Here is an another problem for study : dule_00 wrote: Let $ ABC$ be a triangle with the incenter $ I$ . Denote the points $ \|\begin{array}{c} D\in AB\ ,\ A\in (BD)\ ,\ \boxed {AD = BC} \\ \\ E\in AB\ ,\ B\in (AE)\ ,\ \boxed {BE = AC} \\ \\ P\in AI\cap CB\ ,\ Q\in BI\cap CA\end{array}$ The lines $ \overline {AIP}$ , $ \overline {BIQ}$ meet again the circumcircle $ w$ of $ ABC$ at the points $ M$ , $ N$ respectively. The ray from $ A$ to the circumcenter of $ BME$ and the ray from $ B$ to the circumcenter of $ AND$ intersect at $ X$. Prove that ... (I don't know !) In this case, I found that $ \triangle ADN\ \sim\ \triangle QBA$ , $ \triangle BEM\ \sim\ \triangle PAB$ and $ NB = ND$ , $ MA = ME$ . Denote the circumcircle $ w_a = C(O_a)$ of $ \triangle ADN$ and the circumcircle $ w_b = C(w_b)$ of $ \triangle BCM$ . Thus, $ X\in O_aB\cap O_bA$ . The line $ \overline {BIQN}$ meets again $ w_a$ in $ U$ . The line $ \overline {AIPM}$ meets again $ w_b$ in $ V$ . Then $ \|\begin{array}{c} AU = AB = BV \\ \ AU\parallel CB\ ,\ BV\parallel CA \\ \ \triangle UAD\ \equiv\ \triangle ABC\ \equiv\ \triangle BVE\end{array}$ and the circles $ w$ , $ w_a$ , $ w_b$ are equally a.s.o. (?!)

Virgil Nicula wrote: Then $ AU = AB = BV$ , $ \triangle UAD\ \equiv\ \triangle ABC\ \equiv\ \triangle BVE$ and the circles $ w$ , $ w_a$ , $ w_b$ are equally a.s.o.[/color] And then why does $ CX\perp PQ$, Virgil?

could somebody post the problem truely. which one is right, original one or virgil's one? actually i can't solve both of them therefore i'm looking for a solution also, no matter which one is right.

I agree that Virgil Nicula's problem is interesting, but I posted different problem.

anonymous1173 wrote: could somebody post the problem truely. which one is right, original one or virgil's one? actually i can't solve both of them therefore i'm looking for a solution also, no matter which one is right. You are right ! The my problem is for study. I can't solve the "hard problem" !!

Quote: You are right ! The my problem is for study. I can't solve the "hard problem" !! Don't worry Virgul For this problem there was just one student with 5 and one with 2 points, the rest are zeros. This year Serbian TST was the most difficult one in last years. Top six students received respectively: 25, 18, 16, 13, 13, 12 points. dule_00 is there any place where I can find officila solutions to this year Serbian national olympiad (TST)?

Official solutions should appear soon on http://www.srb.imomath.com, I am waiting for them too. As far as I know, solution to this problem use inversion w.r.t. circumcircle of $ \triangle ABC$.

Inversion Wauu Complex numbers are usual in Serbian competitions but inversion Thank you for link. Salem

Come on, Virgil. I know you can solve any problem in the world!

No one can solve it :

Quote: No one can solve it Question Dule, is this your own problem?

delegat wrote: Quote: No one can solve it Question Dule, is this your own problem? No, it was on Serbian selection test, as you said. It is proposed by Dusan Djukic, I don't know if he created it.

Please delate this post.

Attachments:

But MN isn't parallel to PQ.

Shishkin wrote: Inversion kills this problem. Consider inversion $ I$ in a point $ C$, which transforfs $ \triangle ABC$ into $ \triangle A'B'C = \triangle BAC$. It saves points $ O1$,$ O2$ Gm. Why? I suggest another way without inversion: prove that $ X$ is isogonally conjugate point to the circumcentre $ Y$ of $ CPQ$ w.r.t. triangle $ ABC$. The claim follows immediately. For proving these, consider points $ U$ on $ AM$ and $ V$ on $ AC$ such that $ \angle AEU=\angle AEV=\angle ACB$, so triangle $ AEV$ is similar to $ ACB$ and $ AU$ is a bisector of triangle $ EAV$. Then it is easy to calculate angles and note that $ B$, $ E$, $ U$, $ M$ are concyclic. Also note that $ AB: BE=AB: BC=AV: EV$, hence the circumcentre $ S$ of $ BMUE$ corresponds to the circumcentre $ Y$ of $ CPQ$ in similar triangles $ AEV$ and $ ACB$. Since these triangles are differently oriented and have a common angle $ A$, we conclude that $ AS$ and $ AY$ are symmetric in the bissector of $ \angle CAB$. Analagously, $ BY$ and $ BS_1$ are symmetric in the bissector of $ \angle CBA$, where $ S_1$ is a circumcentre of $ DAN$. Hence $ X$ and $ Y$ are indeed isogonally conjugate w.r.t. $ ABC$.

Thank you for this beautiful solution Fedor Petrov.

Quote: No, it was on Serbian selection test, as you said. It is proposed by Dusan Djukic, I don't know if he created it. Mislio sam da si ti Dusan Finally solution. Thank you Fedor Petrov

Finally, official solutions are published http://srb.imomath.com/zadaci/smo_2008r.pdf I have translated solution (attachment) to this problem (hope that there are no big mistakes)

Attachments:
serbia 2008(problem 2).doc (22kb)

Are there any solutions, that arent writen in cirilica?

Thank every body for nice problem.

Thank for nice solution.

Fedor Petrov wrote: prove that $ X$ is isogonally conjugate point to the circumcentre $ Y$ of $ CPQ$ w.r.t. triangle $ ABC$. . I have another solution for it(Maybe difference from Fedor Petrov ) Let $ S$ be the circumcenter of triangle $ BME$ and we produced the line $ AM$ from $ A$ meets $ (S)$ at $ T$.It is easy to see that $ \angle BET = \angle PCQ = C$ We consider a homothety center at $ A$ which sent points $ T,B,E,S$ to the points $ P,B',E',S'$,respectively and we get $ \angle AE'P = \angle BET = C$ ,since $ \angle PAC = \angle PAE'$ then, triangle $ APC$ and $ APE'$ are congruent From homothety,$ AB' : AB = AP : AT = AE' : AE = AE' : (AB + BC)$ hence $ \frac {AE'}{AB'} = \frac {(AB + BC)}{AB} = 1 + \frac {BC}{AB} = 1 + \frac {CQ}{QA}$ Then, $ \frac {CQ}{QA} = \frac {B'E'}{AB'}$ It follow immediately that triangle $ PB'E'$ and $ PQC$ are congruent Moreover,These two triangle are symmectric to other with respect to a line $ AP$ Hence,$ \angle PAY = \angle PAS'$ $ \therefore AY$ and $ AS$ are isogonal congugately wrt. triangle $ ABC$ but $ X\in AS$,then our claim is true Poon.

The position of X and Y help us have a solution. Let T,J is the circumcenter of DNA and BME , we have two similar triangle YCP and TDA, so BYC and BTD are similar. Similarly, we have X is isogonally conjugate point to Y. We have TA and BS insect at K on (ABC) and CK pass through Y, hence we have Q.E.D

It is really easy if you think to find the AE*AQ whick is equal to bc.Then you use the rootbc-inversion and you're done!If anyone want to see the full answer reply to me and i will write it down.

The problem disappears once you figure out why $D$ and $E$ are defined as they are. Notice that $\sqrt{bc}$ inversion swaps $P$ and $M$ and also $E$ and $Q$. Thus, it swaps $(MBE)$ with $(CPQ)$. Let $O$ be the circumcenter of $(CPQ)$. Then, the reflection of $O$ across the $A$-bisector is collinear with $\overline{AO_1}$; in other words, $\overline{AO}$ and $\overline{AO_1}$ are isogonal. Similarly, $\overline{BO}$ and $\overline{BO_2}$ are isogonal, thus $X$ and $O$ are isogonal conjugates. Now it follows $\overline{CX}$ and $\overline{CO}$ are isogonal conjugates, implying $\overline{CX}$ is the altitude to $\overline{PQ}$ in $CPQ$. This finishes.