Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Find the answer

Triangle $A B C$ is given with $A B=13, B C=14, C A=15$ . Let $E$ and $F$ be the feet of the altitudes from $B$ and $C$ , respectively. Let $G$ be the foot of the altitude from $A$ in triangle $A F E$ . Find $A G$ .

Your answer

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

Solution

By Heron's formula we have $[A B C]=\sqrt{21(8)(7)(6)}=84$ . Let $D$ be the foot of the altitude from $A$ to $B C$ ; then $A D=2 \cdot \frac{84}{14}=12$ . Notice that because $\angle B F C=\angle B E C, B F E C$ is cyclic, so $\angle A F E=90-\angle E F C=90-\angle E B C=\angle C$ . Therefore, we have $\triangle A E F \sim \triangle A B C$ , so $\frac{A G}{A D}=\frac{A E}{A B} ; \frac{1}{2}(B E)(A C)=84 \Longrightarrow B E=\frac{56}{5} \Longrightarrow A E=\sqrt{13^{2}-\left(\frac{56}{5}\right)^{2}}=\sqrt{\frac{65^{2}-56^{2}}{5^{2}}}=\frac{33}{5}$ . Then $A G=A D \cdot \frac{A E}{A B}=12 \cdot \frac{33 / 5}{13}=\frac{396}{65}$ .

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