BMO 2019 Shortlist

We will prove that the desired point of concurrency is the midpoint of $BC$. Assume that $△ABC$ is acute. Let $(ABC{)}^5$ intersect $(AEF)$ at the point $Y^{\prime }$; we will prove that $Y=Y^{\prime }$. Figure 7: G7 Using the fact that $H$ is the incenter of $△DEF$ we get that $D$, $E^{\prime }$, $F$ and $D$, $F^{\prime }$, $E$ are triples of collinear points. Furthermore, $$90^{\circ }={\angle }^{6}AEH=\angle A{F}^{\prime }H=\angle A{E}^{\prime }H=\angle AFH\Rightarrow F^{\prime },E^{\prime },H\in (AEF{Y}^{\prime }).$$ We will now prove that the points $Y^{\prime }$, $B$, $D$, $F^{\prime }$ are concyclic. Indeed, $$\angle Y^{\prime }BD=\angle Y^{\prime }BC=\angle Y^{\prime }AC=\angle Y^{\prime }AE=\angle Y^{\prime }{F}^{\prime }E\Rightarrow (Y^{\prime },B,D,F^{\prime }).$$ Now, as $$\angle F^{\prime }{Y}^{\prime }B=\angle F^{\prime }DC=\angle EDC=\angle CAB=\angle C{Y}^{\prime }B,$$ the points $C$, $F^{\prime }$, $Y^{\prime }$ are collinear. Similarly we get that $B$, $E^{\prime }$, $Y^{\prime }$ are collinear, which implies $$Y^{\prime }=Y=(ABC)\cap (AEF).$$ ⁵$(XYZ)$ denotes the circumcircle of $△XYZ$ ⁶$\angle$ denotes a directed angle modulo $\pi$ ---

Since we proved this property using directed angles, we know that it is also true for obtuse triangles. Notice that the points $A$, $B$, $C$, $H$ form an orthocentric system; in other words $H$ is the orthocenter of $△ABC$ and $A$ is the orthocenter $△HBC$. Furthermore, notice that $F^{\prime }$ is to $△ABC$ as $E^{\prime }$ is to $△HBC$ and that $E^{\prime }$ is to $△ABC$ as $F^{\prime }$ is to $△HBC$. This means that $X$ is to $△HBC$ as $Y$ is to $△ABC$ and, as we know the proven property is also true for obtuse triangles, we get $$X=(HBC)\cap (AEF).$$ By Reflecting the Orthocenter Lemma we know that in a triangle $ABC$, the reflection of its orthocenter over the midpoint of $BC$ is the antipode of $A$ w.r.t. $(ABC)$. Applying this Lemma on the triangles $ABC$ and $HBC$ we get that $YH$ and $AX$ both go through the midpoint of $BC$, thus finishing the solution. □