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Difference between revisions of "2021 IMO Problems/Problem 3"

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==Problem==
 
==Problem==
 
Let <math>D</math> be an interior point of the acute triangle <math>ABC</math> with <math>AB > AC</math> so that <math>\angle DAB= \angle CAD</math>. The point <math>E</math> on the segment <math>AC</math> satisfies <math>\angle ADE= \angle BCD</math>, the point <math>F</math> on the segment <math>AB</math> satisfies <math>\angle FDA= \angle DBC</math>, and the point <math>X</math> on the line <math>AC</math> satisfies <math>CX=BX</math>. Let <math>O_1</math> and <math>O_2</math> be the circumcentres of the triangles <math>ADC</math> and <math>EXD</math> respectively. Prove that the lines <math>BC</math>, <math>EF</math>, and <math>O_1 O_2</math> are concurrent.
 
Let <math>D</math> be an interior point of the acute triangle <math>ABC</math> with <math>AB > AC</math> so that <math>\angle DAB= \angle CAD</math>. The point <math>E</math> on the segment <math>AC</math> satisfies <math>\angle ADE= \angle BCD</math>, the point <math>F</math> on the segment <math>AB</math> satisfies <math>\angle FDA= \angle DBC</math>, and the point <math>X</math> on the line <math>AC</math> satisfies <math>CX=BX</math>. Let <math>O_1</math> and <math>O_2</math> be the circumcentres of the triangles <math>ADC</math> and <math>EXD</math> respectively. Prove that the lines <math>BC</math>, <math>EF</math>, and <math>O_1 O_2</math> are concurrent.
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==Solution==
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<i><b>Lemma</b></i>
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Let <math>AK</math> be bisector of the triangle <math>ABC</math>, point <math>D</math> lies on <math>AK.</math> The point <math>E</math> on the segment <math>AC</math> satisfies <math>\angle ADE= \angle BCD</math>. The point <math>E'</math> is symmetric to <math>E</math> with respect to <math>AK.</math> The point <math>L</math> on the segment <math>AK</math>  satisfies <math>E'L||BC.</math>
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Then <math>EL</math> and <math>BC</math> are antiparallel with respect to the sides of an angle <math>A</math> and <cmath>\frac {AL}{DL} = \frac {AK \cdot DK}{BK \cdot KC}.</cmath>
  
 
==Video solution==
 
==Video solution==
 
https://youtu.be/cI9p-Z4-Sc8 [Video contains solutions to all day 1 problems]
 
https://youtu.be/cI9p-Z4-Sc8 [Video contains solutions to all day 1 problems]

Revision as of 19:35, 9 July 2022

Problem

Let $D$ be an interior point of the acute triangle $ABC$ with $AB > AC$ so that $\angle DAB= \angle CAD$. The point $E$ on the segment $AC$ satisfies $\angle ADE= \angle BCD$, the point $F$ on the segment $AB$ satisfies $\angle FDA= \angle DBC$, and the point $X$ on the line $AC$ satisfies $CX=BX$. Let $O_1$ and $O_2$ be the circumcentres of the triangles $ADC$ and $EXD$ respectively. Prove that the lines $BC$, $EF$, and $O_1 O_2$ are concurrent.

Solution

Lemma

Let $AK$ be bisector of the triangle $ABC$, point $D$ lies on $AK.$ The point $E$ on the segment $AC$ satisfies $\angle ADE= \angle BCD$. The point $E'$ is symmetric to $E$ with respect to $AK.$ The point $L$ on the segment $AK$ satisfies $E'L||BC.$ Then $EL$ and $BC$ are antiparallel with respect to the sides of an angle $A$ and \[\frac {AL}{DL} = \frac {AK \cdot DK}{BK \cdot KC}.\]

Video solution

https://youtu.be/cI9p-Z4-Sc8 [Video contains solutions to all day 1 problems]

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