Is it too hard for you? I would like to see a solution.

I didn't attacked the problem seriously because I was very busy recently. It was the reason to not post a solution. Most of the competitors failed on this problem. Thebault theorem was not allowed to be used. It was Bulgaria 2010 MO last round hardest problem. The level of math education in Bulgaria decreases year by year. I like the problem. It was reason to post the problem.

Ok. It's indeed very hard problem. I can't solve it and will be happy to see solution. Common, mathlinkers!

Dear Mr. Borislav, I think that Sawayama-Thebault's theorem can be used in this problem as a lemma. Here is my proof: Let $X, Y$ be the points of contact of $(I)$ and $(J)$ with $(O)$. If $ABIJ$ is not a trapezoid then $AB$ intersects $IJ$ at exsimilicenter $V$ of $(I)$ and $(J)$. But $X$ and $Y$ are the exsimilicenter of $(I)$ and $(O), (J)$ and $(O)$, respectively so by Monge & D' Alembert 's theorem we obtain $V, X, Y$ are collinear. Therefore $\overline{VX}.\overline{VY}=\overline{VA}.\overline{VB}=\overline{VI}.\overline{VJ}$, which follows that $XYJI$ is a cyclic quadrilateral. (1) Let $M$ be the second intersection of $XY$ and $(J)$. Since $V$ is the estimilicenter of $(I)$ and $(J)$ we get $IX//JM$. This means $\angle IXY=\angle JMY=\angle JYM$. (2) Combining (1) and (2) we claim $XYJI$ is a trapezoid, or $XY//IJ//AB$. Let $E$ be the incenter of triangle $ABC$; $K, H, F$ be the tangencies of $(J), (I), (E)$ and $BC$, respectively. By Sawayama-Thebault's theorem, $E$ lies on $IJ$. But $IJ//AB$ hence $JEFK$ is a rectangle. We deduce that $JE=KF$. On the other side, $EH\perp ID$ and $JD\perp ID$ so $EH//JD$. Moreover, $JE//DH$ thus $JEHD$ is a parallelogram. This means $DH=JE=KF$. By Pythagore 's theorem: $AK=\sqrt{AJ^2-JK^2}=\sqrt{BI^2-IH^2}=BH$. Therefore, $DH+HB=KF+KA$ or $AF=BD$, which is equivalent to $D$ is the tangency of C-excircle and $AB$. Our proof is completed.

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Dear Nguyen, Thank you for the excellent solution! I think it is very interesting problem and I like it. Probably it can be solved without Thebault theorem. If it is not the case I'm not sure the problem is good for olympiad. I think good olympiad problems should seems easy, to have cool application of the theory, to be beatiful statements and to use only well known facts. In other case chances for good presentation for the pupils from small schools and the pupils from specialized schools from big cities are not the same. It is only mine opinion. Am I right?

Yes, you are right. And it will be nice if you post a solution without using Sawayama-Thebault 's theorem. A solution with basic facts is always the best.

The Sawayama-Thebault's can be avoided, also Monge & D' Alembert 's theorem. This can be done only using fakt that if there is a homotheties h1,h2 and h3 qith koeficients k1,k2 and k3 such that h3 = h1 * h2 then k3 = k1 * k2 (this fact dear Borislav is well known from school and can be easily proved). So to avoid Monge & D' Alembert 's theorem: you can use that homoteties : h1(V, k1) (sending (I) to (J)), h2(I, k2) (sending (O) to (I)), h3(J,k3) (sending (O) to (J)). It can easily verified that h1 = h2 * h3 then k1 = k2 * k3 (i'm not taking care of signs of homoteties) then by opposite Minelaus theorem we conclude that X, Y and V are collinear(k1 * k3 * 1/ k1 = 1 are exactly proportions from the opposite Minelaus theorem). Then to avoid Sawayama-Thebault's we use that ABIJ is issosles trapecoid and it easily follows that E, I and J are collinear (because simply distances from them to AB is one and the same).

georgi111 wrote: The Sawayama-Thebault's can be avoided, also Monge & D' Alembert 's theorem. This can be done only using fakt that if there is a homotheties h1,h2 and h3 qith koeficients k1,k2 and k3 such that h3 = h1 * h2 then k3 = k1 * k2 (this fact dear Borislav is well known from school and can be easily proved). So to avoid Monge & D' Alembert 's theorem: you can use that homoteties : h1(V, k1) (sending (I) to (J)), h2(I, k2) (sending (O) to (I)), h3(J,k3) (sending (O) to (J)). It can easily verified that h1 = h2 * h3 then k1 = k2 * k3 (i'm not taking care of signs of homoteties) then by opposite Minelaus theorem we conclude that X, Y and V are collinear(k1 * k3 * 1/ k1 = 1 are exactly proportions from the opposite Minelaus theorem). Then to avoid Sawayama-Thebault's we use that ABIJ is issosles trapecoid and it easily follows that E, I and J are collinear (because simply distances from them to AB is one and the same). Dear georgi111, As you said, we can avoid Monge & D' Alembert 's theorem by using Menelaus's theorem. But you should know that it is only used to reprove Monge & D' Alembert 's theorem. And can you prove for me that E, I and J are collinear by using the distances from them to AB? I think this way is nice

hello livetolove212, I think yes it is not quite clear why these 3 distances from E, I and J are the same. The key fakt to avoid The Sawayama-Thebault's theorem is this : you prove by perfect reasoning that ABIJ is isosless trapezoid (because XYJI is such one - this is proved). You can continue the following way : prove that JK = r (inradius) (it can be done different ways) If this is proved it immediately follows that E, I , J are on the same line. When i have time i'll write this

it's some old problem! (see Sharigin's book or other ...). We have a $r(I)=r(J)$ if and only if $D$ is so.($M\equiv D$).

Redefine the two incenters to be $I_1$ and $I_2$, and let the incircles be $\omega_1$ and $\omega_2$. Let them touch $AB$, $CD$ and $(ABC)$ respectively at $P$, $Q$, $X$ and $R$, $S$, $Y$. Finally, let $I_C$ be the $C$-excenter. Claim 1: $\overline{DI_C} \perp \overline{I_1I_2}$. Proof. This is true regardless of the condition that $A$, $B$, $I_1$, $I_2$ are concyclic. Let $M$, $N$ and $O$ be the midpoints of arc $BC$, $PR$ and $ID$ respectively. Since $DI_C \parallel MO$, it suffices to show that $MO \perp I_1I_2$. In fact, we claim that this line is the radical axis of the two circles. Since $M$ lies on both $\overline{PX}$ and $\overline{RY}$, it follows that \[MP \cdot MX = MA^2 = MR \cdot MY\]and hence $M$ lies on the radical axis. Since $NP^2 = NR^2$, $N$ lies on the radical axis too. Now by the curvilinear incircle lemma, $I$ lies on both $\overline{PQ}$ and $\overline{RS}$. Since $\angle PIR = 180^\circ - \angle I_1DI_2 = 90^\circ$, $\triangle INP$ is isosceles and hence $IN \parallel CD$. Consequently we have \[\frac{IM}{ME} \cdot \frac{EN}{ND} \cdot \frac{DO}{OI} = \frac{MB}{ME} \cdot \frac{EI}{IC} = \frac{AC}{AE}\cdot\frac{EA}{AC} = 1\]and hence $O$, $M$, $N$ are collinear by Menelaus's theorem and the claim is proved. $\square$ By Pappus theorem, we also see that $I_1$, $I$ and $I_2$ are collinear. Claim 2: $I_1I_2 \parallel AB$. Proof. Suppose that this is not true and let them intersect at $K$. Let $Z$ be the intersection of $\overline{DI_C}$ and $\overline{I_1I_2}$. Then this point lies on $(AIB)$. Therefore, \[KI_1 \cdot KI_2 = KA \cdot KB = KI \cdot KZ \Longrightarrow \frac{KZ}{KI_2} = \frac{KI_1}{KI} = \frac{KD}{KR}\]and hence $ZD \parallel I_2R$. But this means that $I_1I_2$ and $AB$ are both perpendicular to $ZD$ which is a contradiction. Hence $I_1I_2 \parallel AB$. $\square$ Now since $D$ is the foot of perpendicular from $I_C$ to $\overline{AB}$ by claim 2, we are done. [asy][asy] size(11cm); defaultpen(fontsize(10pt)); pair C = dir(145); pair A = dir(220); pair B = dir(320); real s = .6; pair D = s*B + (1-s)*A; pair I = incenter(A,B,C); pair M = 2*foot(circumcenter(A,B,C),C,I)-C; pair Ia = 2*M - I; pair P = extension(A,B,I,foot(I,D,bisectorpoint(C,D,A))); pair Q = extension(C,D,I,foot(I,D,bisectorpoint(C,D,A))); pair R = extension(A,B,I,foot(I,D,bisectorpoint(B,D,C))); pair S = extension(C,D,I,foot(I,D,bisectorpoint(B,D,C))); pair I1 = 2*circumcenter(P,Q,D)-D; pair I2 = 2*circumcenter(S,R,D)-D; pair K = extension(I1, I2, A, B); pair Z = extension(Ia, D, I1, I2); pair X = 2*foot(circumcenter(A,B,C),M,P)-M; pair Y = 2*foot(circumcenter(A,B,C),M,R)-M; pair N = midpoint(P--R); pair O = midpoint(I--D); pair E = extension(C,I,A,B); draw(A--B--C--cycle, black+1); draw(C--D); draw(C--Ia); draw(K--I2); draw(K--A); draw(Z--Ia); draw(M--X, dotted); draw(M--Y, dotted); draw(P--Q, dotted); draw(I--R, dotted); draw(I--D); draw(M--O, royalblue+1); draw(circle(I1, abs(I1 - P))); draw(circle(I2, abs(I2 - R))); draw(circumcircle(A,B,C)); draw(circumcircle(A,I,B)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I_1$", I1, dir(90)); dot("$I_2$", I2, dir(90)); dot("$K$", K, dir(180)); dot("$M$", M, dir(270)); dot("$I_C$", Ia, dir(270)); dot("$I$", I, dir(90)); dot("$P$", P, dir(270)); dot("$R$", R, dir(270)); dot("$Q$", Q, dir(60)); dot("$S$", S, dir(60)); dot("$D$", D, dir(225)); dot("$N$", N, dir(270)); dot("$O$", O, dir(180)); dot("$Z$", Z, dir(135)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$E$", E, dir(225)); [/asy][/asy]