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, 2022

Geometry Difficulty 5.1 AIME, harder Prove it United States

Problem:
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 $AP + BP + CP$ .

Solution

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
$\begin{aligned} & \frac{1}{PB} + \frac{1}{PC} = 4, \\ & \frac{1}{PC} + \frac{1}{PA} = 5, \text{ and } \\ & \frac{1}{PA} + \frac{1}{PB} = 7. \end{aligned}$
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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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.