# Geometry Solutions 

Harvard-MIT Math Tournament<br>Written by Anders Kaseorg

1. $m \angle E D X=180^{\circ}-m \angle L A X-m \angle E L A=180^{\circ}-m \angle L A X-\left(180^{\circ}-m \angle A X E\right)=80^{\circ}$.
2. If the angle in radians is $\theta$, then Anne travels $R \theta$ and Lisa travels $(R-r)+r \theta+(R-r)$. Setting these equal yields $R(\theta-2)=r(\theta-2)$, so $\theta=2$ radians.
3. $m \angle M L D=\frac{1}{2} \overparen{A B}=45^{\circ}$.
4. The cylinder has a cross-sectional area $\pi$ times greater than the cube, so the water raises $\frac{1}{\pi}$ times as quickly in the cylinder as it lowers in the cube; that is, at $\frac{1}{\pi} \frac{\mathrm{~cm}}{\mathrm{~s}}$.
5. If the new circle has radius $r$, then the distance from its center to $E$ can be computed either as $1+r$ or $(2-r) \sqrt{2}$. Setting these equal yields $r=\frac{2 \sqrt{2}-1}{\sqrt{2}+1}=5-3 \sqrt{2}$.
6. The central circle has area $\pi\left(\frac{1}{2 \sqrt{3}}\right)^{2}=\frac{\pi}{12}$, and each of the three small triangles are copies of the entire figure dilated by $\frac{1}{3}$. Therefore, the total area is given by $K=\frac{\pi}{12}+3 \cdot\left(\frac{1}{3}\right)^{2} K \Longleftrightarrow$ $\frac{2}{3} K=\frac{\pi}{12} \Longleftrightarrow K=\frac{\pi}{8}$.
7. Let $E=(a, b, 0), A=(-c, b, 0), R=(-c,-d, 0), L=(a,-d, 0), Y=(0,0, h)$, and observe that $E Y^{2}+R Y^{2}=a^{2}+b^{2}+c^{2}+d^{2}+2 h^{2}=A Y^{2}+L Y^{2}$, which can only be satisfied by $E Y=1, A Y=4, R Y=8, L Y=7$ (or the symmetric configurations). Since $E A$ is an integral side of a triangle whose other sides are 1 and 4 , we must have $E A=4$; similarly, $E L=7$. Therefore, the area of rectangle $E A R L$ is 28 . (Such a pyramid may be constructed by taking $a=\frac{1}{8}, b=\frac{1}{14}, c=\frac{31}{8}, d=\frac{97}{14}, h=\frac{\sqrt{3071}}{56}$.)
8. Since $O D \perp A C$ and $\triangle A O C$ is equilateral, we have $\angle A O D=30^{\circ}$. So $A E=\frac{1}{\sqrt{3}}$, and $B H=\sqrt{A B^{2}+A H^{2}}=\sqrt{2^{2}+\left(3-\frac{1}{\sqrt{3}}\right)^{2}}=\sqrt{\frac{40}{3}-2 \sqrt{3}} \approx 3.141533339$.
9. The dilation of ratio $-\frac{2}{3}$ about $T$ sends $C_{2}$ to $C_{1}, O_{2}$ to $O_{1}$, and $S$ to the other intersection of $s$ with $C_{1}$, which we shall call $U$. We can now compute $T R \cdot T S=\frac{3}{2} T R \cdot T U=\frac{3}{2} T P^{2}=$ $\frac{3}{2}\left(O_{1} T^{2}-O_{1} P^{2}\right)=\frac{3}{2}\left(\left(\frac{O_{1} O_{2}}{1+3 / 2}\right)^{2}-O_{1} P^{2}\right)=\frac{3}{2}\left(\left(\frac{20}{5 / 2}\right)^{2}-4^{2}\right)=72$.
10. Suppose that when the ball hits a side of the table, instead of reflecting the ball's path, we reflect the entire table over this side so that the path remains straight. If we repeatedly reflect the table over its sides in all possible ways, we get a triangular grid that tiles the plane. Whenever the path crosses $n$ lines in this grid parallel to $C T$, it will cross $\frac{7}{8} n$ lines parallel to $C H$ and $\frac{15}{8} n$ lines parallel to $H T$. After crosssing $8+7+15=30$ grid lines it will have crossed three lines simultaneously again, which means that the ball will have landed in a pocket after bouncing 27 times. By picturing the grid it is easy to see that the pocket in question is $H$. The distance the ball travels during the $\frac{1}{8}$ of its trip described in the problem is the third side of a triangle with an $120^{\circ}$ angle between two sides 16 and 14 , which is $\sqrt{16^{2}+14^{2}-2 \cdot 16 \cdot 14 \cos 120^{\circ}}=26$, so length of the entire trip is 208 .