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

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==Solution==
 
==Solution==
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{{solution}}
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Assume <math>ABCD</math> is cyclic,
 +
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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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>,
+
<asy>
 +
size(6cm);
 +
draw(circle((0,0),7.07));
 +
draw((-3.7,-6)-- (3.7,-6));
 +
draw((-6.8,-2)-- (6.8,-2));
 +
draw((-5,5)-- (5,5));
 +
draw((-5,5)-- (-3.7,-6));
 +
draw((-5,5)-- (3.7,-6));
 +
draw((-5,5)-- (-6.8,-2));
 +
draw((-5,5)-- (6.8,-2));
 +
draw((5,5)-- (-3.7,-6));
 +
draw((5,5)-- (3.7,-6));
 +
draw((5,5)-- (-6.8,-2));
 +
draw((5,5)-- (6.8,-2));
 +
draw((-3.7,-6)-- (-6.8,-2));
 +
draw((-3.7,-6)-- (6.8,-2));
 +
draw((3.7,-6)-- (-6.8,-2));
 +
draw((3.7,-6)-- (6.8,-2));
 +
label("$A$", (-6.8,-2), SW);
 +
label("$B$", (-3.7,-6), SW);
 +
label("$F$", (3.7,-6), SE);
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label("$C$", (6.8,-2), E);
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label("$E$", (5,5), E);
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label("$D$", (-5,5), W);
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label("$P$", (0,-1.3), N);
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label("$K$", (-1.6,-1.5), E);
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label("$L$", (0.8,-1.5) );                       
 +
</asy>
  
 
<math>\angle PBC=\angle DBA</math>, so <math>AD=CE</math>, and <math>DE//AC</math>.
 
<math>\angle PBC=\angle DBA</math>, so <math>AD=CE</math>, and <math>DE//AC</math>.
 
<math>\angle PDC=\angle BDA</math>, so <math>AB=CF</math>, and <math>AC//BF</math>.
 
<math>\angle PDC=\angle BDA</math>, so <math>AB=CF</math>, and <math>AC//BF</math>.
 +
<math>\angle PLK=\frac12(\overarc{AD}+\overarc{CF})=\frac12(\overarc{CE}+\overarc{AB})=\angle PKL</math>, so <math>\triangle PKL</math> is an isosceles triangle.
 +
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 perpendicular bisector of <math>BF</math>, since <math>ABFC</math> is
 +
an isosceles trapezoid, so <math>P</math> is also on the perpendicular bisector of <math>AC</math>. So <math>PA=PC</math>.
  
−
<math>\angle PLK=\frac12(\texttoptiebar{AD}+\texttoptiebar{CF}=\frac12(\texttoptiebar{CE}+\texttoptiebar{AB}=\angle PKL</math>, so <math>\triangle PKL</math> is an isosceles triangle.
 
−
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
 
−
an isosceles trapezoid, so <math>P</math> is also on the perpendicular bisector of <math>AC</math>. So <math>PA=PC</math>.
 
  
  

Latest revision as of 21:17, 3 November 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

Assume $ABCD$ is cyclic, let $K$ be the intersection of $AC$ and $BE$, let $L$ be the intersection of $AC$ and $DF$,

[asy] size(6cm); draw(circle((0,0),7.07)); draw((-3.7,-6)-- (3.7,-6)); draw((-6.8,-2)-- (6.8,-2)); draw((-5,5)-- (5,5)); draw((-5,5)-- (-3.7,-6)); draw((-5,5)-- (3.7,-6)); draw((-5,5)-- (-6.8,-2)); draw((-5,5)-- (6.8,-2)); draw((5,5)-- (-3.7,-6)); draw((5,5)-- (3.7,-6)); draw((5,5)-- (-6.8,-2)); draw((5,5)-- (6.8,-2)); draw((-3.7,-6)-- (-6.8,-2)); draw((-3.7,-6)-- (6.8,-2)); draw((3.7,-6)-- (-6.8,-2)); draw((3.7,-6)-- (6.8,-2)); label("$A$", (-6.8,-2), SW); label("$B$", (-3.7,-6), SW); label("$F$", (3.7,-6), SE); label("$C$", (6.8,-2), E); label("$E$", (5,5), E); label("$D$", (-5,5), W); label("$P$", (0,-1.3), N); label("$K$", (-1.6,-1.5), E); label("$L$", (0.8,-1.5) ); [/asy]

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

$\angle PLK=\frac12(\overarc{AD}+\overarc{CF})=\frac12(\overarc{CE}+\overarc{AB})=\angle PKL$, so $\triangle PKL$ is an isosceles triangle.
Since $AC//BF$, so $\triangle PBF$ and $\triangle PDE$ are isosceles triangles. So $P$ is on the perpendicular bisector of $BF$, 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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