Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 10: Line 10:
 
The poinx <math>X</math> is inside <math>ABCD,</math> so points <math>E,A,X,C</math> follow in this order.
 
The poinx <math>X</math> is inside <math>ABCD,</math> so points <math>E,A,X,C</math> follow in this order.
  
−
<math>\angle XAB =  \angle XCD \implies  \angle XAE +  </math>\angle XCE = 180^\circ \implies AXCE<math> is cyclic </math>\implies X<math> lie on circle </math>ACE.<math>
+
<math>\angle XAB =  \angle XCD \implies  \angle XAE +  \angle XCE = 180^\circ \implies AXCE</math> is cyclic <math>\implies X</math> lie on circle <math>ACE.</math>
  
 
<i><b>Special case</b></i>
 
<i><b>Special case</b></i>
  
−
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>
  
−
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>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 = ACEF<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 = ACEF</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>
  
−
Let </math>X_0<math> be the point of intersection of the circles </math>\omega<math>  and </math>\Omega.<math>
+
Let <math>X_0</math> be the point of intersection of the circles <math>\omega</math>  and <math>\Omega.</math>
−
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>. This means that point </math>X_0<math> coincides with the point </math>X<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>.
  
−
</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>
+
<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>
  
−
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>
  
−
Similarly, </math>\angle AXD =  90^\circ \implies \angle BXA + \angle DXC = 360^\circ -\angle AXD -\angle CXB = 180^\circ.<math>
+
Similarly, <math>\angle AXD =  90^\circ \implies \angle BXA + \angle DXC = 360^\circ -\angle AXD -\angle CXB = 180^\circ.</math>
  
 
[[File:2018 IMO 6e.png|430px|right]]
 
[[File:2018 IMO 6e.png|430px|right]]
Line 41: Line 41:
 
<i><b>Common case </b></i>
 
<i><b>Common case </b></i>
  
−
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>
+
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>
  
−
The circles </math>BDF<math> and </math>BDE<math> are orthogonal to the circle </math>\omega<math> <i><b>(Claim 2).</b></i>
+
The circles <math>BDF</math> and <math>BDE</math> are orthogonal to the circle <math>\omega</math> <i><b>(Claim 2).</b></i>
  
−
Circles </math>ACF<math> and </math>ACE<math> are symmetric with respect to the circle </math>\omega<math> <i><b>(Lemma).</b></i>
+
Circles <math>ACF</math> and <math>ACE</math> are symmetric with respect to the circle <math>\omega</math> <i><b>(Lemma).</b></i>
  
−
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>
−
Quadrangle </math>BX_0DF<math> is cyclic </math>\implies \angle X_0BC =  \angle X_0DA<math> (see Special case).
+
Quadrangle <math>BX_0DF</math> is cyclic <math>\implies \angle X_0BC =  \angle X_0DA</math> (see Special case).
−
Similarly, quadrangle </math>AX_0CE<math> is cyclic </math>\implies \angle X_0AB =  \angle X_0CD.<math>  
+
Similarly, quadrangle <math>AX_0CE</math> is cyclic <math>\implies \angle X_0AB =  \angle X_0CD.</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>.
  
−
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>
  
−
Quadrangle </math>BYDF<math> is cyclic </math>\implies  \angle CBY =  \angle ADY.<math>
+
Quadrangle <math>BYDF</math> is cyclic <math>\implies  \angle CBY =  \angle ADY.</math>
  
−
Quadrangle </math>AYCF<math> is cyclic </math>\implies  \angle YAD = \angle BCY.<math>  
+
Quadrangle <math>AYCF</math> is cyclic <math>\implies  \angle YAD = \angle BCY.</math>  
  
−
The triangles </math>\triangle YAD \sim \triangle YCB<math> by two angles, so <cmath>\frac {BC}{AD} = \frac {CY}{AY} = \frac {BY} {DY} \hspace{10mm} (1).</cmath>
+
The triangles <math>\triangle YAD \sim \triangle YCB</math> by two angles, so <cmath>\frac {BC}{AD} = \frac {CY}{AY} = \frac {BY} {DY} \hspace{10mm} (1).</cmath>
  
−
The points </math>X<math> and </math>Y<math> are symmetric with respect to the circle </math>\omega<math>, since they lie on the intersection of the circles </math>ACF<math> and </math>ACE<math> symmetric with respect to </math>\omega<math> and the  circle </math>BDF<math> orthogonal to </math>\omega.<math>
+
The points <math>X</math> and <math>Y</math> are symmetric with respect to the circle <math>\omega</math>, since they lie on the intersection of the circles <math>ACF</math> and <math>ACE</math> symmetric with respect to <math>\omega</math> and the  circle <math>BDF</math> orthogonal to <math>\omega.</math>
  
−
The point </math>C<math> is symmetric to itself, the point </math>X<math> is symmetric to </math>Y<math> with respect to </math>\omega \implies \frac{CX}{CY} = \frac {R^2}{OC \cdot OY} , \frac {AX}{AY} = \frac {R^2}{OA \cdot OY}.<math>
+
The point <math>C</math> is symmetric to itself, the point <math>X</math> is symmetric to <math>Y</math> with respect to <math>\omega \implies \frac{CX}{CY} = \frac {R^2}{OC \cdot OY} , \frac {AX}{AY} = \frac {R^2}{OA \cdot OY}.</math>
−
Usung </math>(1)<math> and the equality </math>OA = OC,<math> we get <cmath>\frac{CY}{AY} = \frac {CX}{AX} = \frac{BC}{AD}.</cmath>
+
Usung <math>(1)</math> and the equality <math>OA = OC,</math> we get <cmath>\frac{CY}{AY} = \frac {CX}{AX} = \frac{BC}{AD}.</cmath>
−
The point </math>C<math> is symmetric to itself, the point </math>B<math> is symmetric to </math>D<math> with respect to </math>\omega \implies<math>
+
The point <math>C</math> is symmetric to itself, the point <math>B</math> is symmetric to <math>D</math> with respect to <math>\omega \implies</math>
 
<cmath>\triangle OBC \sim \triangle OCD \implies \frac {OB}{OC} = \frac {BC}{CD} = \frac {OC}{OD},</cmath>
 
<cmath>\triangle OBC \sim \triangle OCD \implies \frac {OB}{OC} = \frac {BC}{CD} = \frac {OC}{OD},</cmath>
 
<cmath>\frac {OB}{OD} = \frac {OB}{OC} \cdot \frac {OC}{OD} = \frac{BC^2}{CD^2} = \frac{BC}{CD} \cdot \frac {AB}{AD}.</cmath>
 
<cmath>\frac {OB}{OD} = \frac {OB}{OC} \cdot \frac {OC}{OD} = \frac{BC^2}{CD^2} = \frac{BC}{CD} \cdot \frac {AB}{AD}.</cmath>
−
The point </math>B<math> is symmetric to </math>D<math> and the point </math>X<math> is symmetric to </math>Y<math> with respect to </math>\omega,<math> hence
+
The point <math>B</math> is symmetric to <math>D</math> and the point <math>X</math> is symmetric to <math>Y</math> with respect to <math>\omega,</math> hence
 
<cmath>\frac {BX}{DY} = \frac {R^2}{OD \cdot OY} ,\frac {DX}{BY} = \frac{R^2}{OB \cdot OY}.</cmath>
 
<cmath>\frac {BX}{DY} = \frac {R^2}{OD \cdot OY} ,\frac {DX}{BY} = \frac{R^2}{OB \cdot OY}.</cmath>
 
<cmath>\frac{BX}{DX} =\frac{DY}{BY} \cdot \frac {OB}{OD} = \frac{AD}{BC} \cdot \frac{BC}{CD} \cdot \frac{AB}{AD} = \frac{AB}{CD}.</cmath>
 
<cmath>\frac{BX}{DX} =\frac{DY}{BY} \cdot \frac {OB}{OD} = \frac{AD}{BC} \cdot \frac{BC}{CD} \cdot \frac{AB}{AD} = \frac{AB}{CD}.</cmath>
Line 77: Line 77:
 
[[File:2018 IMO 6a.png|430px|right]]
 
[[File:2018 IMO 6a.png|430px|right]]
  
−
Denote  </math>\angle XAB =  \angle XCD = \alpha,  \angle BXA = \varphi,  \angle DXC = \psi.<math>
+
Denote  <math>\angle XAB =  \angle XCD = \alpha,  \angle BXA = \varphi,  \angle DXC = \psi.</math>
 
   
 
   
−
By the law of sines for </math>\triangle ABX,<math> we obtain </math>\frac {AB}{\sin \varphi} = \frac{BX}{\sin \alpha}.<math>
+
By the law of sines for <math>\triangle ABX,</math> we obtain <math>\frac {AB}{\sin \varphi} = \frac{BX}{\sin \alpha}.</math>
  
−
By the law of sines for </math>\triangle CDX,<math> we obtain </math>\frac {CD}{\sin \psi} = \frac {DX}{\sin \alpha}.<math>
+
By the law of sines for <math>\triangle CDX,</math> we obtain <math>\frac {CD}{\sin \psi} = \frac {DX}{\sin \alpha}.</math>
  
−
Hence we get </math>\frac{\sin \psi} {\sin \varphi}= \frac {CD}{DX}  \cdot \frac{BX}{AB} = 1.<math>
+
Hence we get <math>\frac{\sin \psi} {\sin \varphi}= \frac {CD}{DX}  \cdot \frac{BX}{AB} = 1.</math>
  
−
If </math>\varphi = \psi,<math> then </math>\triangle XAB \sim  \triangle XCD \implies \frac {CD}{AB} = \frac {BX}{DX} = \frac{AX}{CX} = \frac {AD}{BC}.<math>
+
If <math>\varphi = \psi,</math> then <math>\triangle XAB \sim  \triangle XCD \implies \frac {CD}{AB} = \frac {BX}{DX} = \frac{AX}{CX} = \frac {AD}{BC}.</math>
−
</math>CD \cdot BC = AB \cdot AD \implies AD = CD, AB = BC.<math> This is a special case.  
+
<math>CD \cdot BC = AB \cdot AD \implies AD = CD, AB = BC.</math> This is a special case.  
  
−
In all other cases, the equality of the sines follows </math>\psi = 180° – \varphi \implies \varphi + \psi = 180°.<math>
+
In all other cases, the equality of the sines follows <math>\psi = 180° – \varphi \implies \varphi + \psi = 180°.</math>
  
−
<i><b>Claim 1</b></i> Let </math>A, C,<math> and </math>E<math> be arbitrary points on a circle </math>\omega, l<math> be the perpendicular bisector to the segment </math>AC.<math> Then the straight lines </math>AE<math> and </math>CE<math> intersect </math>l<math> at the points </math>B<math> and </math>D,<math> symmetric with respect to </math>\omega.<math>
+
<i><b>Claim 1</b></i> Let <math>A, C,</math> and <math>E</math> be arbitrary points on a circle <math>\omega, l</math> be the perpendicular bisector to the segment <math>AC.</math> Then the straight lines <math>AE</math> and <math>CE</math> intersect <math>l</math> at the points <math>B</math> and <math>D,</math> symmetric with respect to <math>\omega.</math>
  
  
−
<i><b>Claim 2</b></i> Let points </math>B<math> and </math>D<math> be symmetric with respect to the circle </math>\omega.<math> Then any circle </math>\Omega<math> passing through these points is orthogonal to </math>\omega.<math>
+
<i><b>Claim 2</b></i> Let points <math>B</math> and <math>D</math> be symmetric with respect to the circle <math>\omega.</math> Then any circle <math>\Omega</math> passing through these points is orthogonal to <math>\omega.</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>
+
<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>
  
−
<i><b>Lemma</b></i> The opposite sides of the quadrilateral </math>ABCD<math> intersect at points </math>E<math> and </math>F<math> (</math>E<math> lies on </math>AB<math>). The circle  </math>\omega<math> centered at the point </math>O<math> contains the ends of the diagonal </math>AC.<math> The points </math>B<math> and </math>D<math> are symmetric with respect to the circle </math>\omega<math> (in other words, the inversion with respect to </math>\omega<math> maps </math>B<math> into </math>D).<math> Then the circles </math>ACE<math> and </math>ACF<math> are symmetric with respect to </math>\omega.<math>
+
<i><b>Lemma</b></i> The opposite sides of the quadrilateral <math>ABCD</math> intersect at points <math>E</math> and <math>F</math> (<math>E</math> lies on <math>AB</math>). The circle  <math>\omega</math> centered at the point <math>O</math> contains the ends of the diagonal <math>AC.</math> The points <math>B</math> and <math>D</math> are symmetric with respect to the circle <math>\omega</math> (in other words, the inversion with respect to <math>\omega</math> maps <math>B</math> into <math>D).</math> Then the circles <math>ACE</math> and <math>ACF</math> are symmetric with respect to <math>\omega.</math>
  
−
<i><b>Proof</b></i>  We will prove that the point </math>G,<math> symmetric to the point </math>E<math> with respect to </math>\omega,<math> belongs to the circle </math>ACF.<math> For this, we will prove the equality </math>\angle AGC = \angle AFC.<math>
+
<i><b>Proof</b></i>  We will prove that the point <math>G,</math> symmetric to the point <math>E</math> with respect to <math>\omega,</math> belongs to the circle <math>ACF.</math> For this, we will prove the equality <math>\angle AGC = \angle AFC.</math>
  
−
A circle </math>BDE<math> containing points </math>B<math> and </math>D<math> symmetric with respect to </math>\omega,<math> is orthogonal to </math>\omega<math> (Claim  2) and maps into itself under inversion with respect to the circle </math>\omega.<math> Hence, the point </math>E<math> under this inversion passes to some point </math>G,<math> of the same circle </math>BDE.<math>
+
A circle <math>BDE</math> containing points <math>B</math> and <math>D</math> symmetric with respect to <math>\omega,</math> is orthogonal to <math>\omega</math> (Claim  2) and maps into itself under inversion with respect to the circle <math>\omega.</math> Hence, the point <math>E</math> under this inversion passes to some point <math>G,</math> of the same circle <math>BDE.</math>
  
−
A straight line </math>ABE<math> containing the point </math>A<math> of the circle </math>\omega,<math> under inversion with respect to </math>\omega,<math> maps into the circle </math>OADG.<math> Hence, the inscribed angles of this circle are equal  </math>\angle ADB =  \angle AGE.<math>
+
A straight line <math>ABE</math> containing the point <math>A</math> of the circle <math>\omega,</math> under inversion with respect to <math>\omega,</math> maps into the circle <math>OADG.</math> Hence, the inscribed angles of this circle are equal  <math>\angle ADB =  \angle AGE.</math>
−
</math>\angle OCE =  \angle CGE (CE<math> maps into </math>CG)<math> and  </math>\angle OCE =  \angle BCD (BC<math> maps into </math>DC).<math>  
+
<math>\angle OCE =  \angle CGE (CE</math> maps into <math>CG)</math> and  <math>\angle OCE =  \angle BCD (BC</math> maps into <math>DC).</math>  
−
Consequently, the angles  </math>\angle AFC =  \angle ADB –  \angle FBD = \angle AGE -  \angle CGE =  \angle AGC.<math>
+
Consequently, the angles  <math>\angle AFC =  \angle ADB –  \angle FBD = \angle AGE -  \angle CGE =  \angle AGC.</math>
−
These angles subtend the </math>\overset{\Large\frown} {AC}<math> of the </math>ACF<math> circle, that is, the point </math>G,<math> symmetric to the point </math>E<math> with respect to </math>\omega,<math> belongs to the circle </math>ACF.$
+
These angles subtend the <math>\overset{\Large\frown} {AC}</math> of the <math>ACF</math> circle, that is, the point <math>G,</math> symmetric to the point <math>E</math> with respect to <math>\omega,</math> belongs to the circle <math>ACF.</math>
  
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''

Revision as of 06:00, 23 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 6bb.png
2018 IMO 6.png

We want to find the point $X.$ Let $E$ and $F$ be the intersection points of $AB$ and $CD,$ and $BC$ and $DA,$ respectively. The poinx $X$ is inside $ABCD,$ so points $E,A,X,C$ follow in this order.

$\angle XAB = \angle XCD \implies \angle XAE + \angle XCE = 180^\circ \implies AXCE$ is cyclic $\implies X$ lie on circle $ACE.$

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

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 $ACF.$

vladimir.shelomovskii@gmail.com, vvsss

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_IMO_Problems/Problem_6&oldid=177379"