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Difference between revisions of "2004 IMO Problems/Problem 5"

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Let <math>K</math>  be the intersection of <math>AC</math> and <math>BE</math>, let <math>L</math> be the intersection of <math>AC</math> and <math>DF</math>,
 
Let <math>K</math>  be the intersection of <math>AC</math> and <math>BE</math>, let <math>L</math> be the intersection of <math>AC</math> and <math>DF</math>,
  
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<svg version="1.1" id="Layer_1" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" x="0px" y="0px"
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<math>\angle PBC=\angle DBA</math>, so <math>AD=CE</math>, and <math>DE//AC</math>.
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<math>\angle PDC=\angle BDA</math>, so <math>AB=CF</math>, and <math>AC//BF</math>.
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<text transform="matrix(1 0 0 1 219 680.5)" font-family="'MyriadPro-Regular'" font-size="12">A</text>
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<text transform="matrix(1 0 0 1 258.5 666)" font-family="'MyriadPro-Regular'" font-size="12">K</text>
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<math>\angle PLK=\frac12(\widearc{AD}+\widearc{CF}=\frac12(\widearc{CE}+\widearc{AB}=\angle PKL</math>, so <math>\triangle PKL</math> is an isosceles triangle.
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<text transform="matrix(1 0 0 1 296.4033 650.5)" font-family="'MyriadPro-Regular'" font-size="12">L</text>
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Since <math>AC//BF</math>, so <math>\triangle PBF</math> and <math>\triangle PDE</math> are isosceles triangles. So <math>P</math> is on the angle bisector oof <math>BF</math>, since <math>ABFC</math> is
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<text transform="matrix(1 0 0 1 269.1636 696.5)" font-family="'MyriadPro-Regular'" font-size="12">B</text>
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an isosceles trapezoid, so <math>P</math> is also on the perpendicular bisector of <math>AC</math>. So <math>PA=PC</math>.
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<text transform="matrix(1 0 0 1 345.0273 617.2539)" font-family="'MyriadPro-Regular'" font-size="12">C</text>
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<text transform="matrix(1 0 0 1 276.5273 617.5)" font-family="'MyriadPro-Regular'" font-size="12">P</text>
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~szhangmath
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==See Also==
 
==See Also==
  
 
{{IMO box|year=2004|num-b=4|num-a=6}}
 
{{IMO box|year=2004|num-b=4|num-a=6}}

Revision as of 14:34, 8 February 2024

Problem

In a convex quadrilateral $ABCD$, the diagonal $BD$ bisects neither the angle $ABC$ nor the angle $CDA$. The point $P$ lies inside $ABCD$ and satisfies \[\angle PBC = \angle DBA \text{ and } \angle PDC = \angle BDA.\]

Prove that $ABCD$ is a cyclic quadrilateral if and only if $AP = CP.$

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

Let $K$ be the intersection of $AC$ and $BE$, let $L$ be the intersection of $AC$ and $DF$,

$\angle PBC=\angle DBA$, so $AD=CE$, and $DE//AC$. $\angle PDC=\angle BDA$, so $AB=CF$, and $AC//BF$. 

$\angle PLK=\frac12(\widearc{AD}+\widearc{CF}=\frac12(\widearc{CE}+\widearc{AB}=\angle PKL$ (Error compiling LaTeX. Unknown error_msg), so $\triangle PKL$ is an isosceles triangle.
Since $AC//BF$, so $\triangle PBF$ and $\triangle PDE$ are isosceles triangles. So $P$ is on the angle bisector oof $BF$, since $ABFC$ is 
an isosceles trapezoid, so $P$ is also on the perpendicular bisector of $AC$. So $PA=PC$.


~szhangmath

See Also

2004 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
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