Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.7 National Olympiad Prove it Iran

Two circles $\omega_1(O)$ and $\omega_2$ intersect each other at $A, B$ . And $O$ lies on $\omega_2$ . Let $S$ be the foot of perpendicular line to $AB$ from $O$ . Line $OS$ intersects $\omega_2$ for the second time at $P$ . The bisector of $\widehat{ASP}$ intersects $\omega_1$ at $L$ ( $A$ and $L$ are on the same side of the line $OP$ ). Let $K$ be a point on $\omega_2$ such that $PS = PK$ ( $A$ and $K$ are on the same side of the line $OP$ ). Prove that $SL = KL$ .

Solution

Let $Q$ be the intersection point of $PK$ , $AB$ and let $L'$ be the incenter of triangle $QSP$ . We'll prove that $L \equiv L'$ .

We know that $\widehat{QSP} = \widehat{OKP} = 90^\circ$ therefore $QKSO$ is a cyclic quadrilateral. And since $PS = PK$ , it's also concluded that $PQ = PO$ . Now we have
$\widehat{QLS} = 90^\circ + \frac{1}{2}\widehat{QPS} = 90^\circ + \frac{1}{2}\widehat{QPO} = 180^\circ - \widehat{QOP} = 180^\circ - \widehat{QOS}.$
So $QL'SO$ is a cyclic quadrilateral and points $Q, K, L', S, O$ lie on circle with diameter $OQ$ . Note that
$\begin{aligned} \widehat{QL'K} &= \widehat{QSK} = 90^\circ - \widehat{KSP} = \frac{1}{2}\widehat{KPS} = \widehat{KPL'} \\ \implies QK \cdot QP &= (QL')^2 \end{aligned}$
The LHS of the latest equation is the power of point $Q$ with respect to $\omega_2$ . Since $Q$ lies on the radical axis of circles $\omega_1, \omega_2$ , we have
$(QL')^2 = QK \cdot QP = QO^2 - QA^2 = (QL')^2 + (OL')^2 - OA^2$
$\implies OL' = OA.$
So $L'$ lies on $\omega_1$ which means $L$ and $L'$ are the same points. Now since $L$ lies on the angle bisector of $\widehat{KPS}$ , we finally conclude that $SL = KL$ .

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