Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.4 National Olympiad Prove it Ibero-American Mathematical Olympiad

Problem:

$I$ is the incenter of the triangle $ABC$ . A circle with center $I$ meets the side $BC$ at $D$ and $P$ , with $D$ nearer to $B$ . Similarly, it meets the side $CA$ at $E$ and $Q$ , with $E$ nearer to $C$ , and it meets $AB$ at $F$ and $R$ , with $F$ nearer to $A$ . The lines $EF$ and $QR$ meet at $S$ , the lines $FD$ and $RP$ meet at $T$ , and the lines $DE$ and $PQ$ meet at $U$ . Show that the circumcircles of $DUP$ , $ESQ$ and $FTR$ have a single point in common.

Solution

Solution:

$D$ and $P$ are the reflections of $Q$ and $E$ respectively in the line $CI$ . Hence $PQ$ and $DE$ meet at a point on $CI$ . So $U$ lies on $CI$ . So $\angle PIU = 1/2 \angle PIE = \angle PDE$ ( $I$ is center of circle through $D$ , $P$ , $E$ ) $= \angle PDU$ (same angle). Hence $PDIU$ is cyclic. In other words, $I$ lies on the circumcircle of $DUP$ . Similarly, it lies on the circumcircles of $ESQ$ and $FTR$ .

But the same argument shows that $\angle DPT = \angle DIT$ , so $DPIT$ is cyclic. So $T$ lies on the circle through $D$ , $P$ and $I$ and hence on the circumcircle of $DUP$ . Similarly, for the other circles. So the circumcircles of $CUP$ and $FTR$ meet at $T$ and $I$ . Similarly, the circumcircles of $FTR$ and $ESQ$ meet at $S$ and $I$ , and the circumcircles of $ESQ$ and $DUP$ meet at $U$ and $I$ . So the three circumcircles have just one point in common, namely $I$ .

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