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 $AX_0CF$ is cyclic) and $\angle X_0BC = \angle X_0DA$ (quadrangle $DX_0BF$ is cyclic). This means that $X_0$ coincides with the point $X$ indicated in the condition.

$\angle FCX = \angle BCX$ subtend the arc $\overset{\Large\frown} {XF}$ of $\omega, \angle CBX = \angle XDA$ subtend the arc $\overset{\Large\frown} {XF}$ of $\Omega.$ The sum of these arcs is $180^\circ$ (Claim 3)..

Hence, the sum of the arcs $\overset{\Large\frown} {XF}$ is $180^\circ \implies$

the sum $\angle XCB + \angle XBC = 90^\circ \implies \angle CXB = 90^\circ.$

Similarly, $\angle AXD = 90^\circ \implies \angle BXA + \angle DXC = 180^\circ.$

Claim 1 Let A, C, and E be arbitrary points on a circle ω, l be the middle perpendicular to the segment AC. Then the straight lines AE and CE intersect l at the points B and D, symmetric with respect to ω.

Claim 2 Let points B and D be symmetric with respect to the circle ω. Then any circle Ω passing through these points is orthogonal to ω.

Claim 3 The sum of the arcs between the points of intersection of two perpendicular circles is $180^\circ.$ In the figure they are a blue and red arcs CD, α + β = 180°.

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2018 IMO Problems/Problem 6

Solution