Y^{\prime}$ and $B A C$ are similar, we see that $A B=A C$. Let $X^{\prime \prime}$ be the projection of $Y^{\prime}$ onto $A B$. Since $X^{\prime \prime} B Y^{\prime}$ is similar to $A B C$, and $X^{\prime \prime} Y^{\prime}=r$, we have $X^{\prime \prime} B=r$. It follows that $A B=3 r$, so $r=2$.
![](https://cdn.mathpix.com/cropped/2025_01_24_684c7884f8823da5eea3g-1.jpg?height=386&width=391&top_left_y=2070&top_left_x=908)

Then, the desired area is the area of the quarter circle minus that of the triangle $X^{\prime} O Y^{\prime}$. And the answer is $\frac{1}{4} \pi r^{2}-\frac{1}{2} r^{2}=\pi-2$.
4. [4] In a triangle $A B C$, take point $D$ on $B C$ such that $D B=14, D A=13, D C=4$, and the circumcircle of $A D B$ is congruent to the circumcircle of $A D C$. What is the area of triangle $A B C$ ?
Answer: 108
![](https://cdn.mathpix.com/cropped/2025_01_24_684c7884f8823da5eea3g-2.jpg?height=332&width=438&top_left_y=539&top_left_x=887)

The fact that the two circumcircles are congruent means that the chord $A D$ must subtend the same angle in both circles. That is, $\angle A B C=\angle A C B$, so $A B C$ is isosceles. Drop the perpendicular $M$ from $A$ to $B C$; we know $M C=9$ and so $M D=5$ and by Pythagoras on $A M D, A M=12$. Therefore, the area of $A B C$ is $\frac{1}{2}(A M)(B C)=\frac{1}{2}(12)(18)=108$.
5. [5] A piece of paper is folded in half. A second fold is made such that the angle marked below has measure $\phi\left(0^{\circ}<\phi<90^{\circ}\right)$, and a cut is made as shown below.
![](https://cdn.mathpix.com/cropped/2025_01_24_684c7884f8823da5eea3g-2.jpg?height=185&width=633&top_left_y=1225&top_left_x=784)

When the piece of paper is unfolded, the resulting hole is a polygon. Let $O$ be one of its vertices. Suppose that all the other vertices of the hole lie on a circle centered at $O$, and also that $\angle X O Y=144^{\circ}$, where $X$ and $Y$ are the the vertices of the hole adjacent to $O$. Find the value(s) of $\phi$ (in degrees).

Answer: $81^{\circ}$ Try actually folding a piece of paper. We see that the cut out area is a kite, as shown below. The fold was made on $A C$, and then $B E$ and $D E$. Since $D C$ was folded onto $D A$, we have $\angle A D E=\angle C D E$.
![](https://cdn.mathpix.com/cropped/2025_01_24_684c7884f8823da5eea3g-2.jpg?height=419&width=268&top_left_y=1750&top_left_x=969)

Either $A$ or $C$ is the center of the circle. If it's $A$, then $\angle B A D=144^{\circ}$, so $\angle C A D=72^{\circ}$. Using $C A=D A$, we see that $\angle A C D=\angle A D C=54^{\circ}$. So $\angle E D A=27^{\circ}$, and thus $\phi=72^{\circ}+27^{\circ}=99^{\circ}$, which is inadmissible, as $\phi<90^{\circ}$.
So $C$ is the center of the circle. Then, $\angle C A D=\angle C D A=54^{\circ}, \angle A D E=27^{\circ}$, and $\phi=54^{\circ}+27^{\circ}=81^{\circ}$.
6. [5] Let $A B C$ be a triangle with $\angle A=45^{\circ}$. Let $P$ be a point on side $B C$ with $P B=3$ and $P C=5$. Let $O$ be the circumcenter of $A B C$. Determine the length $O P$.

Answer: $\sqrt{\sqrt{17}}$ Using extended Sine law, we find the circumradius of $A B C$ to be $R=\frac{B C}{2 \sin A}=4 \sqrt{2}$. By considering the power of point $P$, we find that $R^{2}-O P^{2}=P B \cdot P C=15$. So $O P=\sqrt{R^{2}-15}=$ $\sqrt{16 \cdot 2-15}=\sqrt{17}$.
7. [6] Let $C_{1}$ and $C_{2}$ be externally tangent circles with radius 2 and 3 , respectively. Let $C_{3}$ be a circle internally tangent to both $C_{1}$ and $C_{2}$ at points $A$ and $B$, respectively. The tangents to $C_{3}$ at $A$ and $B$ meet at $T$, and $T A=4$. Determine the radius of $C_{3}$.
Answer: 8 Let $D$ be the point of tangency between $C_{1}$ and $C_{2}$. We see that $T$ is the radical center of the three circles, and so it must lie on the radical axis of $C_{1}$ and $C_{2}$, which happens to be their common tangent $T D$. So $T D=4$.
![](https://cdn.mathpix.com/cropped/2025_01_24_684c7884f8823da5eea3g-3.jpg?height=654&width=757&top_left_y=711&top_left_x=725)

We have

$$
\tan \frac{\angle A T D}{2}=\frac{2}{T D}=\frac{1}{2}, \quad \text { and } \quad \tan \frac{\angle B T D}{2}=\frac{3}{T D}=\frac{3}{4} .
$$

Thus, the radius of $C_{3}$ equals to

$$
\begin{aligned}
T A \tan \frac{\angle A T B}{2} & =4 \tan \left(\frac{\angle A T D+\angle B T D}{2}\right) \\
& =4 \cdot \frac{\tan \frac{\angle A T D}{2}+\tan \frac{\angle B T D}{2}}{1-\tan \frac{\angle A T D}{2} \tan \frac{\angle B T D}{2}} \\
& =4 \cdot \frac{\frac{1}{2}+\frac{3}{4}}{1-\frac{1}{2} \cdot \frac{3}{4}} \\
& =8 .
\end{aligned}
$$

8. [6] Let $A B C$ be an equilateral triangle with side length 2 , and let $\Gamma$ be a circle with radius $\frac{1}{2}$ centered at the center of the equilateral triangle. Determine the length of the shortest path that starts somewhere on $\Gamma$, visits all three sides of $A B C$, and ends somewhere on $\Gamma$ (not necessarily at the starting point). Express your answer in the form of $\sqrt{p}-q$, where $p$ and $q$ are rational numbers written as reduced fractions.
Answer: $\sqrt{\frac{28}{3}}-1$ Suppose that the path visits sides $A B, B C, C A$ in this order. Construct points $A^{\prime}, B^{\prime}, C^{\prime}$ so that $C^{\prime}$ is the reflection of $C$ across $A B, A^{\prime}$ is the reflection of $A$ across $B C^{\prime}$, and $B^{\prime}$ is the reflection of $B$ across $A^{\prime} C^{\prime}$. Finally, let $\Gamma^{\prime}$ be the circle with radius $\frac{1}{2}$ centered at the center of $A^{\prime} B^{\prime} C^{\prime}$. Note that $\Gamma^{\prime}$ is the image of $\Gamma$ after the three reflections: $A B, B C^{\prime}, C^{\prime} A^{\prime}$.
![](https://cdn.mathpix.com/cropped/2025_01_24_684c7884f8823da5eea3g-4.jpg?height=302&width=640&top_left_y=234&top_left_x=786)

When the path hits $A B$, let us reflect the rest of the path across $A B$ and follow this reflected path. When we hit $B C^{\prime}$, let us reflect the rest of the path across $B C^{\prime}$, and follow the new path. And when we hit $A^{\prime} C^{\prime}$, reflect the rest of the path across $A^{\prime} C^{\prime}$ and follow the new path. We must eventually end up at $\Gamma^{\prime}$.
It is easy to see that the shortest path connecting some point on $\Gamma$ to some point on $\Gamma^{\prime}$ lies on the line connecting the centers of the two circles. We can easily find the distance between the two centers to be $\sqrt{3^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}}=\sqrt{\frac{28}{3}}$. Therefore, the length of the shortest path connecting $\Gamma$ to $\Gamma^{\prime}$ has length $\sqrt{\frac{28}{3}}-1$. By reflecting this path three times back into $A B C$, we get a path that satisfies our conditions.
9. [7] Let $A B C$ be a triangle, and $I$ its incenter. Let the incircle of $A B C$ touch side $B C$ at $D$, and let lines $B I$ and $C I$ meet the circle with diameter $A I$ at points $P$ and $Q$, respectively. Given $B I=$ $6, C I=5, D I=3$, determine the value of $(D P / D Q)^{2}$.

## Answer: <br> $\qquad$

![](https://cdn.mathpix.com/cropped/2025_01_24_684c7884f8823da5eea3g-4.jpg?height=435&width=386&top_left_y=1273&top_left_x=910)

Let the incircle touch sides $A C$ and $A B$ at $E$ and $F$ respectively. Note that $E$ and $F$ both lie on the circle with diameter $A I$ since $\angle A E I=\angle A F I=90^{\circ}$. The key observation is that $D, E, P$ are collinear. To prove this, suppose that $P$ lies outside the triangle (the other case is analogous), then $\angle P E A=\angle P I A=\angle I B A+\angle I A B=\frac{1}{2}(\angle B+\angle A)=90^{\circ}-\frac{1}{2} \angle C=\angle D E C$, which implies that $D, E, P$ are collinear. Similarly $D, F, Q$ are collinear. Then, by Power of a Point, $D E \cdot D P=D F \cdot D Q$. So $D P / D Q=D F / D E$.
Now we compute $D F / D E$. Note that $D F=2 D B \sin \angle D B I=2 \sqrt{6^{2}-3^{2}}\left(\frac{3}{6}\right)=3 \sqrt{3}$, and $D E=$ $2 D C \sin \angle D C I=2 \sqrt{5^{2}-3^{2}}\left(\frac{3}{5}\right)=\frac{24}{5}$. Therefore, $D F / D E=\frac{5 \sqrt{3}}{8}$.
10. [7] Let $A B C$ be a triangle with $B C=2007, C A=2008, A B=2009$. Let $\omega$ be an excircle of $A B C$ that touches the line segment $B C$ at $D$, and touches extensions of lines $A C$ and $A B$ at $E$ and $F$, respectively (so that $C$ lies on segment $A E$ and $B$ lies on segment $A F$ ). Let $O$ be the center of $\omega$. Let $\ell$ be the line through $O$ perpendicular to $A D$. Let $\ell$ meet line $E F$ at $G$. Compute the length $D G$.
Answer: 2014024 Let line $A D$ meet $\omega$ again at $H$. Since $A F$ and $A E$ are tangents to $\omega$ and $A D H$ is a secant, we see that $D E H F$ is a harmonic quadrilateral. This implies that the pole of $A D$ with respect to $\omega$ lies on $E F$. Since $\ell \perp A D$, the pole of $A D$ lies on $\ell$. It follows that the pole of $A D$ is $G$.
![](https://cdn.mathpix.com/cropped/2025_01_24_684c7884f8823da5eea3g-5.jpg?height=649&width=928&top_left_y=239&top_left_x=645)

Thus, $G$ must lie on the tangent to $\omega$ at $D$, so $C, D, B, G$ are collinear. Furthermore, since the pencil of lines $(A E, A F ; A D, A G)$ is harmonic, by intersecting it with the line $B C$, we see that $(C, B ; D, G)$ is harmonic as well. This means that

$$
\frac{B D}{D C} \cdot \frac{C G}{G B}=-1
$$

(where the lengths are directed.) The semiperimeter of $A B C$ is $s=\frac{1}{2}(2007+2008+2009)=3012$. So $B D=s-2009=1003$ and $C D=s-2008=1004$. Let $x=D G$, then the above equations gives

$$
\frac{1003}{1004} \cdot \frac{x+1004}{x-1003}=1
$$

Solving gives $x=2014024$.
Remark: If you are interested to learn about projective geometry, check out the last chapter of Geometry Revisited by Coxeter and Greitzer or Geometric Transformations III by Yaglom.