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>, | ||
− | < | + | <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>. | |
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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. |
− | < | + | 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>. |
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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 , the diagonal bisects neither the angle nor the angle . The point lies inside and satisfies
Prove that is a cyclic quadrilateral if and only if
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
Let be the intersection of and , let be the intersection of and ,
, so , and . , so , and .
$\angle PLK=\frac12(\widearc{AD}+\widearc{CF}=\frac12(\widearc{CE}+\widearc{AB}=\angle PKL$ (Error compiling LaTeX. Unknown error_msg), so is an isosceles triangle.
Since , so and are isosceles triangles. So is on the angle bisector oof , since is
an isosceles trapezoid, so is also on the perpendicular bisector of . So .
~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 |