Let $ABC$ be an acute triangle. Let $D$, $E$ and $F$ be the feet of the altitudes from $A$, $B$ and $C$ respectively and let $H$ be the orthocenter of $\triangle ABC$. Let $X$ be an arbitrary point on the circumcircle of $\triangle DEF$ and let the circumcircles of $\triangle EHX$ and $\triangle FHX$ intersect the second time the lines $CF$ and $BE$ second at $Y$ and $Z$, respectively. Prove that the line $YZ$ passes through the midpoint of $BC$.
Problem
Source: 2021 Macedonian Balkan MO TST - Problem 1
Tags: geometry, cyclic quadrilateral
L567
18.08.2021 18:40
Note that $\angle FXE = 180 - \angle FDE = 180 - (180 - 2A) = 2A$ Since $\angle FXZ = \angle FHB = A$ and $\angle EPY = \angle EHY = A$, we have that $X,Y,Z$ are collinear. Also, since $XYZ$ is the angle bisector in $\triangle EXF$, it passes through the midpoint of arc, which is $M$, the midpoint of $BC$. So, $YZ$ passes through $M$. $\blacksquare$
yuheng305
18.08.2021 18:58
Consider $\mathcal{I}_{H}^k$ be the inversion with center $H$, $k=P_{H/(EDF)}$ $\mathcal{I}_{H}^k: M \longleftrightarrow M', X \longleftrightarrow X',Y \longleftrightarrow Y', Z \longleftrightarrow Z' $ Since: $\angle Z'HY'=\dfrac{1}{2}(\stackrel\frown{FI}+\stackrel\frown{EK})=\dfrac{1}{2}(\stackrel\frown{DI}+\stackrel\frown{DK})=\dfrac{1}{2}\stackrel\frown{IK}=\angle Z'X'Y'$ Thus, $Z',H,X',Y'$ are concyclic. Similarity, we have: $Z',M',H,X'$ are concyclic $\Rightarrow M',X',Y',Z',H$ are concyclic $\Rightarrow X,Y,Z,M$ are collinear.
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VicKmath7
18.08.2021 21:08
This is ELMO 2018 SL G1.