37th Iranian Mathematical Olympiad

Let $P$ be the $A$-mixtilinear touchpoint with $\Gamma$.

---

We know that $P,I$ and $T$ are collinear. First suppose that we have $IJ\parallel AP$. We have $$\angle ITX=\angle PTX=\angle PAX=\angle IJX⟹\angle ITX=\angle IJX.$$ So, $IJTX$ is cyclic and it suffices to show that $IJ\parallel AP$. We know that $EF,AT$ and $BC$ are concurrent at some point $Q$. Let $D$ be the intersection point of lines $AI,BC$ and $N$ be the intersection point of line $AI$ and circumcircle of $△ABC$. Suppose that $QN$ intersects the circumcircle of $△ABC$ at $X^{\prime }(X^{\prime }\ne N)$ and let $J^{\prime }$ be the intersection point of the lines $A{X}^{\prime }$ and $EF$. We have $$(A{X}^{\prime },BC)\stackrel{\{}{N}\}\{=\}(DQ,BC)\stackrel{\{}{A}\}\{=\}-1.$$ Therefore, $X^{\prime }$ lies on the $A$-symmedian of $△ABC$. Since $S$ is where the circumcircles of two triangles $△ABC$ and $△AEF$ meet for the second time, we know that $S$ is the Miquel's point of the quadrilateral $BCEF$ and quadrilateral $SQBF$ is cyclic. So $$\angle SQF=\angle SBF=\angle SNA⟹\angle SQK=\angle SNK,$$ and $SQNK$ is cyclic. Therefore \begin\{aligned\} \angle AJK &= \angle ASK = \angle AS'K = \angle ASN - \angle NSK = \angle AX'N - \angle NQK \\ &= \angle AJ'K. \end\{aligned\} Which gives us $\angle AJK=\angle A{J}^{\prime }K$. So $J$ and $J^{\prime }$ are symmetrical with respect to line $AI$. Now let $X_A$ be the touchpoint of $A$-excircle with side $BC$. Since --- ## IRN_ABooklet_2020 — Page 53 $J$ and $J^{\prime }$ are symmetrical, it suffices to show that $A{X}_A\parallel I{J}^{\prime }$ to get $AP\parallel IJ$. If $M$ is the midpoint of $BC$, we know that $MI\parallel A{X}_A$. So we just need to prove that $M,I$ and $J^{\prime }$ are collinear which is true since $SQNK$ is cyclic and $$(EF,J^{\prime }Q)=A(EF,JQ)=(BC,MQ).$$