62nd International Mathematical Olympiad / IMO 2021 / | Problem 3 - Solution - 21 July 2021

Problem 3. Let D be an interior point of the acute triangle ABC with AB > AC so that ∠DAB = ∠CAD. The point E on the segment AC satisfies ∠ADE = ∠BCD, the point F on the segment AB satisfies ∠FDA = ∠DBC, and the point X on the line AC satisfies CX = BX. Let O₁ and O₂be the circumcentres of the triangles ADC and EXD, respectively. Prove that the lines BC, EF, and O₁O₂ are concurrent.

proposed by Mykhailo Shtandenko, Ukraine

Solution

$[asy]size(18cm);pair A, B, C, D, E, F, G, M, T, W, X, O1, O2, I, P;A = dir(120); B = dir(210); C = dir(330);I = incenter(A, B, C);D = I * 1.2 - A *.2;G = extension(B, foot(D, I, B) * 2 - D, C, foot(D, I, C) *2- D);F = foot(2 * circumcenter (D, B, G) - B, A, B);E = foot(2 * circumcenter (D, G, C) - C, A, C);T = extension(E, F, B, C);M = foot(extension(E, B, F, C), A, T);W = extension(B, M, A, C);X = extension(A, C, origin, (B+C)/2);P = foot(circumcenter(D, B, M) * 2 - D, W, D);draw(circumcircle(A, D, C), cyan);draw(circumcircle(M, D, B), cyan);draw(circumcircle(E, D, C), purple);draw(circumcircle(F, D, B), purple);draw(circumcircle(A, B, C));draw(circumcircle(D, E, X), cyan);draw(circumcircle(E, F, B));draw(circumcircle(M,B,X), heavygreen);draw(A--B--C--cycle);draw(T--B);draw(T--A);draw(A--D);draw(W--B);draw(W--A);draw(F--D--B);draw(E--D--C);draw(W--D);draw(T--E);dot("$A$", A, dir(A)*2*dir(340));dot("$B$", B, dir(B));dot("$C$", C, dir(C));dot("$T$", T, dir(T));dot("$W$", W, dir(W));dot("$D$", D, dir(270)*2);dot("$E$", E, dir(E)*2*dir(340));dot("$F$", F, dir(F));dot("$X$", X, dir(90)*2);dot("$M$", M, dir(M));dot("$P$", P, -2*dir(P));[/asy]$

We assume AB>AC .

Let $G$ be the isogonal conjugate of $D$ with respect to $\triangle ABC$ . We know that $G$ lies on $AD$ , and observe that
$\angle FBG = \angle DBC = \angle FDA = 180 - \angle FDG$ This proves that $(BFDG)$ is cyclic. Similarly, $(EDGC)$ is cyclic, so
$AF\cdot AB = AD\cdot AG = AE\cdot AC$ Therefore, $(BFEC)$ is cyclic.

Let $T = FE\cap BC$ . I claim that $(DBC)$ is tangent to $TD$ . This is because, observe that
$\angle FDE = \angle DBC + \angle DCB = 180 - \angle BDC$ which means $C$ and $F$ are isogonal with respect to $\angle BDE$ . Now, since $TB\cap AE$ and $TE\cap AB$ are isogonal with respect to $\angle BDE$ , by the second isogonality lemma, $T$ and $A$ are isogonal with respect to $\angle BDE$ . Finally, this means
$\angle TDB = \angle ADE = \angle DCA$ Therefore, $TD$ is tangent to $(BDC)$ . Now, we have $TD^{2} = TB\cdot TC$ .

Define $M$ as the miquel point of $EFBC$ . By miquel properties, we have $M\in (ABC)$ . We have
$\angle BME = 180 - \angle TMB - \angle AME = 180 - \angle C - \angle C = 180 - 2\angle C = 180 - \angle AXB$ This shows that $(AXBM)$ is cyclic. Next, let $W = AC\cap MB$ . We have
$WM\cdot WB = WA\cdot WC = WE\cdot WX$ by power of a point. This means that $W$ lies on the radical axis of $(DEX)$ and $(MDB)$ , and $W$ lies on the radical axis of $(MDB)$ and $(ADC)$ . Therefore, $WD$ is the radical axis of all three circles, and since all three circles pass through $D$ , they therefore must pass through a second point. Let's call this point $P$ . Finally, by inversion at $T$ with radius $\sqrt{TF\cdot TE}$ , we get $(MDB)$ gets sent to $(ADC)$ , so $(MDB)\cap (ADC)$ is fixed, which means $P$ is fixed. Therefore, $TP = TD$ , and since $DP$ is the radical axis of $(DEX)$ and $(DAC)$ , we have $O_{1}, O_{2}, T$ collinear.