Let $H$ be the orthocenter of triangle $ABC$, and $AD$, $BE$, $CF$ be the three altitudes of triangle $ABC$. Let $G$ be the orthogonal projection of $D$ onto $EF$, and $DD'$ be the diameter of the circumcircle of triangle $DEF$. Line $AG$ and the circumcircle of triangle $ABC$ intersect again at point $X$. Let $Y$ be the intersection of $GD'$ and $BC$, while $Z$ be the intersection of $AD'$ and $GH$. Prove that $X$, $Y$, and $Z$ are collinear. Proposed by Li4 and Untro368.
Problem
Source: 2023 Taiwan TST Round 3 Independent Study 1-G
Tags: geometry
keglesnit
26.04.2023 09:39
Claim. $X,H,D'$ are colinear. Proof. Let $I=DG\cap (DEF)\neq D$, $H'=AH\cap (ABC)\neq A$, $S=AD\cap(DEF)\neq D$, and $X'$ be the image of $D'$ under inversion in $H$ with radius $\sqrt{-AH\cdot HD}$. Note that $ID'\parallel EF$. Now consider $$(B,C;H',X)\overset{A}{=}(F,E;AD\cap EF,G)\overset{D}{=}(F,E;S,I)\overset{\infty_{EF}}{=}(E,F;S,D')=(B,C;H',X')$$where the last equality follows by inversion in $H$ with radius $\sqrt{-AH\cdot HD}$. We have $X=X'$ and thus the claim is proven. Note that $S$ is the midpoint of arc $ID'$ so $\overline{DA}$ bisects $\angle GDD'$. This, combined with the fact that $\angle YDH=90^{\circ}$ gives $(D',G;DH\cap EF,Y)=-1$ so points $X$, $Y$, and $Z$ must be colinear due to harmonic bundles in a complete quad.
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VicKmath7
26.04.2023 17:44
Let $HD' \cap (ABC)=A'$ and let $M$ be the midpoint of $AH$. Since the homothety at $H$ with coefficient $2$ sends the nine-point circle to the circumcircle, we have $D'$ is midpoint of $HA'$ and thus $D'M$ is midline in $\triangle HAA'$, so $AA' \parallel BC$. Note that by inversion at $A$ with radius $\sqrt{AH.AD}$ we have that $GHDX$ is cyclic (if $X$ is the image of $G$, then $X$ lies on $(ABC)$ as $G$ lies on $EF$). Then $\angle GXH=\angle GDH=\beta-\gamma=\angle GXA'$, so $X, H, A'$ are collinear. In particular, $X, H, D'$ are collinear.
Let $AH \cap D'G=T$; then $(YT; GD')=-1$ due to the isogonality of $DD'$ and $DG$ in $\angle EDF$. Now by Ceva-Menelaus in $\triangle AXZ$ we get that $X, Y, Z$ are collinear, done.
axolotlx7
01.05.2024 18:00
This problem is a "paper tiger" - you can pretty easily work backwards and reduce the statement further and further until you end up with something trivial. Note that $DH$ bisects $\angle GDD'$, so $A(H, Y; G, D') = D(H, Y; G, D') = -1$, so by harmonic bundle properties it suffices to show that $D',H,X$ collinear. But since $D$ is the Miquel point of $AGHD'$, it suffices to prove that $GHDX$ is cyclic, which is evident by considering the inversion at $A$ swapping $EF$ with $(ABC)$.