Difference between revisions of "1985 IMO Problems/Problem 1"
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=== Solution 2 === | === Solution 2 === | ||
− | Let <math>\displaystyle T</math> be the point on <math>\displaystyle AB </math> such that <math> \displaystyle AT = AD </math>. Then <math> \displaystyle \angle DTA = \frac{ \pi - \angle DAB}{2} = \angle DCO</math>, so <math> \displaystyle DCOT </math> is a cyclic quadrilateral and <math> \displaystyle T </math> is in fact the <math> \displaystyle T</math> of the previous solution. The conclusion follows. | + | Let <math> \displaystyle O </math> be the center of the circle mentioned in the problem, and let <math>\displaystyle T</math> be the point on <math>\displaystyle AB </math> such that <math> \displaystyle AT = AD </math>. Then <math> \displaystyle \angle DTA = \frac{ \pi - \angle DAB}{2} = \angle DCO</math>, so <math> \displaystyle DCOT </math> is a cyclic quadrilateral and <math> \displaystyle T </math> is in fact the <math> \displaystyle T</math> of the previous solution. The conclusion follows. |
=== Solution 3 === | === Solution 3 === | ||
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We use the notation of the previous solution. Let <math>\displaystyle X</math> be the point on the ray <math>\displaystyle AD</math> such that <math> \displaystyle AX = AO</math>. We note that <math>\displaystyle OF = OG = r </math>; <math> \angle OFC = \angle OGX = \frac{\pi}{2} </math>; and <math> \angle FCO = \angle GXO = \frac{\pi - \angle BAD}{2}</math>; hence the triangles <math>\displaystyle OFC, OGX</math> are congruent; hence <math> \displaystyle GX = FC = CE </math> and <math> \displaystyle AO = AG + GX = AG + CE</math>. Similarly, <math> \displaystyle OB = EB + GD </math>. Therefore <math> \displaystyle AO + OB = AG + GD + CE + EB </math>, Q.E.D. | We use the notation of the previous solution. Let <math>\displaystyle X</math> be the point on the ray <math>\displaystyle AD</math> such that <math> \displaystyle AX = AO</math>. We note that <math>\displaystyle OF = OG = r </math>; <math> \angle OFC = \angle OGX = \frac{\pi}{2} </math>; and <math> \angle FCO = \angle GXO = \frac{\pi - \angle BAD}{2}</math>; hence the triangles <math>\displaystyle OFC, OGX</math> are congruent; hence <math> \displaystyle GX = FC = CE </math> and <math> \displaystyle AO = AG + GX = AG + CE</math>. Similarly, <math> \displaystyle OB = EB + GD </math>. Therefore <math> \displaystyle AO + OB = AG + GD + CE + EB </math>, Q.E.D. | ||
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== Resources == | == Resources == |
Revision as of 14:32, 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 center of the circle mentioned in the problem, and 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.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.