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

Revision as of 21:28, 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.

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 $AC//BF$ .

, so  is an isosceles triangle.
Since , so  and  are isosceles triangles. So  is on the perpendicular bisector of , since  is 
an isosceles trapezoid, so  is also on the perpendicular bisector of . So .

~szhangmath

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

Page

Toolbox

Search

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

Revision as of 21:28, 8 February 2024

Problem

Solution

See Also

@@ Line 9: / Line 9: @@
 {{solution}}
-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>,
+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>,
+<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>
+<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 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>.
+~szhangmath
 ==See Also==
 {{IMO box|year=2004|num-b=4|num-a=6}}