Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.3 National Olympiad Prove it Romania

Let $P$ be a point in the interior of the acute triangle $ABC$ and let $D$ , $E$ , $F$ be respectively the intercepts of lines $AP$ , $BP$ , $CP$ with the sides $BC$ , $CA$ , $AB$ .

a) Prove that the area of the triangle $DEF$ is not greater than a quarter of the area of triangle $ABC$ .

b) Prove that the inradius of $DEF$ is not greater than a quarter of the inradius of triangle $ABC$ .

Solution

a) Put $\frac{BD}{CD} = x$ , $\frac{CE}{AE} = y$ , $\frac{AF}{BF} = z$ to obtain $\frac{S_{AEF}}{S_{ABC}} = \frac{AF}{AB} \cdot \frac{AE}{AC} = \frac{z}{(z+1)(y+1)}$ and the similar ones. As by Ceva's theorem $xyz = 1$ , it follows $\frac{S_{DEF}}{S_{ABC}} = 1 - \sum \frac{z}{(z+1)(y+1)} = \frac{xyz+1}{(x+1)(y+1)(z+1)} = \frac{2}{(x+1)(y+1)(z+1)}$ (The symbol $S_{MNP}$ denotes the area of the triangle $MNP$ ). By the AM inequality $\frac{2}{(x+1)(y+1)(z+1)} \le \frac{2}{8\sqrt{xyz}} = \frac{1}{4}$ , concluding thus the proof.

b) One knows that the perimeter $DEF$ is at least the perimeter of the ortic triangle $A'B'C'$ . We get $r_{DEF} = \frac{S_{DEF}}{p_{DEF}} \le \frac{S_{ABC}}{4p_{A'B'C'}}$ .

$2R \sin A \sin B \sin C$ and $S_{ABC} = 2R^2 \sin A \sin B \sin C$ the conclusion follows.

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