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, 2020

Geometry Difficulty 5.8 AIME, harder Find the answer United States

Problem:

In acute triangle $A B C$ , let $H$ be the orthocenter and $D$ the foot of the altitude from $A$ . The circumcircle of triangle $B H C$ intersects $A C$ at $E \neq C$ , and $A B$ at $F \neq B$ . If $B D = 3$ , $C D = 7$ , and $\frac{A H}{H D} = \frac{5}{7}$ , the area of triangle $A E F$ can be expressed as $\frac{a}{b}$ , where $a, b$ are relatively prime positive integers. Compute $100 a + b$ .

Your answer

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

Solution

Solution:

Let $A H$ intersect the circumcircle of $\triangle A B C$ again at $P$ , and the circumcircle of $\triangle B H C$ again at $Q$ . Because $\angle B H C = 180 - \angle A = \angle B P C$ , $P$ is the reflection of $H$ over $D$ . Thus, we know that $P D = H D$ . From power of a point and $A D = \frac{12 H D}{7}$ ,
$B D \cdot C D = A D \cdot P D = \frac{12 H D^{2}}{7}$
From this, $H D = \frac{7}{2}$ and $A H = \frac{5}{2}$ . Furthermore, because $\triangle B H C$ is the reflection of $\triangle B P C$ over $B C$ , the circumcircle of $\triangle B H C$ is the reflection of the circumcircle of $\triangle A B C$ over $B C$ . Then, $A Q = 2 A D = 12$ . Applying Power of a Point,
$A C \cdot A E = A B \cdot A F = A H \cdot A Q = 30$
We can compute $A C = \sqrt{85}$ and $A B = 3 \sqrt{5}$ , which means that $A E = \frac{6 \sqrt{85}}{17}$ and $A F = 2 \sqrt{5}$ . Also, $[A B C] = \frac{B C \cdot A D}{2} = 30$ . Therefore,
$[A E F] = \frac{A E \cdot A F}{A C \cdot A B} \cdot [A B C] = \frac{4}{17} \cdot 30 = \frac{120}{17}$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.