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Irish Mathematical Olympiad

Geometry Difficulty 6.7 National Olympiad Prove it Ireland

$A$ , $B$ and $C$ are points on the circumference of a circle with centre $O$ , such that $\triangle ABC$ is not a right-angled triangle. The point $P$ lies on the circumcircle $\Gamma_1$ of the triangle $OAB$ such that $OP$ is a diameter of $\Gamma_1$ . The point $Q$ lies on the circumcircle $\Gamma_2$ of the triangle $OAC$ such that $OQ$ is a diameter of $\Gamma_2$ . Tangents are drawn to the circles $\Gamma_1$ and $\Gamma_2$ at $P$ and $Q$ respectively; these two tangents intersect at $K$ . The line $CA$ meets the circle $\Gamma_1$ at $A$ and $X$ . Prove that $X$ lies on the line $KO$ .

Solution

Case 1: $X$ and $P$ lie on the same side of the line $AO$ .

Step 1: $\angle QAO = 90^\circ$ (angle in a semicircle) and similarly $\angle PAO = 90^\circ$ .
Therefore $PAQ$ is a straight line and is a tangent to the circumcircle of $\triangle ABC$ at the point $A$ .

Step 2: Note also that $CX \perp OQ$ since $OA = OC$ and $OQ$ is a diameter.

Step 3: Now
$\begin{aligned} \angle AXO &= \angle APO \text{ (subtending same arc)} \\ &= 90^\circ - \angle KPQ \text{ since } KP \text{ is a tangent at } P. \end{aligned}$
Also $\angle AXO = \angle CXO = 90^\circ - \angle XOQ$ since $OQ \perp CX$ , and therefore $\angle KPQ = \angle XOQ$ .

Step 4: Now $\angle KPO + \angle KQO = 90^\circ + 90^\circ = 180^\circ$ , so $KPOQ$ is a cyclic quadrilateral.

Step 5: Therefore $\angle KOQ = \angle KPQ$ .

Step 6: Therefore $\angle KOQ = \angle XOQ$ .

Step 7: Therefore $KO$ passes through $X$ .

Case 2: $X$ and $P$ lie on opposite sides of the line $AO$ .

We have
$\angle AXO + \angle APO = 180^\circ$

and so
$\begin{aligned} \angle AXO &= 180^\circ - \angle APO \\ &= 180^\circ - (90^\circ - \angle KPQ) = 90^\circ + \angle KPQ. \end{aligned}$
Also
$\begin{aligned} \angle AXO &= 180^\circ - \angle CXO \\ &= 180^\circ - (90^\circ - \angle XOQ) \\ &= 90^\circ + \angle XOQ. \end{aligned}$
Therefore $\angle KPQ = \angle XOQ$ .

Step 4: Now $\angle KPO + \angle KQO = 90^\circ + 90^\circ = 180^\circ$ , so $KPOQ$ is a cyclic quadrilateral.

Step 5: Therefore $\angle KOQ = \angle KPQ$ .

Step 6: Therefore $\angle KOQ = \angle XOQ$ .

Step 7: Therefore $KO$ passes through $X$ .

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