Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.2 National olympiad Prove it Saudi Arabia

Let $ABC$ be a triangle with circumcenter $O$ . Points $P$ and $Q$ are interior to sides $CA$ and $AB$ , respectively. Circle $\mathcal{C}$ passes through the midpoints of segments $BP$ , $CQ$ , $PQ$ . Prove that if line $PQ$ is tangent to circle $\mathcal{C}$ , then $OP = OQ$ .

Solution

Let $K$ , $L$ , $M$ , $B'$ , $C'$ be the midpoints of $BP$ , $CQ$ , $PQ$ , $CA$ , and $AB$ , respectively. Since $CA \parallel LM$ , we have $\widehat{LMP} = \widehat{QPA}$ . Since $\mathcal{C}$ touches the segment $PQ$ at $M$ , we find $\widehat{LMP} = \widehat{LKM}$ . It follows
$\widehat{QPA} = \widehat{LKM} \tag{1}$

Similarly, from $AB \parallel MK$ we get
$\widehat{PQA} = \widehat{KLM} \tag{2}$
From (1) and (2) we obtain that triangles $APQ$ and $MKL$ are similar, hence
$\frac{AP}{AQ} = \frac{MK}{ML} = \frac{\frac{QB}{2}}{\frac{PC}{2}} = \frac{QB}{PC} \tag{3}$
Now (3) is equivalent to $AP \cdot PC = AQ \cdot QB$ which means that the power of points $P$ and $Q$ with respect to the circumcircle of $\triangle ABC$ are equal, hence $OP = OQ$ .

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