IMO 2021 Problem 3

Proposed by: -
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 49s

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 circumcenters of the triangles $ADC$ and $EXD$ respectively. Prove that the lines $BC$, $EF$, and $O_1 O_2$ are concurrent.

Exploration

The configuration is driven by angle equalities that force repeated isogonal relationships at the vertices $A$, $B$, and $C$. The condition $\angle DAB=\angle CAD$ places $D$ on the internal bisector of $\angle BAC$, so the point $D$ is symmetric with respect to the rays $AB$ and $AC$.

The definitions of $E$ and $F$ encode two further angle equalities that connect $D$ simultaneously to $B$ and $C$. This strongly suggests that $E$ and $F$ are determined by isogonal conjugation constraints inside triangles involving $D$, and that a hidden cyclic structure is present. The point $X$ defined by $BX=CX$ places $X$ on the perpendicular bisector of $BC$ and simultaneously on line $AC$, which makes it the intersection of $AC$ with the symmetry axis of $BC$.

The appearance of circumcenters $O_1$ and $O_2$ indicates that the final concurrency is likely a consequence of a common power or spiral similarity argument between the circumcircles $(ADC)$ and $(EXD)$. The most delicate part is relating the line $EF$ to this circumcenter line, since $E$ and $F$ are defined asymmetrically with respect to different vertices.

The expected structure is that a single point on $BC$ is simultaneously characterized as an intersection with $EF$ and as a center of a spiral similarity sending $(ADC)$ to $(EXD)$.

Problem Understanding

This is a Type A problem, since it asks to prove that three lines are concurrent.

We are given a triangle $ABC$ with an interior point $D$ satisfying $\angle DAB=\angle CAD$, so $AD$ is the internal angle bisector of $\angle BAC$. Points $E$ and $F$ are constructed on $AC$ and $AB$ using angle equalities involving $D$, and $X$ is the intersection of line $AC$ with the perpendicular bisector of $BC$. The circumcenters of triangles $ADC$ and $EXD$ are $O_1$ and $O_2$.

The task is to prove that lines $BC$, $EF$, and $O_1O_2$ meet at a single point.

The core difficulty is that the construction mixes two different geometric mechanisms. The point $D$ is tied to angle bisectors at $A$, while $E$ and $F$ are defined using angles at $D$ relative to $B$ and $C$, producing a coupled system that is not immediately cyclic. The line $O_1O_2$ comes from two unrelated circumcenters, so one must identify a hidden transformation, typically a spiral similarity or homothety, linking $(ADC)$ and $(EXD)$ through a point on $BC$.

Proof Architecture

The first structural step is to show that $\angle DAB=\angle CAD$ implies $AD$ is the internal bisector of $\angle BAC$. This will be proved directly from angle decomposition at $A$.

The second step is to prove that $A$, $D$, $C$, $E$ are concyclic after translating the condition $\angle ADE=\angle BCD$ into an angle relation involving chords of a common circle. This establishes a cyclic quadrilateral involving $A$, $D$, $C$, and $E$.

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