Let $ BC\cap B_{1}C_{1}=X, CA\cap C_{1}A_{1}=Y, AB\cap A_{1}B_{1}=Z$. It is easy to show that $ BCB_{1}C_{1}$ is cyclic, so radical center of $ (ABC), (BCB_{1}C_{1}), (B_{1}C_{1}A_{2})$ is $ X$. $ \Rightarrow$ $ XA_{2}$ is tangent to $ (ABC)$ at $ A_{2}$. WTS : $ \frac{BA_{2}}{A_{2}C}\frac{CB_{2}}{B_{2}A}\frac{AC_{2}}{C_{2}B}=1$. But $ \frac{BA_{2}}{A_{2}C}=\frac{XA_{2}}{XC}=\frac{\sqrt{XB\cdot XC}}{XC}=\sqrt{\frac{XB}{XC}}$, $ \frac{BX}{XC}\frac{CB_{1}}{B_{1}A}\frac{AC_{1}}{C_{1}B}=1 \Rightarrow \frac{XB}{XC}=\frac{B_{1}A}{CB_{1}}\frac{C_{1}B}{AC_{1}}$. $ \Rightarrow \frac{BA_{2}}{A_{2}C}\frac{CB_{2}}{B_{2}A}\frac{AC_{2}}{C_{2}B}=\frac{AC_{1}}{C_{1}B}\frac{BA_{1}}{A_{1}C}\frac{CB_{1}}{B_{1}A}=1$. Is it right? BTW, Problems in this olympiad are very interesting for me. Is there any official homepage of it?

It can be also solved with the following nice lemma: In the triangle $ ABC$ let $ D, E, F$ be the points of tangency of the incircle with $ BC, CA, AB$. Let $ X, Y, Z$ be the points on incircle. Then $ XD, YE, ZF$ are concurrent iff $ AX, BY, CZ$ are concurrent. Proof is an application of a trigonometric Ceva, I can post it in case anybody is interested, it was probably on the forum already. Let $ l, k, m$ be the tangents at $ A_{2}, B_{2}, C_{2}$ and we denote by $ P, Q, R$ intersection of lines $ k, m$, $ m, l$, $ l, k$ . As Little Gauss has already mentioned line $ l$, line $ B_{1}C_{1}$ and line $ BC$ concurr at one point and similary for $ B_{2}$, $ C_{2}$. Since $ AA_{1}, BB_{1}, CC_{1}$ concurr in one point from the Desaurges theorem the intersections $ B_{1}C_{1}\cap BC$, $ C_{1}A_{1}\cap CA$, $ A_{1}B_{1}\cap AB$ are collinear (one can omit the Desaurges here and see that these intersections lie on radical axis of circumcircle of $ ABC$ and circumcircle of $ A_{1}B_{1}C_{1}$). Once more from the Desaurges theorem we conclude that lines $ AP, BQ, CR$ concurr in one point. But note that the circumcircle of $ ABC$ is an incircle of $ PQR$ so from the lemma we are done.

Actually, this lemma is called Steinbart's Theorem !

Nice! Let $A_0B_0C_0$ be a triangle whose intouch triangle is $ABC$. Let $T_C=A_1B_1 \cap AB$ and define other analogues of it. Clearly, $\ell=T_AT_BT_C$ is the tripolar of $H$ and these three points are collinear. Now, by the radical axis theorem, one can see that $T_CC_2$ is tangent to $(ABC)$. Therefore, the pole of $\ell$ in the circumcircle $(O)$ of $ABC$ is the intersection point of the three lines $C_0C_2,B_0B_2,A_0A_2$. We see that by the Steinbart theorem, the lines $AA_2,BB_2,CC_2$ are concurrent. The result holds.

TomciO wrote: It can be also solved with the following nice lemma: In the triangle $ ABC$ let $ D, E, F$ be the points of tangency of the incircle with $ BC, CA, AB$. Let $ X, Y, Z$ be the points on incircle. Then $ XD, YE, ZF$ are concurrent iff $ AX, BY, CZ$ are concurrent. Shouldn't this lemma have some more assumptions,? Let's take three cevians concurrent in $P$ inside the incircle and not lying on any of $AD$, $BE$, $CF$. Let $AP$ intersect incircle in $X$ and $X'$. Let one of intersection points of $BP$ with incircle be $Y$ and of $CP$ be $Z$. Now triples $AX $, $BY $, $CZ $ and $AX' $, $BY $, $CZ $ can't be both concurrent.

Little Gauss wrote: Let $ BC\cap B_{1}C_{1}=X, CA\cap C_{1}A_{1}=Y, AB\cap A_{1}B_{1}=Z$. It is easy to show that $ BCB_{1}C_{1}$ is cyclic, so radical center of $ (ABC), (BCB_{1}C_{1}), (B_{1}C_{1}A_{2})$ is $ X$. $ \Rightarrow$ $ XA_{2}$ is tangent to $ (ABC)$ at $ A_{2}$. WTS : $ \frac{BA_{2}}{A_{2}C}\frac{CB_{2}}{B_{2}A}\frac{AC_{2}}{C_{2}B}=1$. Hello, how does $\frac{BA_{2}}{A_{2}C}\frac{CB_{2}}{B_{2}A}\frac{AC_{2}}{C_{2}B}=1$ prove the concurrency? Isn't Ceva supposed to be on the sides?

Dear Mathlinkers, for the Steinbart's theorem, you can see and more on http://jl.ayme.pagesperso-orange.fr/Docs/Les%20points%20de%20Steinbart%20et%20de%20Rabinowitz.pdf Sincerely Jean-Louis

jayme wrote: Dear Mathlinkers, for the Steinbart's theorem, you can see and more on http://jl.ayme.pagesperso-orange.fr/Docs/Les%20points%20de%20Steinbart%20et%20de%20Rabinowitz.pdf Sincerely Jean-Louis do you have this article in english version???

pohoatza wrote: $ AA_{1}$, $ BB_{1}$, $ CC_{1}$ are altitudes of an acute triangle $ ABC$. A circle passing through $ A_{1}$ and $ B_{1}$ touches the arc $ AB$ of its circumcircle at $ C_{2}$. The points $ A_{2}$, $ B_{2}$ are defined similarly. Prove that the lines $ AA_{2}$, $ BB_{2}$, $ CC_{2}$ are concurrent.

Little Gauss wrote: Let $ BC\cap B_{1}C_{1}=X, CA\cap C_{1}A_{1}=Y, AB\cap A_{1}B_{1}=Z$. It is easy to show that $ BCB_{1}C_{1}$ is cyclic, so radical center of $ (ABC), (BCB_{1}C_{1}), (B_{1}C_{1}A_{2})$ is $ X$. $ \Rightarrow$ $ XA_{2}$ is tangent to $ (ABC)$ at $ A_{2}$. WTS : $ \frac{BA_{2}}{A_{2}C}\frac{CB_{2}}{B_{2}A}\frac{AC_{2}}{C_{2}B}=1$. But $ \frac{BA_{2}}{A_{2}C}=\frac{XA_{2}}{XC}=\frac{\sqrt{XB\cdot XC}}{XC}=\sqrt{\frac{XB}{XC}}$, $ \frac{BX}{XC}\frac{CB_{1}}{B_{1}A}\frac{AC_{1}}{C_{1}B}=1 \Rightarrow \frac{XB}{XC}=\frac{B_{1}A}{CB_{1}}\frac{C_{1}B}{AC_{1}}$. $ \Rightarrow \frac{BA_{2}}{A_{2}C}\frac{CB_{2}}{B_{2}A}\frac{AC_{2}}{C_{2}B}=\frac{AC_{1}}{C_{1}B}\frac{BA_{1}}{A_{1}C}\frac{CB_{1}}{B_{1}A}=1$. Is it right? BTW, Problems in this olympiad are very interesting for me. Is there any official homepage of it? Ya its correct