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Romanian Mathematical Olympiad - District Round

Geometry Difficulty 5.8 AIME, harder Prove it Romania

Let $ABC$ be a triangle and let points $D, E \in (BC)$ , $F, G \in (CA)$ , $H, I \in (AB)$ such that $BD = CE$ , $CF = AG$ and $AH = BI$ . Consider $M, N, P$ the midpoint of the segments $GH$ , $DI$ , $EF$ respectively and let $M'$ be the intersection point of the lines $AM$ and $BC$ .

a. Show that $\frac{BM'}{CM'} = \frac{AG}{AH} \cdot \frac{AB}{AC}$ .

b. Prove that lines $AM$ , $BN$ and $CP$ are concurrent.

Solution

a. Set $m = \frac{BM'}{CM'}$ . We have $\overrightarrow{AM} = \frac{1}{2}\overrightarrow{AH} + \frac{1}{2}\overrightarrow{AG} = \frac{1}{2}\frac{AH}{AB} \cdot \overrightarrow{AB} + \frac{1}{2}\frac{AG}{AC} \cdot \overrightarrow{AC}$ and $\overrightarrow{AM'} = \frac{1}{m+1}\overrightarrow{AB} + \frac{m}{m+1}\overrightarrow{AC}$ . Points $A$ , $M$ , $M'$ are collinear, hence $\frac{AH}{AB} \cdot \frac{m}{m+1} = \frac{AG}{AC} \cdot \frac{1}{m+1}$ and the claim follows.

b. Define similarly the points $N'$ , $P'$ . Notice that $\frac{BM'}{CM'} \cdot \frac{CN'}{AN'} \cdot \frac{AP'}{BP'} = 1$ and apply Ceva's theorem to reach the conclusion.

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