IMO 2015 Team Selection Tests

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

$T^{\prime }$ be an inside point of the quadrilateral $PEDF$ such that $\angle DE{T}^{\prime }=\angle ABC$ and $\angle DF{T}^{\prime }=\angle ACB$. We need to show that $T^{\prime }\equiv T$. We have $$\begin{array}{rl}\angle PED& =\angle PEA+\angle FEA+\angle FED\\ \angle PBA+\angle ABC+\angle PBC& =2\angle ABC=2\angle DE{T}^{\prime },\end{array}$$ so $E{T}^{\prime }$ is the bisector of the angle $PED$. Similarly, $F{T}^{\prime }$ is the bisector of the angle $PFD$. Note that $$\angle FEA=\angle ABC=\angle ABP+\angle PBC=\angle AEP+\angle FEQ.$$ Hence, $\angle DE{T}^{\prime }=\angle PEA+\angle FED$. On the other hand, $\angle DE{T}^{\prime }=\angle FE{T}^{\prime }+\angle FED$. These imply that $\angle T^{\prime }EF=\angle PAE$. Therefore, $T^{\prime }E$ and $AE$ are isogonal conjugation in the triangle $PEF$. Similarly, $T^{\prime }F$ and $AF$ are isogonal conjugation in the triangle $PEF$. Hence, $AP$ and $T^{\prime }P$ are isogonal conjugation in the same triangle. Since $AP$ is the bisector of $\angle EPF$, $T^{\prime }P$ is also the bisector of this angle, or $AP$ goes through $T^{\prime }$. Hence, $T^{\prime }$ is the center of the inscribed circle of $PEDF$. This implies that $T^{\prime }$ belongs to the bisector of $EDF$, or $T^{\prime }\equiv T$.

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 $(J)$ ($DY\ne DF$). 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.