Difference between revisions of "1985 IMO Problems/Problem 1"

Revision as of 14:18, 3 April 2016

Problem

A circle has center on the side $\displaystyle AB$ of the cyclic quadrilateral $\displaystyle ABCD$ . The other three sides are tangent to the circle. Prove that $\displaystyle AD + BC = AB$ .

Solutions

Solution 1

Let $O$ be the center of the circle mentioned in the problem. Let $T$ be the second intersection of the circumcircle of $CDO$ with $AB$ . By measures of arcs, $\angle DTA = \angle DCO = \frac{\angle DCB}{2} = \frac{\pi}{2} - \frac{\angle DAB}{2}$ . It follows that $AT = AD$ . Likewise, $TB = BC$ , so $AD + BC = AB$ , as desired.

Solution 2

Let $\displaystyle O$ be the center of the circle mentioned in the problem, and let $\displaystyle T$ be the point on $\displaystyle AB$ such that $\displaystyle AT = AD$ . Then $\displaystyle \angle DTA = \frac{ \pi - \angle DAB}{2} = \angle DCO$ , so $\displaystyle DCOT$ is a cyclic quadrilateral and $\displaystyle T$ is in fact the $\displaystyle T$ of the previous solution. The conclusion follows.

Solution 3

Let the circle have center $\displaystyle O$ and radius $\displaystyle r$ , and let its points of tangency with $\displaystyle BC, CD, DA$ be $\displaystyle E, F, G$ , respectively. Since $\displaystyle OEFC$ is clearly a cyclic quadrilateral, the angle $\displaystyle COE$ is equal to half the angle $\displaystyle GAO$ . Then

$\begin{matrix} {CE} & = & r \tan(COE) \\ & = & \displaystyle r \left( \frac{1 - \cos (GAO)}{\sin(GAO)} \right) \\ & = & AO - AG \\ \end{matrix}$

Likewise, $\displaystyle DG = OB - EB$ . It follows that

$\displaystyle {EB} + CE + DG + GA = AO + OB$ ,

Q.E.D.

Solution 4

We use the notation of the previous solution. Let $\displaystyle X$ be the point on the ray $\displaystyle AD$ such that $\displaystyle AX = AO$ . We note that $\displaystyle OF = OG = r$ ; $\angle OFC = \angle OGX = \frac{\pi}{2}$ ; and $\angle FCO = \angle GXO = \frac{\pi - \angle BAD}{2}$ ; hence the triangles $\displaystyle OFC, OGX$ are congruent; hence $\displaystyle GX = FC = CE$ and $\displaystyle AO = AG + GX = AG + CE$ . Similarly, $\displaystyle OB = EB + GD$ . Therefore $\displaystyle AO + OB = AG + GD + CE + EB$ , Q.E.D.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1985 IMO (Problems) • Resources
Preceded by First question	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
All IMO Problems and Solutions

@@ Line 7: / Line 7: @@
 === 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.
+Let <math>O</math> be the center of the circle mentioned in the problem.  Let <math>T</math> be the second intersection of the circumcircle of <math>CDO </math> with <math>AB </math>.  By measures of arcs, <math> \angle DTA = \angle DCO = \frac{\angle DCB}{2} = \frac{\pi}{2} - \frac{\angle DAB}{2} </math>.  It follows that <math>AT = AD </math>.  Likewise, <math>TB = BC</math>, so <math>AD + BC = AB </math>, as desired.
 === Solution 2 ===

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