Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.5 National olympiad Prove it

18. Let $M$ be a point inside $\triangle ABC$ , and let $P, Q, R$ be the points of intersection of line $AM$ with $BC$ , line $BM$ with $AC$ , and line $CM$ with $AB$ , respectively. Prove that: $\frac{A M}{M P} \cdot \frac{B M}{M Q} \cdot \frac{C M}{M R} \geqslant 8$ . (1997 Macedonian Mathematical Olympiad

Solution

18. From the area relationship of the triangle, we have
$\begin{array}{l} \frac{M P}{A P}+\frac{M Q}{B Q}+\frac{M R}{C R}=\frac{S_{\triangle M B C}}{S_{\triangle A B C}}+\frac{S_{\triangle M C A}}{S_{\triangle A B C}}+\frac{S_{\triangle M A B}}{S_{\triangle A B C}}= \\ \frac{S_{\triangle A B C}}{S_{\triangle A B C}}=1 \\ \text { Let } x=\frac{M P}{A P}, y=\frac{M Q}{B Q}, z=\frac{M R}{C R} \text {, then } x+y+z=1 \text {, } \end{array}$
$\begin{array}{l} \frac{1-x}{x} \cdot \frac{1-y}{y} \cdot \frac{1-z}{z}= \\ \frac{y+z}{x} \cdot \frac{z+x}{y} \cdot \frac{x+y}{z} \geqslant \frac{2 \sqrt{y z}}{x} \cdot \frac{2 \sqrt{z x}}{y} \cdot \frac{2 \sqrt{x y}}{z}=8 \end{array}$

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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.