HMMT November

Geometry Difficulty 4.9 AIME Find the answer

Suppose point $P$ is inside triangle $ABC$ . Let $AP, BP$ , and $CP$ intersect sides $BC, CA$ , and $AB$ at points $D, E$ , and $F$ , respectively. Suppose $\angle APB=\angle BPC=\angle CPA, PD=\frac{1}{4}, PE=\frac{1}{5}$ , and $PF=\frac{1}{7}$ . Compute $AP+BP+CP$ .

Your answer

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

Solution

The key is the following lemma: Lemma: If $\angle X=120^{\circ}$ in $\triangle XYZ$ , and the bisector of $X$ intersects $YZ$ at $T$ , then $\frac{1}{XY}+\frac{1}{XZ}=\frac{1}{XT}$ . Proof of the Lemma. Construct point $W$ on $XY$ such that $\triangle XWT$ is equilateral. We also have $TW \parallel XZ$ . Thus, by similar triangles, $\frac{XT}{XZ}=\frac{YT}{YX}=1-\frac{XT}{XY}$ implying the conclusion. Now we can write $\frac{1}{PB}+\frac{1}{PC}=4$ , $\frac{1}{PC}+\frac{1}{PA}=5$ , and $\frac{1}{PA}+\frac{1}{PB}=7$ . From here we can solve to obtain $\frac{1}{PA}=4, \frac{1}{PB}=3, \frac{1}{PC}=1$ , making the answer $\frac{19}{12}$ .

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