2018 IMO Problems/Problem 6

A convex quadrilateral $ABCD$ satisfies $AB\cdot CD=BC \cdot DA.$ Point $X$ lies inside $ABCD$ so that $\angle XAB = \angle XCD$ and $\angle XBC = \angle XDA.$ Prove that $\angle BXA + \angle DXC = 180^{\circ}$

Solution

Special case

We construct point $X_0$ and prove that $X_0$ coincides with the point $X.$ Let $AD = CD$ and $AB = BC \implies AB \cdot CD = BC \cdot DA.$ Let $E$ and $F$ be the intersection points of $AB$ and $CD,$ and $BC$ and $DA,$ respectively. The points $B$ and $D$ are symmetric with respect to the circle $\omega = EACF$ (Claim 1). The circle $\Omega = FBD$ is orthogonal to the circle $\omega$ (Claim 2). Let $X_0$ be the point of intersection of the circles $\omega$ and $\Omega.$ $\angle X_0AB = \angle X_0CD$ (quadrilateral AX0CF is cyclic) and ∠X0BC = ∠X0DA (quadrangle DX0BF is cyclic) are equal to the property of opposite angles of cyclic quadrangles. This means that X0 coincides with the point X indicated in the condition. The angle ∠FCX = ∠BCX subtend the arc XF of ω, ∠CBX = ∠XDA subtend the arc XF of Ω. The sum of these arcs is 180° (Claim 3 for orthogonal circles). Hence, the sum of the arcs XF is 180°, the sum of the angles ∠XСВ + ∠XВС = 90°, ∠СХВ = 90°. Similarly, ∠AXD = 90°, that is, ∠BXA + ∠DXC = 180°.

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2018 IMO Problems/Problem 6

Solution