Difference between revisions of "1985 IMO Problems/Problem 1"
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== Problem == | == Problem == | ||
− | A circle has center on the side <math>\displaystyle AB</math> of the [[cyclic quadrilateral]] <math>\displaystyle ABCD</math>. The other three sides are tangent to the circle. Prove that <math>\displaystyle AD + BC = AB</math>. | + | A [[circle]] has center on the side <math>\displaystyle AB</math> of the [[cyclic quadrilateral]] <math>\displaystyle ABCD</math>. The other three sides are [[tangent]] to the circle. Prove that <math>\displaystyle AD + BC = AB</math>. |
== Solutions == | == Solutions == |
Revision as of 14:12, 5 November 2006
Contents
[hide]Problem
A circle has center on the side of the cyclic quadrilateral . The other three sides are tangent to the circle. Prove that .
Solutions
Solution 1
Let be the center of the circle mentioned in the problem. Let be the second intersection of the circumcircle of with . By measures of arcs, . It follows that . Likewise, , so , as desired.
Solution 2
Let be the point on such that . Then , so is a cyclic quadrilateral and is in fact the of the previous solution. The conclusion follows.
Solution 3
Let the circle have center and radius , and let its points of tangency with be , respectively. Since is clearly a cyclic quadrilateral, the angle is equal to half the angle . Then
Likewise, . It follows that
,
Q.E.D.
Solution 4
We use the notation of the previous solution. Let be the point on the ray such that . We note that ; ; and ; hence the triangles are congruent; hence and . Similarly, . Therefore , Q.E.D.