Triangle $ABC$ is inscribed in circle $\Omega$ with center $O$. The incircle of $ABC$ is tangent to $BC$, $AC$, $AB$ at $D$, $E$, $F$ respectively, and its center is $I$. The reflection of the tangent line to $\Omega$ at $A$ with respect to $EF$ will be denoted $\ell_A$. We similarly define $\ell_B$, $\ell_C$. Show that the orthocenter of the triangle with sides $\ell_A$, $\ell_B$, $\ell_C$ lies on $OI$.
Problem
Source: 2024 Israel TST Test 8 P2
Tags: geometry, geometric transformation, reflection, incircle
Geometrineq
10.05.2024 20:22
Who is the author of this problem?
huoxy1623
10.05.2024 20:48
Let A' be reflection of A wrt EF.Similarly,we have B',C'. 1.Not hard to prove that A'B'C' and pedal triangle D'E'F' of Bevan point V of ABC wrt ABC are homothetic(also congruent) 2.Finally,construct Spieker point Sp,orthocenter H' of DEF and J,reflection of H' through Sp.We see D'E'F'J and A'B'C'I are homothetic.The rest is obvious
gvole
24.05.2024 17:31
This is the same as above, but looks pretty I guess... The desired triangle is homothetic to $\triangle ABC$ due to angles. Now let $A'$ be the reflection of $A$ over $EF$ and $D'$ the reflection of $D$ over the midpoint of $BC$. Now $D'-A'=(A+B+C)-(D+E+F)=\lambda$, so $\triangle D'E'F'$ is just $\triangle A'B'C'$ translated by $\lambda$. This means that the desired triangle is just $\triangle ABC $ translated by $\lambda$, and so the desired orthocenter is just $H'=H-\lambda$, so $$\overrightarrow{OH'}=\overrightarrow{OH}-\lambda=\overrightarrow{OD}+\overrightarrow{OE}+\overrightarrow{OF}=3\overrightarrow{OI}+\overrightarrow{ID}+\overrightarrow{IE}+\overrightarrow{IF}$$But $\overrightarrow{ID}+\overrightarrow{IE}+\overrightarrow{IF}$ is parallel to $OI$ since it's a sum of unit vectors orthogonal to the sides of $\triangle ABC$, and $I$ is the orthocenter of the minor arc midpoints.