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NMO Selection Tests for BMO and IMO

Geometry Difficulty 8.4 Shortlist Prove it Romania

Let $ABC$ be a triangle such that $CA \ne CB$ , and let $D, F$ , and $G$ be the midpoints of the sides $AB, AC$ , and $BC$ , respectively. A circle $\gamma$ through $C$ and tangent to $AB$ at $D$ meets the segments $AF$ and $BG$ at $H$ and $I$ , respectively. Reflect $H$ and $I$ across $F$ and $G$ , respectively, to obtain $H'$ and $I'$ , respectively. The line $H'I'$ meets the lines $CD$ and $FG$ at $Q$ and $M$ , respectively, and the line $CM$ meets $\gamma$ again at $P$ . Prove that the triangle $CPQ$ is isosceles.
IMO 2015 Shortlist

Solution

Letting $DF$ and $DG$ meet $\gamma$ again at $R$ and $S$ , respectively, we claim that $R$ and $S$ both lie on the line $H'I'$ .

Notice that $\angle H'CQ = \angle SDC = \angle SRC$ and $\angle QCI' = \angle CDR = \angle CSR$ to deduce that $(CH'Q, RCQ)$ and $(CI'Q, SCQ)$ are pairs of similar triangles, so $QH' \cdot QR = QC^2 = QI' \cdot QS$ .
Finally, invert from $Q$ with radius $QC$ (Figure 2) and notice that $R, C$ and $S$ are mapped to $H', C$ and $I'$ , respectively, to infer that $\gamma$ , the circumcircle of the triangle $RCS$ , is mapped to $\omega$ , the circumcircle of the triangle $H'CI'$ . Since $P$ and $C$ both lie on either circle, and the latter is fixed under the inversion, so is the former. Consequently, $QP^2 = QC^2$ , i.e., $QP = QC$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.