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Team Selection Test for IMO, 2023

Geometry Difficulty 8.3 Shortlist Prove it Turkey

For the points $A$ , $B$ , $K$ , $L$ , $X$ located on the circle $\Gamma$ in the given order the lengths of arcs $\widehat{BK}$ and $\widehat{KL}$ are equal. The circle which passes through $A$ and is tangent to $BK$ at $B$ intersects the line segment $KX$ at points $P$ and $Q$ . The second intersection point of the circle which passes through $A$ and is tangent to $BL$ at $B$ with the line segment $BX$ is $T$ . Show that $\angle PTB = \angle XTQ$ .

Solution

The oriented angle equalities $\angle QXA = \angle KBA = \angle BPA$ and $\angle XQA = \angle PBA$ imply the oriented similarity $AXQ \sim APB$ , and the oriented angle equalities $\angle XTA = \angle LBA$ and $\angle TXA = \angle BLA$ imply the oriented similarity $AXT \sim ALB$ . If we put a C-coordinate system on the plane such that $A$ is at the origin, these similarities would be expressed as $bx = pq = tl$ . Then letting $s := tb/q = tp/x$ , one gets the oriented similarities $SBP \sim TQX$ and $STP \sim BQL$ . Consequently one gets $\angle PST = \angle LBQ = \angle LBK + \angle KBQ = \angle KXB + \angle BPQ = \angle PBT$ , hence $S$ is on the circle $(BPT)$ , hence $\angle XTQ = \angle PSB = \angle PTB$ .

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