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Problem 1894

IMO P2/P5; hard shortlist

Geometry Difficulty 9.1 Prove it IMO 2015 Team Selection Tests · Vietnam · 2015

Given an acute, non-isosceles triangle $ABC$ , and a point $P$ inside the triangle such that $\angle APB = \angle APC = \alpha$ with $\alpha > 80^\circ - \angle BAC$ . The circle $(APB)$ intersects the line $AC$ at $E$ , the circle $(APC)$ intersects the line $AB$ at $F$ . Let $Q$ be the inside point of the triangle $AEF$ such that $\angle AQE = \angle AQF = \alpha$ . Let $D$ be the symmetric point of $Q$ through $EF$ . The bisector of $\angle EDF$ intersects $AP$ at $T.

a) Prove that $\angle DET = \angle ABC$ , $\angle DFT = \angle ACB$ .

b) The line $AP$ intersects the line $DE$ , $DF$ respectively at $M$ , $N$ . Let $I$ and $J$ be centers of incircles of $PEM$ and $PFN$ respectively. Let $K$ be the center of $(DIJ)$ . The line $DT$ intersects $(K)$ at $H$ . Prove that $HK$ goes through the incircle center of the triangle $DMN$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

a) Since the quadrilaterals $APBE$ and $APCF$ are inscribed, we have
$\angle BEC = \angle BEA = \angle CFA = \angle CFB = 180^\circ - \alpha.$
Hence, the quadrilateral $BCFE$ is inscribed in a circle. We have two triangles $ABC$ and $AEF$ are similar, so the transformation from $ABC \to AEF$ , mapping $P$ into $Q$ , then $\angle PBC = \angle QEF = \angle DEF$ and $\angle PCB = \angle QFE = \angle DFE$ . Let

$T'$ be an inside point of the quadrilateral $PEDF$ such that $\angle DET' = \angle ABC$ and $\angle DFT' = \angle ACB$ . We need to show that $T' \equiv T$ .
We have
$\begin{aligned} \angle PED &= \angle PEA + \angle FEA + \angle FED \\ \angle PBA + \angle ABC + \angle PBC &= 2\angle ABC = 2\angle DET', \end{aligned}$
so $ET'$ is the bisector of the angle $PED$ . Similarly, $FT'$ is the bisector of the angle $PFD$ . Note that
$\angle FEA = \angle ABC = \angle ABP + \angle PBC = \angle AEP + \angle FEQ.$
Hence, $\angle DET' = \angle PEA + \angle FED$ . On the other hand, $\angle DET' = \angle FET' + \angle FED$ . These imply that $\angle T'EF = \angle PAE$ . Therefore, $T'E$ and $AE$ are isogonal conjugation in the triangle $PEF$ . Similarly, $T'F$ and $AF$ are isogonal conjugation in the triangle $PEF$ . Hence, $AP$ and $T'P$ are isogonal conjugation in the same triangle. Since $AP$ is the bisector of $\angle EPF$ , $T'P$ is also the bisector of this angle, or $AP$ goes through $T'$ . Hence, $T'$ is the center of the inscribed circle of $PEDF$ . This implies that $T'$ belongs to the bisector of $EDF$ , or $T' \equiv T$ .

b) Let $L$ be the incenter of the triangle $DMN$ . Note that $DL$ and $DT$ are bisectors of two complement angles $\angle EDF$ and $\angle MDF$ so $\angle KDH = 90^\circ$ . We only need to show that $L$ lies on the circle $(DIJ)$ .
Since $IL$ goes through $M$ , we have
$\angle ILJ = \angle LNM + \angle LMN = \frac{1}{2}(\angle DMN + \angle DNM).$
We need to show that
$\angle IDJ = \frac{1}{2}(\angle DMN + \angle DNM) = \frac{1}{2}\angle EDF.$
Draw a tangent line $DY$ at $Y$ to the circle $(J)$ ( $DY \neq DF$ ). Let $X$ be the intersection between $DY$ and $PN$ . In the complete quadrilateral formed by the lines $FD$ , $FP$ , $DX$ , $PX$ , we have $DF + PX = PF + DX$ or $DF - PF = DX - PX$ . On the other hand, there is a circle inscribed in the quadrilateral $PEDF$ so $DF - PF = DE - PF$ . These imply that $DX - PX = DE - PE$ or $DX + PE = DE + PX$ , or there is a circle inscribed inside $PEDX$ . Hence, $DY$ is tangent to $(I)$ . By the property of the tangent lines, $DI$ and $DJ$ are bisectors of $\angle EDX$ and $\angle FDX$ . The second part of the problem follows.

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