Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.7 AIME Prove it United States

Problem:

Altitudes $B E$ and $C F$ of acute triangle $A B C$ intersect at $H$ . Suppose that the altitudes of triangle $E H F$ concur on line $B C$ . If $A B = 3$ and $A C = 4$ , then $B C^{2} = \frac{a}{b}$ , where $a$ and $b$ are relatively prime positive integers. Compute $100 a + b$ .

Solution

Solution:

Let $P$ be the orthocenter of $\triangle E H F$ . Then $E H \perp F P$ and $E H \perp A C$ , so $F P$ is parallel to $A C$ . Similarly, $E P$ is parallel to $A B$ . Using similar triangles gives
$1 = \frac{B P}{B C} + \frac{C P}{B C} = \frac{A E}{A C} + \frac{A F}{A B} = \frac{A B \cos A}{A C} + \frac{A C \cos A}{A B}$
so $\cos A = \frac{12}{25}$ . Then by the law of cosines, $B C^{2} = 3^{2} + 4^{2} - 2(3)(4)\left(\frac{12}{25}\right) = \frac{337}{25}$ .

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