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Selection tests for the Gulf Mathematical Olympiad, 2013

Geometry Difficulty 8.0 Shortlist Prove it Saudi Arabia

An acute triangle $ABC$ is inscribed in circle $\omega$ centered at $O$ . Line $BO$ and side $AC$ meet at $B_1$ . Line $CO$ and side $AB$ meet at $C_1$ . Line $B_1C_1$ meets circle $\omega$ at $P$ and $Q$ . If $AP = AQ$ , prove that $AB = AC$ .

Solution

Assume $AP = AQ$ . Because $OP = OQ$ , the line $AO$ is perpendicular to $PQ$ . But since $\angle C_1AO = 90^{\circ} - \angle ACB$ then $\angle B_1C_1A = \angle ACB$ , and therefore, quadrilateral $BCB_1C_1$ is cyclic.

Using the power of the point $O$ with respect to the circumcircle of $BCB_1C_1$ we get
$OB_1 = \frac{OB_1 \cdot OB}{R} = \frac{OC_1 \cdot OC}{R} = OC_1,$
where $R$ is the circumradius of triangle $ABC$ . Therefore, $BB_1 = CC_1$ , that is the minors $\widehat{BB_1}$ and $\widehat{CC_1}$ of the circumcircle of $BCB_1C_1$ have the same length. This is equivalent to saying that $\angle CBC_1 = \angle B_1CB$ , and therefore $AB = AC$ .

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