SinaQane wrote: The altitudes $AA_1$ and $CC_1$ of an acute-angled triangle $ABC$ intersect at point $H$. A straight line passing through $H$ parallel to line $A_1C_1$ intersects the circumscribed circles of triangles $AHC_1$ and $CHA_1$ at points $X$ and $Y$, respectively. Prove that points $X$ and $Y$ are equidistant from the midpoint of segment $BH$. Here is my solution: Let $D$ be the midpoint of $BH$. It's easy to see that $D$ is the center of $(BA_{1}C_{1})$. We also have $\triangle{BAC}$ ~ $\triangle{BA_{1}C_{1}}$. Hence, there exists a similarity with center $B$ whichs turns $A$ into $A_{1}$, $C$ into $C_{1}$, $D$ into $O$. Thus, $\angle{DA_{1}C_{1}} = \angle{OCA} = \angle{C_{1}CA_{1}} = \angle{HYA_{1}}$. We only need to prove that $D, A_{1},Y$ are collinear. Which is true if and only if $\angle{BA_{1}D} = \angle{CA_{1}Y} \Leftrightarrow \angle{DBA_{1}} = \angle{YHC} \Leftrightarrow \angle{A_{1}C_{1}H} = \angle{YHC}$ which is obviously true since $HY / /A_{1}C_{1}$. So $\triangle{DC_{1}A_{1}}$ ~ $\triangle{DXY}$. Because $DA_{1} = DC_{1}$, we deduce that $DX=DY$. End of the solution.

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SinaQane wrote: The altitudes $AA_1$ and $CC_1$ of an acute-angled triangle $ABC$ intersect at point $H$. A straight line passing through $H$ parallel to line $A_1C_1$ intersects the circumscribed circles of triangles $AHC_1$ and $CHA_1$ at points $X$ and $Y$, respectively. Prove that points $X$ and $Y$ are equidistant from the midpoint of segment $BH$. Can I know about the source of this problem? What is "239 2013 J3"?

Akatsuki1010 wrote: Can I know about the source of this problem? What is "239 2013 J3"? It's an open olympiad held by a school named 239 in Saint Petersburg. Its full name is "Open Math Olympiad of 239 Presidential Physics and Mathematics Lyceum in Saint-Petersburg". You can see its problems here.

Let $M$ be the midpoint of $\overline{BH}$ and $B_1$ be the foot of $\overline{BH}$ on $\overline{AC}$, so $AB_1C_1H$ and $CB_1A_1H$ are cyclic. It suffices to show that $M,C_1,X$ and $M,A_1,Y$ are collinear, since we know $MC_1=MA_1$ and can thus finish with a homothety. To that end, define $X'=\overline{MC_1} \cap (AB_1C_1H) \neq C_1$. Then because $M$ lies on the perpendicular bisector of $\overline{C_1H}$, $B_1HC_1X$ is an isosceles trapezoid, hence $\measuredangle C_1HX'=\measuredangle B_1XH=\measuredangle B_1AH=\measuredangle HC_1A_1$, hence $\overline{X'H} \parallel \overline{A_1C_1}$. If $Y'$ is defined similarly, we also have $\overline{Y'H} \parallel \overline{A_1C_1}$, hence $X',H,Y'$ are collinear with $\overline{X'HY'} \parallel \overline{A_1C_1}$, i.e. $X'=X,Y'=Y$, as desired. $\blacksquare$

Let $M,N$ be the midpoints of $AC,BH$ respectively. $AX\perp \overline{XHY}\perp CY$ Also $MN\perp A_1C_1$ since $A_1C_1$ the radical axis of $(AC)$ and $(BH)$. And $M$ is on the perpendicular bisector of $XY$ since $MA=MC$ and $AX\parallel CY$. Thus $N$ is also on the perpendicular bisector of $XY$ which gives that $NX=NY$ as desired.$\blacksquare$