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2018 IMO Problems/Problem 6

Revision as of 08:58, 22 August 2022 by Vvsss (talk | contribs) (→‎Solution)

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

2018 IMO 6.png

Suppose point X is unique. We will construct point $X_0$ and prove that $X_0$ coincides with the point $X.$

Special case

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 $\omega = ACEF$ (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 $\implies$ \[\angle X_0AB = \frac {\overset{\Large\frown} {X_0CE}}{2} = \frac {360^\circ -\overset{\Large\frown} {X_0AFE}}{2} = 180^\circ - \angle X_0CE = \angle X_0CD.\]

Similarly, quadrangle $DX_0BF$ is cyclic $\implies \angle X_0BC = \angle X_0DA$. This means that point $X_0$ coincides with the point $X$.

$\hspace{10mm} \angle FCX = \angle BCX = \frac {\overset{\Large\frown} {XAF}}{2}$ of $\omega.$ $\hspace{10mm} \angle CBX = \angle XDA = \frac {\overset{\Large\frown} {XBF}}{2}$ 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, $\angle AXD = 90^\circ \implies \angle BXA + \angle DXC = 360^\circ -\angle AXD -\angle CXB = 180^\circ.$

2018 IMO 6e.png
2018 IMO 6bb.png
2018 IMO 6c.png
2018 IMO 6d.png

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 $OA.$

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 $\implies \angle X_0AB = \angle X_0CD.$

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

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 $\implies \angle YAD = \angle BCY.$

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 $C$ is symmetric to itself, the point $X$ is symmetric to $Y$ with respect to $\omega \implies \frac{CX}{CY} = \frac {R^2}{OC \cdot OY} , \frac {AX}{AY} = \frac {R^2}{OA \cdot OY}.$ Usung $(1)$ and the equality $OA = OC,$ we get \[\frac{CY}{AY} = \frac {CX}{AX} = \frac{BC}{AD}.\] The point $C$ is symmetric to itself, 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 \[\frac {BX}{DY} = \frac {R^2}{OD \cdot OY} ,\frac {DX}{BY} = \frac{R^2}{OB \cdot OY}.\] \[\frac{BX}{DX} =\frac{DY}{BY} \cdot \frac {OB}{OD} = \frac{AD}{BC} \cdot \frac{BC}{CD} \cdot \frac{AB}{AD} = \frac{AB}{CD}.\]

2018 IMO 6 angles.png
2018 IMO 6 Claim 3.png
2018 IMO 6a.png
2018 IMO 60.png

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 $\psi = 180° – \varphi \implies \varphi + \psi = 180°.$

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 $ACF.$ For this, we will prove the equality $\angle AGC = \angle AFC.$

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 $ACF.$

Attention The uniqueness of the point $X$ has not been proven. In the figure, the blue and green curves show the locus points with equal pairs of angles for one case. In the general case, one can divide the plane into parts and prove that the point $X$ cannot be outside $ABCD.$ It's not done here.

vladimir.shelomovskii@gmail.com, vvsss

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