Nice problem. I will reformulate it as follows. Let $A_1A_2A_3$ be a non-equilateral triangle with circumcentre $O$ and incentre $I$, and let $T_1$, $T_2$, $T_3$ be the tangency points of the incircle of $A_1A_2A_3$ with $A_2A_3$, $A_3A_1$, $A_1A_2$ respectively. Let $S_1S_2S_3$ be the orthic triangle of $T_1T_2T_3$ (that is, the triangle formed by the feet of the altitudes of $T_1T_2T_3$). Let $H$ be the orthocentre of triangle $T_1T_2T_3$. a) Prove that the lines $A_1S_1$, $A_2S_2$ and $A_3S_3$ are concurrent. b) Prove that $O$, $I$ and $H$ are collinear. Solution. a) By an easy angle chase we deduce that $A_1A_2 \parallel S_1S_2$, $A_2A_3 \parallel S_2S_3$ and $A_3A_1 \parallel S_3S_1$. Hence the triangles $A_1A_2A_3$ and $S_1S_2S_3$ are homothetic. This means that the lines $A_1S_1$, $A_2S_2$ and $A_3S_3$ are concurrent at some point $P$ (which is the centre of the homothety mapping $A_1A_2A_3$ to $S_1S_2S_3$.) b) Notice that $H$ is the incentre of $S_1S_2S_3$ (it is wel known that the orthocentre of a triangle is the incentre of the orthic triangle). The homothety with centre $P$ which maps $S_1S_2S_3$ to $A_1A_2A_3$ maps the incentre of $S_1S_2S_3$ to the incentre of $A_1A_2A_3$, so $H$ is mapped to $I$. Hence, $P$, $H$ and $I$ are collinear. Now, since $I$ and $H$ are the circumcentre and orthocentre of $T_1T_2T_3$ respectively, the points $P$, $H$ and $I$ lie on the Euler line of $T_1T_2T_3$. It is well-known that the circumcentre $W$ of $S_1S_2S_3$, which is the centre of the nine-point circle of $S_1S_2S_3$, lies on this Euler line as well (in fact, $W$ is the midpoint of $IH$). It follows that the points $P$, $H$, $I$ and $W$ are all collinear. The homothety mentioned above, mapping $S_1S_2S_3$ to $A_1A_2A_3$, maps the circumcentre of $S_1S_2S_3$ to the circumcentre of $A_1A_2A_3$, so $W$ is mapped to $O$. Hence $P$, $W$ and $O$ are collinear. Since $P$, $H$, $I$ and $W$ are collinear, we can conclude that $O$, $I$, $H$, $W$ and $P$ lie on the Euler line of $S_1S_2S_3$. Hence result.

I noticed that this problem appeared in the Bulgarian Mathematics Olympiad (2002) as well.

Are you serious??? Well, this problem was supposed to be original from Aarón (he is my instructor ) but anyway...

arent BL,CM AND BN thE ALTITUDES OF ABC?

Jutaro wrote: Let $ABC$ be a triangle. $P$, $Q$ and $R$ are the points of contact of the incircle with sides $AB$, $BC$ and $CA$, respectively. Let $L$, $M$ and $N$ be the feet of the altitudes of the triangle $PQR$ from $R$, $P$ and $Q$, respectively. a) Show that the lines $AN$, $BL$ and $CM$ meet at a point. b) Prove that this points belongs to the line joining the orthocenter and the circumcenter of triangle $PQR$. This problem is due to Aarón Ramírez (El Salvador) The problem is hyper-well-known; it was given to us at the German IMO training seminar in Oberwolfach 2004, just to mention one of its occurences. Of course, it's possible that Aarón Ramírez independently discovered this fact, as one often discovers old things when doing geometry. In http://www.mathlinks.ro/Forum/viewtopic.php?t=6228 post #8, I proved that the triangles NLM and ABC are homothetic; this immediately yields the concurrence of the lines AN, BL and CM, so that part a) is proven. I denoted the center of homothety - i. e., the point of concurrence of the lines AN, BL and CM - by S, and then showed that this point S lies on the Euler line of triangle PQR, i. e. on the line joining the orthocenter and the circumcenter of triangle PQR. So we have also proven part b). @ Marko: No, the lines AN, BL and CM are not the altitudes of triangle ABC (else, the orthocenter of triangle ABC would lie on the Euler line of triangle PQR, what is wrong ). The lines AN, BL and CM concur at the so-called isogonal Mitten point of triangle ABC; this is $X_{57}$ in Kimberling's ETC. If this says something to you. Darij

thanks darij its just that inmy figure it looks like it was true. anyway the centroamerican olympiad is a contest for people less than 15 years and though dariij´s solution is okay there has to be a more geometrical way because the arguments darij´s uses are most of the time not known bye students who participate in this competition. so anyone??

Arne wrote: Nice problem. I will reformulate it as follows. Let $A_1A_2A_3$ be a non-equilateral triangle with circumcentre $O$ and incentre $I$, and let $T_1$, $T_2$, $T_3$ be the tangency points of the incircle of $A_1A_2A_3$ with $A_2A_3$, $A_3A_1$, $A_1A_2$ respectively. Let $S_1S_2S_3$ be the orthic triangle of $T_1T_2T_3$ (that is, the triangle formed by the feet of the altitudes of $T_1T_2T_3$). Let $H$ be the orthocentre of triangle $T_1T_2T_3$. a) Prove that the lines $A_1S_1$, $A_2S_2$ and $A_3S_3$ are concurrent. b) Prove that $O$, $I$ and $H$ are collinear. Solution. a) By an easy angle chase we deduce that $A_1A_2 \parallel S_1S_2$, $A_2A_3 \parallel S_2S_3$ and $A_3A_1 \parallel S_3S_1$. Hence the triangles $A_1A_2A_3$ and $S_1S_2S_3$ are homothetic. This means that the lines $A_1S_1$, $A_2S_2$ and $A_3S_3$ are concurrent at some point $P$ (which is the centre of the homothety mapping $A_1A_2A_3$ to $S_1S_2S_3$.) b) Notice that $H$ is the incentre of $S_1S_2S_3$ (it is wel known that the orthocentre of a triangle is the incentre of the orthic triangle). The homothety with centre $P$ which maps $S_1S_2S_3$ to $A_1A_2A_3$ maps the incentre of $S_1S_2S_3$ to the incentre of $A_1A_2A_3$, so $H$ is mapped to $I$. Hence, $P$, $H$ and $I$ are collinear. Now, since $I$ and $H$ are the circumcentre and orthocentre of $T_1T_2T_3$ respectively, the points $P$, $H$ and $I$ lie on the Euler line of $T_1T_2T_3$. It is well-known that the circumcentre $W$ of $S_1S_2S_3$, which is the centre of the nine-point circle of $S_1S_2S_3$, lies on this Euler line as well (in fact, $W$ is the midpoint of $IH$). It follows that the points $P$, $H$, $I$ and $W$ are all collinear. The homothety mentioned above, mapping $S_1S_2S_3$ to $A_1A_2A_3$, maps the circumcentre of $S_1S_2S_3$ to the circumcentre of $A_1A_2A_3$, so $W$ is mapped to $O$. Hence $P$, $W$ and $O$ are collinear. Since $P$, $H$, $I$ and $W$ are collinear, we can conclude that $O$, $I$, $H$, $W$ and $P$ lie on the Euler line of $S_1S_2S_3$. Hence result. I don't understand, isn't the idea to prove that $P$, $H$ and $I$ are collinear ?

marko avila wrote: thanks darij its just that inmy figure it looks like it was true. anyway the centroamerican olympiad is a contest for people less than 15 years and though dariij´s solution is okay there has to be a more geometrical way because the arguments darij´s uses are most of the time not known bye students who participate in this competition. so anyone?? It's basically the same, let $ H $ the orthocenter of the triangle $PQR $. Let $ I $ the incenter of the triangle $ ABC $. Notice that $ I $ is the circumcenter of the triangle $ PQR $. Part a) Since $LMN$ is the triangle orthic of the triangle $PQR$, then $ \angle PLN = \angle PRQ $. Then, as $ I $ is the circumcenter of the triangle $PQR$, then $\angle PIQ = 2 \angle PRQ $, but $ IB $ is bisecting the angle $ \angle PIQ $, then $ \angle BIQ = \angle PRQ $. But see that $BQIP$ is cyclic ($ \angle BPI + \angle IQB = 90 +90 = 180 $) therefore $ \angle BPQ = \angle BIQ $. From the above, $ \angle BPL = \angle BPQ = \angle $ PLN and thus $ BA \parallel LN$. Similarly one has $ NM \parallel AC $ and $ RQ \parallel AB$. Then $LMN$ is similar to the triangle $ ABC $. Let $ X $ the intersection of the lines $ AN $ y $ CM $. We will show that $ A, L, X $ are collinear, and thus lines must be satisfied in one spot. To do this, see that being $ MN \parallel AC $, then triangles $XAC$ and $XNM $ are similar. In turn, as the triangle $ LNM $ is similar to the triangle $ ABC $, then $ \displaystyle k = \frac{LN}{AB} = \frac{NM}{AC} = \frac{XN}{AX}$. In turn, $ \angle LNX = \angle BAX$ because $LN \parallel AB$. This approach implies that $\triangle LNX \ sim \triangle BAX$. From the above, $ \ angle AXB = \angle NXL = \angle AXL $ where $ L $ is over $ BX $ and conclude that $ A, L, X $ are collinear. Part b) It is well known that the orthocenter of a triangle is the incenter of its orthic triangle. Also that the circuncenter of the triangle $ PQR $ is the incenter of the triangle $ ABC $. Thus $ H $ is the incenter of the triangle $ LMN$. See triangles $ ABC $ and $ LMN $, for his likeness, $ H $ in $ \triangle LMN$ would be $ I $ in $ \triangle ABC $, but $ X $ is the same in both. Of this, $ X, H, I $ are collinear and this follows the result. $QWEXT$

From here, we get that $\Delta LMN$ and $\Delta ABC$ are homological, hence, the conclusion follows from Desargues' Theorem For the second part, Let the concurrency be $X$, then, $\Delta XLM \sim \Delta XBC$ and cyclically, hence, $X$ is the homothety center mapping $\implies$ $X$ maps the incircle of $\Delta LMN$ to $\Delta ABC$ $\implies $ hence done!