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Difference between revisions of "2018 IMO Problems/Problem 6"

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Quadrilateral <math>AX_0CF</math> is cyclic <math>\implies</math> <cmath>\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.</cmath>
 
Quadrilateral <math>AX_0CF</math> is cyclic <math>\implies</math> <cmath>\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.</cmath>
  
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Analogically, quadrangle <math>DX_0BF</math> is cyclic <math>\implies \angle X_0BC = \angle X_0DA</math>.
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Similarly, quadrangle <math>DX_0BF</math> is cyclic <math>\implies \angle X_0BC = \angle X_0DA</math>.
  
 
This means that point <math>X_0</math> coincides with the point <math>X</math>.
 
This means that point <math>X_0</math> coincides with the point <math>X</math>.
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In the figure they are a blue and red arcs <math>\overset{\Large\frown} {CD}, \alpha + \beta = 180^\circ.</math>
 
In the figure they are a blue and red arcs <math>\overset{\Large\frown} {CD}, \alpha + \beta = 180^\circ.</math>
  
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[[File:2018 IMO 6a.png|490px|right]]
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[[File:2018 IMO 6a.png|430px|right]]
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[[File:2018 IMO 6bb.png|430px|right]]
 
<i><b>Common case </b></i>
 
<i><b>Common case </b></i>
  
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Denote by <math>X_0</math> the point of intersection of the circles <math>BDF</math> and <math>ACE.</math>
 
Denote by <math>X_0</math> the point of intersection of the circles <math>BDF</math> and <math>ACE.</math>
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Quadrangle <math>BX_0DF</math> is cyclic, hence,  <math>\angle X_0BC =  \angle X_0DA.</math>  
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Quadrangle <math>BX_0DF</math> is cyclic <math>\implies \angle X_0BC =  \angle X_0DA</math> (see Special case).
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Quadrangle <math>AX_0CE</math> is cyclic, hence,  <math>\angle X_0AB =  \angle X_0CD = \alpha.</math>  
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Similarly, quadrangle <math>AX_0CE</math> is cyclic <math>\implies \angle X_0AB =  \angle X_0CD = \alpha.</math>  
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The required point <math>X = X_0</math>  is constructed.
 
The required point <math>X = X_0</math>  is constructed.
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 +
Denote by <math>Y</math> the point of intersection of circles <math>BDF</math> and <math>ACF.</math>
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Quadrangle <math>BYDF</math> is cyclic <math>\implies  \angle CBY =  \angle ADY.</math>
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Quadrangle <math>AYCD</math> is cyclic <math>\implies  \angle YAD = \angle BCY.</math>
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The triangles <math>\triangle YAD \sim \triangle YCD</math> by two angles, so <cmath>frac {BC}{AD} = \frac {CY}{AY} = \frac {BY} {DY}.</cmath>

Revision as of 08:04, 19 August 2022

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
2018 IMO 6 Claim 3.png

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

2018 IMO 6a.png
2018 IMO 6bb.png

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 \[frac {BC}{AD} = \frac {CY}{AY} = \frac {BY} {DY}.\]

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