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Harvard-MIT Mathematics Tournament

Geometry Difficulty 5.1 AIME, harder Prove it United States

Problem:

In $\triangle ABC$ , $D$ is the midpoint of $BC$ , $E$ is the foot of the perpendicular from $A$ to $BC$ , and $F$ is the foot of the perpendicular from $D$ to $AC$ . Given that $BE = 5$ , $EC = 9$ , and the area of triangle $ABC$ is $84$ , compute $|EF|$ .

Solution

Solution:

There are two possibilities for the triangle $ABC$ based on whether $E$ is between $B$ and $C$ or not. We first consider the former case.

We find from the area and the Pythagorean theorem that $AE = 12$ , $AB = 13$ , and $AC = 15$ . We can then use Stewart's theorem to obtain $AD = 2\sqrt{37}$ .

Since the area of $\triangle ADC$ is half that of $ABC$ , we have $\frac{1}{2} AC \cdot DF = 42$ , so $DF = 14/5$ . Also, $DC = 14/2 = 7$ so $ED = 9 - 7 = 2$ .

Notice that $AEDF$ is a cyclic quadrilateral. By Ptolemy's theorem, we have $EF \cdot 2\sqrt{37} = (28/5) \cdot 12 + 2 \cdot (54/5)$ . Thus $EF = \frac{6\sqrt{37}}{5}$ as desired.

The latter case is similar.

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