Didier wrote: Altitudes $AH1$ and $AH2$ of acute triangle $ABC$ intersect at $H$. Let $w1$ be the circle that goes through $H2$ and touches the line $BC$ at $H1$, and let $w2$ be the circle that goes through $H1$ and touches the line $AC$ at $H2$. Prove, that the intersection point of two other tangent lines $BX$ and $CY$( $X$ and $Y$ are different from $H1$ and $H2$) to circles $w1$ and $w2$ respectively, lies on the circumcircle of triangle $HXY$. How are there 2 altitudes?

Oh, sorry. Thanks, edited.

I'm a bit confused as the problem statement appears to be incorrect? Judging by a diagram the circle $(HXY)$ does not pass through the intersection of $BX$ and $CY$ (which I called $P$ in my diagram). Here's my diagram below. Is there something wrong with it? [asy][asy] size(400); import olympiad; pair A,B,C,H,H1,H2,X,Y,P,O1,O2; A=(0,0); B=(4,-3.5); C=(-2,-3.5); H1=foot(A,B,C); H2=foot(B,A,C); H=extension(A,H1,B,H2); pair M,N; M=midpoint(H1--H2); N=rotate(90)*(H1-M)+M; O1=extension(A,H1,M,N); O2=extension(B,H2,M,N); path w1,w2; w1=circle(O1,arclength(O1--H1)); w2=circle(O2,arclength(O2--H1)); Y=reflect(O2,C)*H2; X=reflect(O1,B)*H1; P=extension(C,Y,B,X); draw(A--B--C--cycle); draw(w1,blue); draw(w2,blue); draw(C--P--X); draw(A--H1); draw(B--H2); dot(P,red+linewidth(3)); draw(circumcircle(H,X,Y),red); label("$A$",A,dir(90)); label("$B$",B,E); label("$C$",C,SW); label("$H_1$",H1,S); label("$H_2$",H2,NW); label("$X$",X,dir(90)); label("$Y$",Y,S); label("$P$",P,SE); label("$H$",H,dir(120)); label("$\omega_1$",w1); label("$\omega_2$",w2); [/asy][/asy]

Sorry, again(. In original text it was exactly like I`ve written. But now I understand there was some typo (in the original)). Thanks, edited again. ( It should be $AY$ instead of $CY$)

If someone has any ideas, PLEASE, post it.)

Is it supposed to be easy?

For IMO geometers it is supposed to be medium or even easy. On the contest this problem was solved by two people: Fedir Yudin and Ihor Pylaiev.

lahmacun wrote: Is it supposed to be easy? In the official solution was used only simple angle chasing+trigonometry. I`m looking for alternative solutions, that`s why I posted the problem.

I solved this by proving $H_2HY$ is similar to $XHH_1$ But used trigonometry as well. Any solution without trigonometry?

If anyone is interested at the official solution in Ukrainian : https://matholymp.com.ua/wp-content/uploads/2020/10/1-tur-rozvyazannya-19-20.pdf P.S Too long to translate

teddy8732 wrote: I solved this by proving $H_2HY$ is similar to $XHH_1$ But used trigonometry as well. Any solution without trigonometry? I`m still interested at your solution. Can you post it, please?

I solved this problem by 3steps. step1) Let $H_1X$ and $H_2Y$ intersects at $T$. Then, $T$ lies on $CH$. use the Desargues' theorem. step2)$ (H_1,H,Y,T)$ and $(H_2,H,X,T)$ is cyclic easy angle chasing. step3) the intersection of $BX$ and $AY$ lies on the circumcircle of $\triangle {HXY}$. also easy angle chasing.

This is the whole solution.. $D$ is the intersection of $CH$ and $AB.$ step1) Let $H_1X$ and $H_2Y$ intersects at $T$. Then, $T$ lies on $CH$. Let $AH_1$ intersects $w_2$ and $YH_2$ at $R,V, BH_2$ intersects $w_1$ and $XH_1$ at $Q,U.$ since $BQ*BH_2=BH_1^2$ and $BH*BH_2=BD*BA-->\frac{BH}{BQ}=\frac{BD*BA}{BH_1^2}=\frac{CB*BH_1}{BH_1^2}=\frac{CB}{BH_1}$ So, $\frac{QH}{BQ}=\frac{CH_1}{BH_1}.$ by the same way, $\frac{RH}{AR}=\frac{CH_2}{AH_2}.$ Let $Z$ the intersection of $H_1H_2$ and $AB.$ Then, $\frac{QH}{BQ}*\frac{RA}{HR}*\frac{ZB}{ZA}=\frac{CH_1}{BH_1}*\frac{AH_2}{CH_2}*\frac{DB}{DA}=1---> (Z,Q,R)$colinear. since $(B,U;Q,H_2)=(A,V;R,H_1)=-1--> (Z,R,Q)$colinear. So, $\triangle{AH_2V}$ and $\triangle{BH_1U}$ is perspective$--->(C,H,T)colinear$. step2)$ (H_1,H,Y,T)$ and $(H_2,H,X,T)$ is cyclic $\angle{H_2HT}=180-\angle{CHH_2}=180-\angle{CH_1H_2}=180-\angle{H_2XH_1}=\angle{H_2XT}-->(H_2,H,X,T)$cyclic by the same way, $(H_2,H,X,T)$ cyclic step3) the intersection of $BX$ and $AY$ lies on the circumcircle of $\triangle {HXY}$. Just simple angle chasing...(i don't have the energy to write it