Difference between revisions of "2004 IMO Problems/Problem 5"
Szhangmath (talk | contribs) (→Solution) |
Szhangmath (talk | contribs) (→Solution) |
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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>, | ||
+ | |||
[asy] | [asy] | ||
− | |||
size(10cm); | size(10cm); | ||
draw(circle((0,0),7.07)); | draw(circle((0,0),7.07)); | ||
Line 21: | Line 21: | ||
draw((-5,5)-- (-6.8,-2)); | draw((-5,5)-- (-6.8,-2)); | ||
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)-- (3.7,-6)); | draw((5,5)-- (3.7,-6)); | ||
Line 38: | Line 37: | ||
label("<math>P</math>", (0,-1.3), N); | label("<math>P</math>", (0,-1.3), N); | ||
label("<math>K</math>", (-1,-1.6), E); | label("<math>K</math>", (-1,-1.6), E); | ||
− | label("<math>L</math>", (0.7,-1.6) ); | + | label("<math>L</math>", (0.7,-1.6) ); |
− | |||
[/asy] | [/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>. |
Revision as of 16:12, 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 ,
[asy] size(10cm); 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("", (-6.8,-2), SW); label("", (-3.7,-6), SW); label("", (3.7,-6), SE); label("", (6.8,-2), E); label("", (5,5), E); label("", (-5,5), W); label("", (0,-1.3), N); label("", (-1,-1.6), E); label("", (0.7,-1.6) ); [/asy]
, so , and . , so , and .
, 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 |