Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Find the answer

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 $100 a+b$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

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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