IMO Shortlist

Geometry Difficulty 8.4 Shortlist Prove it IMO

Let $ABC$ be a triangle with $AC > BC$ . Let $\omega$ be the circumcircle of triangle $ABC$ and let $r$ be the radius of $\omega$ . Point $P$ lies on segment $AC$ such that $BC = CP$ and point $S$ is the foot of the perpendicular from $P$ to line $AB$ . Let ray $BP$ intersect $\omega$ again at $D$ and let $Q$ lie on line $SP$ such that $PQ = r$ and $S, P, Q$ lie on the line in that order. Finally, let the line perpendicular to $CQ$ from $A$ intersect the line perpendicular to $DQ$ from $B$ at $E$ .
Prove that $E$ lies on $\omega$ .

Solutions — 2

Solution 1

Solution 1 (Similar Triangles).

First observe that
$\angle DPA = \angle BPC \stackrel{CP=CB}{=} \angle CBP = \angle CBD = \angle CAD = \angle PAD$
so $DP = DA$ . Thus there is a symmetry in the problem statement swapping $(A, D) \leftrightarrow (B, C)$ .
Let $O$ be the centre of $\omega$ and let $E$ be the reflection of $P$ in $CD$ which, by
$\angle CED = \angle DPC = 180^{\circ} - \angle CPB \stackrel{CP=CB}{=} 180^{\circ} - \angle PBC = 180^{\circ} - \angle DBC$
lies on $\omega$ . We claim the two lines concur at $E$ . By the symmetry noted above, it suffices to prove that $BE \perp DQ$ and then $AE \perp CQ$ will follow by symmetry.
We have $AO = PQ$ , $AD = DP$ and
$\angle DAO = 90^{\circ} - \angle ABD \stackrel{PQ \perp AB}{=} \angle DPQ.$
Hence $\triangle AOD \cong \triangle PQD$ . Thus
$\angle QDB + \angle DBE = \angle ODA + \angle DAE \stackrel{DE=DA}{=} \angle ODA + \angle AED = (90^{\circ} - \angle AED) + \angle AED = 90^{\circ}$ giving $BE \perp DQ$ as required.

Solution 2

Solution 2 (Second Circle).

As in Solution 1, we prove that $DA = DP$ and note the symmetry in the problem statement swapping $(A, D) \leftrightarrow (B, C)$ .
Let $\Gamma$ be the circumcircle of $\triangle PCD$ . Since $DP = DA$ and $\angle ACD = \angle PCD$ , the radius of $\Gamma$ is equal to that of $\omega$ . We have that
$\angle DPQ = \angle BPS = 90^{\circ} - \angle ABD = 90^{\circ} - \angle PCD.$
This, combined with $PQ$ being equal to the common circumradius of $\Gamma$ and $\omega$ , means that $Q$ is the circumcentre of $\Gamma$ .
Let the perpendiculars to $CQ, DQ$ from $A, B$ intersect at $E$ then we have
$\begin{aligned} & \angle EAC = 90^{\circ} - \angle ACQ \stackrel{QC=QP}{=} 90^{\circ} - \angle QPC = 90^{\circ} - \angle SPA = \angle CAB \Longrightarrow \angle EAB = 2\angle PAB \\ & \angle DBE = 90^{\circ} - \angle QDP \stackrel{QD=QP}{=} 90^{\circ} - \angle DPQ = 90^{\circ} - \angle BPS = \angle ABD \Longrightarrow \angle ABE = 2\angle ABP . \end{aligned}$
Combining these
$\angle BEA = 180^{\circ} - 2(\angle PAB + \angle ABP) = 180^{\circ} - 2\angle APD \stackrel{DA=DP}{=} \angle BDA$
which gives that $E$ lies on $\omega$ .

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