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 61: Line 61:
 
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>AYCD</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 YCD</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 orthogonal <math>\omega</math> circle <math>BDE.</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>
Line 81: Line 81:
 
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>
  
−
We make transformation and get <cmath>{\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 \psi.</cmath>
+
We make transformation and get <cmath>{\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.</cmath>
 +
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.</math> Given that <math>AB \cdot CD = BC \cdot DA \implies BC \cdot (CD – AD) = AB \cdot (AD – CD)\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>
  
 
<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>

Revision as of 11:24, 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

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 $\omega = ACEF$ (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
2018 IMO 6 Claim 3.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 $AYCF$ 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 $\omega.$

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_IMO_Problems/Problem_6&oldid=177219"