HMMT February

Geometry Difficulty 5.3 AIME, harder Find the answer

Let $A B C D$ be a quadrilateral inscribed in a unit circle with center $O$ . Suppose that $\angle A O B=\angle C O D=135^{\circ}, B C=1$ . Let $B^{\prime}$ and $C^{\prime}$ be the reflections of $A$ across $B O$ and $C O$ respectively. Let $H_{1}$ and $H_{2}$ be the orthocenters of $A B^{\prime} C^{\prime}$ and $B C D$ , respectively. If $M$ is the midpoint of $O H_{1}$ , and $O^{\prime}$ is the reflection of $O$ about the midpoint of $M H_{2}$ , compute $O O^{\prime}$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Put the diagram on the complex plane with $O$ at the origin and $A$ at 1. Let $B$ have coordinate $b$ and $C$ have coordinate $c$ . We obtain easily that $B^{\prime}$ is $b^{2}, C^{\prime}$ is $c^{2}$ , and $D$ is $b c$ . Therefore, $H_{1}$ is $1+b^{2}+c^{2}$ and $H_{2}$ is $b+c+b c$ (we have used the fact that for triangles on the unit circle, their orthocenter is the sum of the vertices). Finally, we have that $M$ is $\frac{1}{2}\left(1+b^{2}+c^{2}\right)$ , so the reflection of $O$ about the midpoint of $M H_{2}$ is $\frac{1}{2}\left(1+b^{2}+c^{2}+2 b+2 c+2 b c\right)=\frac{1}{2}(b+c+1)^{2}$ , so we just seek $\frac{1}{2}|b+c+1|^{2}$ . But we know that $b=\operatorname{cis} 135^{\circ}$ and $c=\operatorname{cis} 195^{\circ}$ , so we obtain that this value is $\frac{1}{4}(8-\sqrt{6}-3 \sqrt{2})$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.