2021 IMO Problems/Problem 3

Problem

Let $D$ be an interior point of the acute triangle $ABC$ with $AB > AC$ so that $\angle DAB= \angle CAD$ . The point $E$ on the segment $AC$ satisfies $\angle ADE= \angle BCD$ , the point $F$ on the segment $AB$ satisfies $\angle FDA= \angle DBC$ , and the point $X$ on the line $AC$ satisfies $CX=BX$ . Let $O_1$ and $O_2$ be the circumcentres of the triangles $ADC$ and $EXD$ respectively. Prove that the lines $BC$ , $EF$ , and $O_1 O_2$ are concurrent.

Solution

By statement point $D$ is located on the bisector $AK$ of $\triangle ABC.$ Let $P$ be the intersection point of the tangent to the circle $\omega_2 = BDC$ at the point $D$ and the line $BC, A'$ is inverse to $A$ with respect to the circle $\Omega_0$ centered at $P$ with radius $PD.$ Then the pairs of points $F$ and $E, B$ and $C$ are inverse with respect to $\Omega_0$ , so the points $F, E,$ and $P$ are collinear. Quadrilaterals containing the pairs of inverse points $B$ and $C, E$ and $F, A$ and $A'$ are inscribed, $FE$ is antiparallel to $BC$ with respect to angle $A$ $(\boldsymbol{Lemma \hspace{3mm}1})$ .

Consider the circles $\omega = ACD$ centered at $O_1, \omega' = A'BD,$ $\omega_1 = ABC, \Omega = EXD$ centered at $O_2 , \Omega_1 = A'BX,$ and $\Omega_0.$

Denote $\angle ACB = \gamma$ . Then $\angle BXC = \angle BXE = \pi – 2\gamma,$ $\angle AA'B = \gamma (AA'CB$ is cyclic), $\angle AA'E = \pi – \angle AFE = \pi – \gamma (AA'EF$ is cyclic, $FE$ is antiparallel), $\angle BA'E = \angle AA'E – \angle AA'B = \pi – 2\gamma = \angle BXE \implies$

$\hspace{13mm}E$ is the point of the circle $\Omega_1.$

Let Y be the radical center of the circles $\omega, \omega',$ and $\omega_1.$ Let $T$ be the point of intersection $\omega \cap \omega',$ let $T'$ be the point of intersection $\omega \cap \Omega.$ Since the circles $\omega$ and $\omega'$ are inverse with respect to $\Omega_0,$ then $T$ lies on $\Omega_0,$ and $P$ lies on the perpendicular bisector of $DT.$ The power of a point $Y$ with respect to the circles $\omega, \omega',$ and $\Omega$ are the same $(\boldsymbol{Lemma\hspace{3mm}2}),$ so $DL\cdot DT = DL \cdot DT' \implies$ the points $T$ and $T'$ coincide.

The centers of the circles $\omega$ and $\Omega$ ( $O_1$ and $O_2$ ) are located on the perpendicular bisector $DT'$ , the point $P$ is located on the perpendicular bisector $DT$ and, therefore, the points $P, O_1,$ and $O_2$ lie on a line, that is, the lines $BC, EF,$ and $O_1 O_2$ are concurrent.

$\boldsymbol{Lemma \hspace{3mm}1}$

Let $AK$ be bisector of the triangle $ABC$ , point $D$ lies on $AK.$ The point $E$ on the segment $AC$ satisfies $\angle ADE= \angle BCD$ . The point $E'$ is symmetric to $E$ with respect to $AK.$ The point $L$ on the segment $AK$ satisfies $E'L||BC.$ Then $EL$ and $BC$ are antiparallel with respect to the sides of an angle $A$ and $\frac {AL}{DL} = \frac {AK \cdot DK}{BK \cdot KC}.$ Proof

Symmetry of points $E$ and $E'$ with respect bisector $AK$ implies $\angle AEL = \angle AE'L.$ $\angle DCK = \angle E'DL, \angle DKC = \angle E'LD \implies$ $\triangle DCK \sim \triangle E'DL \implies \frac {E'L}{KD}= \frac {DL}{KC}.$ $\triangle ALE' \sim \triangle AKB \implies \frac {E'L}{BK}= \frac {AL}{AK}\implies$ $\frac {AL}{DL} = \frac {AK \cdot DK}{BK \cdot KC}.$ Corollary

In the given problem $EF$ and $BC$ are antiparallel with respect to the sides of an angle $A,$ quadrangle $BCEF$ is concyclic.

$\boldsymbol{Lemma \hspace{3mm}2}$

Let the point $Y$ be the radical center of the circles $\omega, \omega', \omega_1.$ It has the same power $\nu$ with respect to these circles. The common chords of the pairs of circles $A'B, AC, DT$ intersect at this point.