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Difference between revisions of "2018 IMO Problems/Problem 6"
(→‎Solution)
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==Solution==
 
==Solution==
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[[File:2018 IMO 6.png|490px|right]]
 
<i><b>Special case</b></i>
 
<i><b>Special case</b></i>
  
 
We construct point <math>X_0</math> and prove that <math>X_0</math> coincides with the point <math>X.</math>
 
We construct point <math>X_0</math> and prove that <math>X_0</math> coincides with the point <math>X.</math>
−
Let <math>AD = CD</math> and <math>AB = BC \implies  AB \cdot CD = BC \cdot DA.</math> Let <math>E</math> and <math>F</math> be the intersection points of <math>AB</math> and <math>CD,</math> and <math>BC</math> and <math>DA,</math> respectively.
+
 
 +
Let <math>AD = CD</math> and <math>AB = BC \implies  AB \cdot CD = BC \cdot DA.</math>
 +
 
 +
Let <math>E</math> and <math>F</math> be the intersection points of <math>AB</math> and <math>CD,</math> and <math>BC</math> and <math>DA,</math> respectively.
 +
 
 
The points <math>B</math> and <math>D</math> are symmetric with respect to the circle <math>\omega = EACF</math> <i><b>(Claim 1).</b></i>
 
The points <math>B</math> and <math>D</math> are symmetric with respect to the circle <math>\omega = EACF</math> <i><b>(Claim 1).</b></i>
 
The circle <math>\Omega = FBD</math> is orthogonal to the circle <math>\omega</math> <i><b>(Claim 2).</b></i>
 
The circle <math>\Omega = FBD</math> is orthogonal to the circle <math>\omega</math> <i><b>(Claim 2).</b></i>
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Let <math>X_0</math> be the point of intersection of the circles <math>\omega</math>  and <math>\Omega.</math> <math>\angle X_0AB = \angle X_0CD</math> (quadrilateral AX0CF is cyclic) and ∠X0BC = ∠X0DA (quadrangle DX0BF is cyclic) are equal to the property of opposite angles of cyclic quadrangles. This means that X0 coincides with the point X indicated in the condition.
+
Let <math>X_0</math> be the point of intersection of the circles <math>\omega</math>  and <math>\Omega.</math> <math>\angle X_0AB = \angle X_0CD</math> (quadrilateral <math>AX_0CF</math> is cyclic) and <math>\angle X_0BC = \angle X_0DA</math> (quadrangle <math>DX_0BF</math> is cyclic). This means that <math>X_0</math> coincides with the point <math>X</math> indicated in the condition.
−
The angle ∠FCX = ∠BCX subtend the arc XF of ω, ∠CBX = ∠XDA subtend the arc XF of Ω. The sum of these arcs is 180° (Claim 3 for orthogonal circles). Hence, the sum of the arcs XF is 180°, the sum of the angles ∠XСВ + ∠XВС = 90°, ∠СХВ = 90°. Similarly, ∠AXD = 90°, that is, ∠BXA + ∠DXC = 180°.
+
 
 +
<math>\angle FCX = \angle BCX</math> subtend the arc <math>\overset{\Large\frown} {XF}</math> of <math>\omega, \angle CBX = \angle XDA</math> subtend the arc <math>\overset{\Large\frown} {XF}</math> of <math>\Omega.</math> The sum of these arcs is <math>180^\circ</math>  <i><b>(Claim 3).</b></i>.
 +
 
 +
Hence, the sum of the arcs XF is 180°, the sum of the angles ∠XСВ + ∠XВС = 90°, ∠СХВ = 90°. Similarly, ∠AXD = 90°, that is, ∠BXA + ∠DXC = 180°.
 +
 
 +
<i><b>Claim 1</b></i>
 +
 
 +
Let A, C, and E be arbitrary points on a circle ω, 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 ω.

Revision as of 13:04, 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

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 XF is 180°, the sum of the angles ∠XСВ + ∠XВС = 90°, ∠СХВ = 90°. Similarly, ∠AXD = 90°, that is, ∠BXA + ∠DXC = 180°.

Claim 1

Let A, C, and E be arbitrary points on a circle ω, 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 ω.

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