Difference between revisions of "1985 IMO Problems/Problem 1"
Line 1: | Line 1: | ||
== 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 == |
− | Let the circle have center <math> \displaystyle O </math> and radius <math> \displaystyle r </math>, and let its points of tangency with <math> \displaystyle BC, CD, DA </math> be <math> \displaystyle E, F, G </math>, respectively. Since <math> \displaystyle OEFC </math> is clearly a | + | === Solution 1 === |
+ | |||
+ | Let <math> \displaystyle O</math> be the center of the circle mentioned in the problem. Let <math> \displaystyle T</math> be the second intersection of the circumcircle of <math> \displaystyle CDO </math> with <math> \displaystyle AB </math>. By measures of arcs, <math> \angle DTA = \angle CDO = \frac{\angle DCB}{2} = \frac{\pi}{2} - \frac{\angle DAB}{2} </math>. It follows that <math> \displaystyle AT = AD </math>. Likewise, <math>\displaystyle TB = BC</math>, so <math> \displaystyle AD + BC = AB </math>, as desired. | ||
+ | |||
+ | === 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. | ||
+ | |||
+ | === Solution 3 === | ||
+ | |||
+ | Let the circle have center <math> \displaystyle O </math> and radius <math> \displaystyle r </math>, and let its points of tangency with <math> \displaystyle BC, CD, DA </math> be <math> \displaystyle E, F, G </math>, respectively. Since <math> \displaystyle OEFC </math> is clearly a cyclic quadrilateral, the angle <math> \displaystyle COE </math> is equal to half the angle <math> \displaystyle GAO </math>. Then | ||
<center> | <center> | ||
Line 25: | Line 35: | ||
Q.E.D. | Q.E.D. | ||
+ | |||
+ | === Solution 4 === | ||
+ | |||
+ | 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. | ||
+ | |||
+ | == Resources == | ||
+ | |||
+ | * [[1985 IMO Problems]] | ||
+ | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366584#p366584 Discussion on AoPS/MathLinks] | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Revision as of 14:02, 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.