Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.5 AIME, harder Find the answer

Example 4 Let $P$ be a point inside $\triangle ABC$ , and draw perpendiculars from $P$ to the three sides $BC$ , $CA$ , and $AB$ , with the feet of the perpendiculars being $D$ , $E$ , and $F$ respectively. For what position of $P$ is the value of $\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$ minimized? (22nd IMO problem)

Your answer

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

Solution

Solution: Let $B C=a, C A=b$ , $A B=c, P D=d_{1}, P E=d_{2}$ , $P F=d_{3},$ and the area of $\triangle A B C$ be $\Delta$ . Clearly, we have $2 \Delta=a d_{1}+b d_{2}+c d_{3}$ .
$\begin{array}{c} \therefore \frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F}=\frac{a}{d_{1}}+ \\ \frac{b}{d_{2}}+\frac{c}{d_{3}} \geqslant \frac{(a+b+c)^{2}}{\left(a d_{1}+b d_{2}+c d_{3}\right)}=\frac{(a+b+c)^{2}}{2 \Delta} \text {. When } \end{array}$

and only when $d_{1}=d_{2}=d_{3}$ , the equality holds, i.e., when $P$ is the incenter of $\triangle A B C$ , $\frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F}$ has the minimum value
$\frac{(a+b+c)^{2}}{2 \Delta}$

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.