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 $\Omega.$ Quadrilateral $AX_0CF$ is cyclic $\implies$ $\angle X_0AB = \frac {1}{2}\overset{\Large\frown} {X_0CE} = \frac {1}{2} (360^\circ -\overset{\Large\frown} {X_0AE}) = 180^\circ - \angle X_0CE = \angle X_0CD.$

Similarly, quadrangle $DX_0BF$ is cyclic $\implies \angle X_0BC = \angle X_0DA$ .

This means that point $X_0$ coincides with the point $X$ .

$\hspace{10mm} \angle FCX = \angle BCX = \frac {1}{2} \overset{\Large\frown} {XAF}$ of $\omega.$

$\hspace{10mm} \angle CBX = \angle XDA = \frac {1}{2} \overset{\Large\frown} {XBF}$ of $\Omega.$

The sum $\overset{\Large\frown} {XAF} + \overset{\Large\frown} {XBF} = 180^\circ$ (Claim 3) $\implies$

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

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 $\omega, l$ 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 $\omega.$

Claim 2 Let points $B$ and $D$ be symmetric with respect to the circle $\omega.$ Then any circle $\Omega$ passing through these points is orthogonal to $\omega.$

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 $\overset{\Large\frown} {CD}, \alpha + \beta = 180^\circ.$

Common case

Denote by $O$ the intersection point of the perpendicular bisector of $AC$ and $BD.$ Let $\omega$ be a circle (red) with center $O$ and radius $OA.$

The points $B$ and $D$ are symmetric with respect to the circle $\omega$ (Claim 1).

The circles $BDF$ and $BDE$ are orthogonal to the circle $\omega$ (Claim 2).

Circles $ACF$ and $ACE$ are symmetric with respect to the circle $\omega$ (Lemma).

Denote by $X_0$ the point of intersection of the circles $BDF$ and $ACE.$ Quadrangle $BX_0DF$ is cyclic $\implies \angle X_0BC = \angle X_0DA$ (see Special case). Similarly, quadrangle $AX_0CE$ is cyclic $\implies \angle X_0AB = \angle X_0CD = \alpha.$

The required point $X = X_0$ is constructed.

Denote by $Y$ the point of intersection of circles $BDF$ and $ACF.$

Quadrangle $BYDF$ is cyclic $\implies \angle CBY = \angle ADY.$

Quadrangle $AYCD$ is cyclic $\implies \angle YAD = \angle BCY.$

The triangles $\triangle YAD \sim \triangle YCD$ by two angles, so $\frac {BC}{AD} = \frac {CY}{AY} = \frac {BY} {DY}.$

The points $X$ and $Y$ are symmetric with respect to the circle $\omega$ , since they lie on the intersection of the circles $ACF$ and $ACE$ symmetric with respect to $\omega$ and the orthogonal $\omega$ circle $BDE.$

The point $C$ is symmetric to itself, the point $X$ is symmetric to $Y$ with respect to $\omega \implies \frac{CX}{CY} = \frac {R^2}{OC \cdot OY} , \frac {AX}{AY} = \frac {R^2}{OA \cdot OY}.$

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2018 IMO Problems/Problem 6

Solution