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

We want to find the point $X.$ Let $E$ and $F$ be the intersection points of $AB$ and $CD,$ and $BC$ and $DA,$ respectively. The poinx $X$ is inside $ABCD,$ so points $E,A,X,C$ follow in this order.

$\angle XAB = \angle XCD \implies \angle XAE + \angle XCE = 180^\circ$ $\implies AXCE$ is cyclic $\implies X$ lie on circle $ACE.$

Similarly, $X$ lie on circle $BDF.$

Point $X$ is the point of intersection of circles $ACE$ and $\Omega = BDF.$

Special case

Let $AD = CD$ and $AB = BC \implies AB \cdot CD = BC \cdot DA.$

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

The circle $BDF$ is orthogonal to the circle $\theta$ (Claim 2).

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

$\overset{\Large\frown} {XAF} + \overset{\Large\frown} {XBF} = 180^\circ$ (Claim 3) $\implies$ $\angle XCB + \angle XBC = 90^\circ \implies \angle CXB = 90^\circ.$

Similarly, $\angle AXD = 90^\circ \implies$ $\angle BXA + \angle DXC = 360^\circ -\angle AXD -\angle CXB = 180^\circ.$

Common case

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

We will prove $\sin\angle BXA =\sin \angle DXC$ using point $Y$ symmetric to $X$ with respect to $\omega.$

The points $B$ and $D$ are symmetric with respect to $\omega$ (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 $Y$ the point of intersection of circles $BDF$ and $ACF.$

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

Quadrangle $AYCF$ is cyclic $\implies \angle YAD = \angle YCB.$

The triangles $\triangle YAD \sim \triangle YCB$ by two angles, so $\frac {BC}{AD} = \frac {CY}{AY} = \frac {BY} {DY} \hspace{10mm} (1).$

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 circle $BDF$ orthogonal to $\omega.$

The point $B$ is symmetric to $D$ with respect to $\omega \implies$ $\triangle OBC \sim \triangle OCD \implies \frac {OB}{OC} = \frac {BC}{CD} = \frac {OC}{OD},$ $\frac {OB}{OD} = \frac {OB}{OC} \cdot \frac {OC}{OD} = \frac{BC^2}{CD^2} = \frac{BC}{CD} \cdot \frac {AB}{AD}.$ The point $B$ is symmetric to $D$ and the point $X$ is symmetric to $Y$ with respect to $\omega,$ hence $\frac {BX}{DY} = \frac {R^2}{OD \cdot OY} ,\frac {DX}{BY} = \frac{R^2}{OB \cdot OY}.$ $\frac{BX}{DX} =\frac{DY}{BY} \cdot \frac {OB}{OD} = \frac{AD}{BC} \cdot \frac{BC}{CD} \cdot \frac{AB}{AD} = \frac{AB}{CD}.$

Denote $\angle XAB = \angle XCD = \alpha, \angle BXA = \varphi, \angle DXC = \psi.$

By the law of sines for $\triangle ABX,$ we obtain $\frac {AB}{\sin \varphi} = \frac{BX}{\sin \alpha}.$

By the law of sines for $\triangle CDX,$ we obtain $\frac {CD}{\sin \psi} = \frac {DX}{\sin \alpha}.$

Hence we get $\frac{\sin \psi} {\sin \varphi}= \frac {CD}{DX} \cdot \frac{BX}{AB} = 1.$

If $\varphi = \psi,$ then $\triangle XAB \sim \triangle XCD \implies \frac {CD}{AB} = \frac {BX}{DX} = \frac{AX}{CX} = \frac {AD}{BC}.$ $CD \cdot BC = AB \cdot AD \implies AD = CD, AB = BC.$ This is a special case.

In all other cases, the equality of the sines follows $\varphi + \psi = 180^\circ .$

Claim 1 Let $A, C,$ and $E$ be arbitrary points on a circle $\omega, l$ be the perpendicular bisector 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 $\overset{\Large\frown} {CD}, \alpha + \beta = 180^\circ.$

Lemma The opposite sides of the quadrilateral $ABCD$ intersect at points $E$ and $F$ ( $E$ lies on $AB$ ). The circle $\omega$ centered at the point $O$ contains the ends of the diagonal $AC.$ The points $B$ and $D$ are symmetric with respect to the circle $\omega$ (in other words, the inversion with respect to $\omega$ maps $B$ into $D).$ Then the circles $ACE$ and $ACF$ are symmetric with respect to $\omega.$

Proof We will prove that the point $G,$ symmetric to the point $E$ with respect to $\omega,$ belongs to the circle $ACF$ becouse $\angle AGC = \angle AFC.$

A circle $BDE$ containing points $B$ and $D$ symmetric with respect to $\omega,$ is orthogonal to $\omega$ (Claim 2) and maps into itself under inversion with respect to the circle $\omega.$ Hence, the point $E$ under this inversion passes to some point $G,$ of the same circle $BDE.$

A straight line $ABE$ containing the point $A$ of the circle $\omega,$ under inversion with respect to $\omega,$ maps into the circle $OADG.$ Hence, the inscribed angles of this circle are equal $\angle ADB = \angle AGE.$ $\angle OCE = \angle CGE (CE$ maps into $CG)$ and $\angle OCE = \angle BCD (BC$ maps into $DC).$ Consequently, the angles $\angle AFC = \angle ADB – \angle FBD = \angle AGE - \angle CGE = \angle AGC.$ These angles subtend the $\overset{\Large\frown} {AC}$ of the $ACF$ circle, that is, the point $G,$ symmetric to the point $E$ with respect to $\omega,$ belongs to the circle $ACF.$

vladimir.shelomovskii@gmail.com, vvsss

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2018 IMO Problems/Problem 6

Solution