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

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(→‎Solution)
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==Solution==
 
==Solution==
 
[[File:2018 IMO 6.png|490px|right]]
 
[[File:2018 IMO 6.png|490px|right]]
 +
Suppose point X is unique. We will construct point <math>X_0</math> and prove that <math>X_0</math> coincides with the point <math>X.</math>
  
 
<i><b>Special case</b></i>
 
<i><b>Special case</b></i>
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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>AD = CD</math> and <math>AB = BC \implies  AB \cdot CD = BC \cdot DA.</math>
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Quadrilateral <math>AX_0CF</math> is cyclic <math>\implies</math> <cmath>\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.</cmath>
 
Quadrilateral <math>AX_0CF</math> is cyclic <math>\implies</math> <cmath>\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.</cmath>
  
−
Similarly, quadrangle <math>DX_0BF</math> is cyclic <math>\implies \angle X_0BC = \angle X_0DA</math>.
+
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>.
 
−
 
 
−
<math>\hspace{10mm} \angle FCX =  \angle BCX  =  \frac {\overset{\Large\frown} {XAF}}{2}</math> of <math>\omega.</math>
 
  
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<math>\hspace{10mm} \angle CBX = \angle XDA =  \frac {\overset{\Large\frown} {XBF}}{2}</math> of <math>\Omega.</math>
+
<math>\hspace{10mm} \angle FCX = \angle BCX  =  \frac {\overset{\Large\frown} {XAF}}{2}</math> of <math>\omega.</math> <math>\hspace{10mm} \angle CBX = \angle XDA =  \frac {\overset{\Large\frown} {XBF}}{2}</math> of <math>\Omega.</math>
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The sum <math>\overset{\Large\frown} {XAF} + \overset{\Large\frown} {XBF} = 180^\circ</math> <i><b>(Claim 3)</b></i> <math>\implies</math>
 
  
 +
The sum <math>\overset{\Large\frown} {XAF} + \overset{\Large\frown} {XBF} = 180^\circ</math>  <i><b>(Claim 3)</b></i> <math>\implies</math>
 
<math>\angle XCB + \angle XBC = 90^\circ \implies \angle CXB = 90^\circ.</math>
 
<math>\angle XCB + \angle XBC = 90^\circ \implies \angle CXB = 90^\circ.</math>
  
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<i><b>Common case </b></i>
 
<i><b>Common case </b></i>
  
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Denote by <math>O</math> the intersection point of the perpendicular bisector of <math>AC</math> and <math>BD.</math> Let <math>\omega</math> be a circle (red) with center <math>O</math> and radius <math>OA.</math>
+
Denote by <math>O</math> the intersection point of <math>BD</math> and the perpendicular bisector of <math>AC.</math> Let <math>\omega</math> be a circle (red) with center <math>O</math> and radius <math>OA.</math>
  
 
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>
 
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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Similarly, quadrangle <math>AX_0CE</math> is cyclic <math>\implies \angle X_0AB =  \angle X_0CD = \alpha.</math>  
 
Similarly, quadrangle <math>AX_0CE</math> is cyclic <math>\implies \angle X_0AB =  \angle X_0CD = \alpha.</math>  
  
−
The required point <math>X = X_0</math> is constructed.
+
This means that point <math>X_0</math> coincides with the point <math>X</math>.
  
 
Denote by <math>Y</math> the point of intersection of circles <math>BDF</math> and <math>ACF.</math>
 
Denote by <math>Y</math> the point of intersection of circles <math>BDF</math> and <math>ACF.</math>

Revision as of 11:54, 21 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

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 6a.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 $\implies \angle X_0AB = \angle X_0CD = \alpha.$

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 \[\frac {BX}{DY} = \frac {R^2}{OD \cdot OY} ,\frac {DX}{BY} = \frac{R^2}{OB \cdot OY}.\] \[\frac{BX}{DX} =\frac{DY}{BY} \cdot \frac {OB}{OD} = \frac{AD}{BC} \cdot \frac{BC}{CD} \cdot \frac{AB}{AD} = \frac{AB}{CD}.\] 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 $\frac {CD}{\sin \psi} = \frac {DX}{\sin \alpha}.$

We make transformation and get \[{\sin \psi} = \frac {CD}{DX} {\sin \alpha} = \frac{CD}{DX} \cdot {\frac{DX \cdot AB}{BX \cdot CD}} {\sin \alpha} = \frac {AB}{BX}\sin \alpha = \sin \varphi.\] If $\varphi = \psi,$ then $\triangle XAB \sim \triangle XCD \implies \frac {CD}{AB} = \frac {BX}{DX} = \frac{AX}{CX} = \frac {AD}{BC}.$ $CD \cdot BC = AB \cdot AD.$ Given that $AB \cdot CD = BC \cdot DA \implies BC \cdot (CD – AD) = AB \cdot (AD – CD)\implies AD = CD, AB = BC.$ 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 $\omega.$

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

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 $ACF.$

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