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

Lemma

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.

Shelomovskii, vvsss, www.deoma-cmd.ru

Video solution

https://youtu.be/cI9p-Z4-Sc8 [Video contains solutions to all day 1 problems]

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2021 IMO Problems/Problem 3

Problem

Solution

Video solution