Let $D$ and $E$ be the midpoints of sides $[AB]$ and $[AC]$ of the triangle $ABC$. The circle of diameter $[AB]$ intersects the line $DE$ on the opposite side of $AB$ than $C$, in $X$. The circle of diameter $[AC]$ intersects $DE$ on the opposite side of $AC$ than $B$ in $Y$ . Let $T$ be the intersection of $BX$ and $CY$. Prove that the orthocenter of triangle $XY T$ lies on $BC$.
Problem
Source: 2014 Romania JBMO TST 1.5
Tags: geometry, orthocenter, circles, midpoint
dangkhuong
31.05.2020 19:36
What is the point T?
parmenides51
31.05.2020 19:41
Let $T$ be the intersection of $BX$ and $CY$
Ricochet
01.06.2020 22:39
First notice that $AXTY$ is cyclic, now let $H$ be the foot of the perpendicular from $T$ onto $BC$ and let $H'= TH \cap (AXTY)$. If we prove that $H'$ is the reflection of $H$ across $XY$ we are done, indeed see that $\angle AH'T=\angle AXT=90^{\circ}$, so $AH'\|XY\|BC$ since $D, E$ are midpoints of $AB, BC$ resp. its easy to conclude that $M=HH' \cap XY$ is the midpoint of $HH'$ and we are done.
dgreenb801
02.06.2020 11:34
Since $DE$ is a midsegment of $\triangle ABC$, the distance from $A$ to line $DE$ (which is line $XY$) equals the distance from line $BC$ to line $XY$. Thus, there is a point $G$ on $BC$ such that $AXGY$ is a parallelogram. Then I claim $G$ is the orthocenter of $\triangle XYT$. Indeed, since $GX \parallel AY$ and $AY \perp YC$ (since $AC$ is a diameter), we have $GX \perp YT$. Similarly $GY \perp XT$. Thus, $G$ is the orthocenter of $\triangle XYT$ and lies on $BC$. I may add a video later