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

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[[File:2018 IMO 6.png|490px|right]]
 
[[File:2018 IMO 6.png|490px|right]]
 
[[File:2018 IMO 6 Claim 3.png|370px|right]]
 
[[File:2018 IMO 6 Claim 3.png|370px|right]]
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[[File:2018 IMO 6a.png|490px|right]]
 
<i><b>Special case</b></i>
 
<i><b>Special case</b></i>
  
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<i><b>Claim 3</b></i> The sum of the arcs between the points of intersection of two perpendicular circles is <math>180^\circ.</math>
 
<i><b>Claim 3</b></i> The sum of the arcs between the points of intersection of two perpendicular circles is <math>180^\circ.</math>
 
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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<i><b>Common case </b></i>
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Denote by <math>O</math> the intersection point of the midpoint perpendicular of the segment <math>AC</math> and the line <math>BD.</math> Let <math>\omega</math> be a circle (red) with center <math>O</math> and radius <math>OA.</math>
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The points <math>B</math> and <math>D</math> are symmetric with respect to the circle <math>\omega</math> <i><b>(Claim 1).</b></i>
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The circles <math>BDF</math> and <math>BDE</math> are orthogonal to the circle <math>\omega</math> <i><b>(Claim 2).</b></i>
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Circles <math>ACF</math> and <math>ACE</math> are symmetric with respect to the circle <math>\omega</math> <i><b>(Lemma).</b></i>

Revision as of 14:51, 17 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
2018 IMO 6a.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 $\Omega.$ $\angle X_0AB = \angle X_0CD$ (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 $\omega, l$ 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 midpoint perpendicular of the segment $AC$ and the line $BD.$ 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).

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