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Singapore Mathematical Olympiad (SMO)

Geometry Difficulty 4.8 AIME Prove it Singapore

The incircle of $\triangle ABC$ touches the sides $BC$ , $CA$ , $AB$ , at $D$ , $E$ , $F$ respectively. A circle through $A$ and $B$ encloses $\triangle ABC$ and intersects the line $DE$ at points $P$ and $Q$ . Prove that the midpoint of $AB$ lies on the circuncircle of $\triangle PQF$ .

Solution

Let $M$ be the midpoint of $AB$ . If $DE \parallel AB$ , then $\triangle ABC$ is isosceles with $CA = CB$ , and $F$ coincides with $M$ .

Consider the case where $DE \parallel AB$ . Let the lines $DE$ and $AB$ intersect at $X$ . By Menelaus theorem, $\frac{AX}{XB} = \frac{BD}{CD} = \frac{CE}{EA} = 1$ .

$\therefore\ XA \cdot BF = XB \cdot AF$ (since $CD = CE$ , $BD = BF$ , $AE = AF$ .)

$\therefore\ XA \cdot (XB - XF) = XB \cdot (XF - XA)$ (since $BF = XB - XF$ , $AF = XF - XA$ .)

$\therefore\ 2XA \cdot XB = (XA + XB) \cdot XF = 2XM \cdot XF$

$\therefore\ XM \cdot XF = XA \cdot XB = X'P \cdot XQ \Rightarrow P, Q, F, M$ are concyclic.

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