Really nice one, but slightly easy to be in the Chinese TST. Just a simple application of a bit generalized Pythagorean theorem applied plenty of times. In fact the point $ I$ does not play a great role here, the only thing matters is this point to lie on $ OI$. Image not found

Can you post a solution because I cannot solve it.Thanks.

Only calculate: $ AP\perp BC\leftrightarrow \frac{1}{4}(AB^2-AC^2)=AM^2-AN^2=PM^2-PN^2$ $ =PK^2+KM^2-PL^2-NL^2=PI^2-IK^2+KM^2-(PI^2-IL^2)-NL^2$ $ =IL^2-IK^2+KM^2-NL^2=IE^2-EN^2-ID^2+DM^2=(IS^2+ES^2)-(IR^2+RD^2)-EN^2+DM^2$ $ =ES^2-RD^2-EN^2+DM^2=(a-(p-c))^2-(a-(p-b))^2-(a-b/2)^2+(a-c/2)^2$ It's true! We end the proof

Erken wrote: In fact the point $ I$ does not play a great role here, the only thing matters is this point to lie on $ OI$. From here, we have $ OI\perp DE$ so for every point $ I'$ lies on $ OI$: $ I'D^2-I'E^2=ID^2-IE^2$ and we get the result.

My solution : From $ OI \perp DE $ (well-known) $ \Longrightarrow \triangle ADE $ and $ \triangle INM $ are orthogonal , so the perpendicular from $ A, D, E $ to $ NM, IM, IN $, resp are concurrent $ \Longrightarrow AP \perp MN $ . i.e. $ AP \perp BC $ Q.E.D

Dear Mathlinkers, see also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=395892 Sincerely Jean-Louis

Let $P, Q$ be the points of tangency of the incircle $\omega$ with sides $\overline{AB}, \overline{AC}$, respectively. Let $r$ denote the radius of $\omega$, and let $\Gamma_b, \Gamma_c, \gamma$ be the circle centered at $M$ with radius $MA$, the circle centered at $N$ with radius $NA$, and the circle centered at $I$ with radius $R := \sqrt{r^2 + s^2 - a^2 + bc}$, respectively. We claim that $P$ is the radical center of $\Gamma_b, \Gamma_c, \gamma.$ This will imply the desired result, because the radical axis of $\Gamma_b, \Gamma_c$ is just the $A$-altitude of $\triangle ABC.$ To see this, we apply the Pythagorean Theorem to $\triangle IPD$ to obtain \[ID^2 = IP^2 + PD^2 = r^2 + (BD - BP)^2 = r^2 + (s - c)^2.\] Meanwhile, $MA^2 = \tfrac{1}{4}c^2$ and $MD^2 = |BD - BM|^2 = |a - \tfrac{c}{2}|^2.$ Upon expansion, it is straightforward to see that $MD^2 - MA^2 = ID^2 - R^2.$ Therefore, $D$ has equal power w.r.t. $\Gamma_b$ and $\gamma.$ Then because $PD \perp MI$, it follows that $PD$ is the radical axis of $\Gamma_b$ and $\gamma.$ Similarly, $PE$ is the radical axis of $\Gamma_c$ and $\gamma$, and therefore $P$ is the radical center of these three circles. $\square$

TelvCohl wrote: My solution : From $ OI \perp DE $ (well-known) $ \Longrightarrow \triangle ADE $ and $ \triangle INM $ are orthogonal Excuse me, but what does it for two triangles to be orthogonal to each other? Having orthic centre?

Another name for the relation is for the triangles to be orthologic; two triangles $ABC$ and $DEF$ are orthologic if the perpendiculars from $A$ to $EF$, $B$ to $DF$, and $C$ to $DE$ concur. By Carnot's Theorem, this is equivalent to the reverse relation, that the perpendiculars from $D$ to $BC$, $E$ to $AC$, and $C$ to $AB$ concur, so the relation is symmetric.

Basically, we want to prove that $ADE$ and $IMN$ are orthologic triangles since we want perpendiculars from $E$ to $IM$, $D$ to $IN$ and $A$ to $MN$ concurrent. It is equivalent that perpendiculars from $M$ to $AE$ and from $N$ to $AD$ and $I$ to $DE$ are concurrent. Thus we want line joining mid point of $AH$ to $I$ perpendicular to $DE$. By Linearity of power of a point applied to incircle and circumcircle we get that $OI \perp DE$ which is what we wanted.

Here is my bary solution.

@above you left out bunch of nasty calculations i don't think you're done. It suffice to prove that $\triangle IMN$ and $\triangle ADE$ are orthologic but as this is an symmetry relation it' sufficient to prove that perpendicular bisectors of $AC,AB$ and the line thru $I$ perpendicular to $DE$ are concurrent.But $IO\perp DE$ and the result follows.

nikolapavlovic wrote: @above you left out bunch of nasty calculations i don't think you're done. It suffice to prove that $\triangle IMN$ and $\triangle ADE$ are orthologic but as this is an symmetry relation it' sufficient to prove that perpendicular bisectors of $AC,AB$ and the line thru $I$ perpendicular to $DE$ are concurrent.But $IO\perp DE$ and the result follows. Why?

Generally if you are bashing, bash right to the end.Solution like "we just do this and that " usually get 1 or zero .

nikolapavlovic wrote: Generally if you are bashing, bash right to the end.Solution like "we just do this and that " usually get 1 or zero . I dont post My solution for asking How many points can I get.Also The last part can be done by everyone who knows what computation is.If you dont know how to do computations (last part)I can write for you and for everyone who doesnt know computations. You can be sure that I will write completeley my solution at exam if I can bash.

Murad.Aghazade wrote: nikolapavlovic wrote: Generally if you are bashing, bash right to the end.Solution like "we just do this and that " usually get 1 or zero . I dont post My solution for asking How many points can I get.Also The last part can be done by everyone who knows what computation is.If you dont know how to do computations (last part)I can write for you and for everyone who doesnt know computations. You can be sure that I will write completeley my solution at exam if I can bash. Heyy Don't get offended ! I am sure what he meant was... That practice solving solutions to the end That is all Of course posting or not is your own choice

Let incircle tangents $AB,CA at $$X,Y$ respectively. $AK \perp MN,EL \perp IL,DJ \perp MI$ and$ K \in MN,L \in IN,J \in MI $ Applying carnot theorem about$ \triangle MIN$ $AK,EL,JD$ are concurrent $\Leftrightarrow 0=(MK^2-KN^2)+(LN^2-LI^2)+(IJ^2-JM^2)$ $\Leftrightarrow ((\frac{AB}{2})^2-(\frac{AC}{2})^2)+(EN^2-EI^2)+(DI^2-DM^2)=0$ $\Leftrightarrow ((\frac{c}{2})^2-(\frac{b}{2})^2)+(EN^2-EY^2)+(XD^2-DM^2)=0$ $\Leftrightarrow ((\frac{c}{2})^2-(\frac{b}{2})^2)+((a-\frac{b}{2})^2-(\frac{b+c-a}{2})^2)+((a-\frac{c}{2})^2-(\frac{a+b-c}{2})^2)=0 $We are done