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

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==Problem==
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In a convex quadrilateral <math>ABCD</math>, the diagonal <math>BD</math> bisects neither the angle <math>ABC</math>
 
In a convex quadrilateral <math>ABCD</math>, the diagonal <math>BD</math> bisects neither the angle <math>ABC</math>
 
nor the angle <math>CDA</math>. The point <math>P</math> lies inside <math>ABCD</math> and satisfies <cmath>\angle PBC = \angle DBA \text{ and } \angle PDC = \angle BDA.</cmath>  
 
nor the angle <math>CDA</math>. The point <math>P</math> lies inside <math>ABCD</math> and satisfies <cmath>\angle PBC = \angle DBA \text{ and } \angle PDC = \angle BDA.</cmath>  
  
 
Prove that <math>ABCD</math> is a cyclic quadrilateral if and only if <math>AP = CP.</math>
 
Prove that <math>ABCD</math> is a cyclic quadrilateral if and only if <math>AP = CP.</math>
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==Solution==
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{{solution}}
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==See Also==
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{{IMO box|year=2004|num-b=4|num-a=6}}

Revision as of 23:54, 18 November 2023

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

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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