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Difference between revisions of "2017 JBMO Problems/Problem 3"

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== Problem ==
 
== Problem ==
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Let  <math>ABC </math> be an acute triangle such that <math>AB\neq AC</math> ,with circumcircle <math> \Gamma</math> and circumcenter <math>O</math>. Let <math>M</math> be the midpoint of <math>BC</math> and <math>D</math> be a point on <math> \Gamma</math> such that <math>AD \perp  BC</math>. let <math>T</math> be a point such that <math>BDCT</math> is a parallelogram  and <math>Q</math> a point on the same side of  <math>BC</math> as <math>A</math> such that <math>\angle{BQM}=\angle{BCA}</math> and  <math>\angle{CQM}=\angle{CBA}</math>. Let the line <math>AO</math> intersect <math> \Gamma</math> at <math>E</math> <math>(E\neq A)</math> and let the circumcircle of <math>\triangle ETQ</math>  intersect <math> \Gamma</math> at point <math>X\neq E</math>. Prove that the point <math>A,M</math> and <math>X</math> are collinear .
  
 
== Solution ==
 
== Solution ==
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{{JBMO box|year=2017|num-b=2|num-a=4|five=}}
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[[Category:Intermediate Geometry Problems]]

Latest revision as of 14:39, 17 September 2017

Problem

Let $ABC$ be an acute triangle such that $AB\neq AC$ ,with circumcircle $\Gamma$ and circumcenter $O$. Let $M$ be the midpoint of $BC$ and $D$ be a point on $\Gamma$ such that $AD \perp BC$. let $T$ be a point such that $BDCT$ is a parallelogram and $Q$ a point on the same side of $BC$ as $A$ such that $\angle{BQM}=\angle{BCA}$ and $\angle{CQM}=\angle{CBA}$. Let the line $AO$ intersect $\Gamma$ at $E$ $(E\neq A)$ and let the circumcircle of $\triangle ETQ$ intersect $\Gamma$ at point $X\neq E$. Prove that the point $A,M$ and $X$ are collinear .

Solution

See also

2017 JBMO (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4
All JBMO Problems and Solutions
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