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

 
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== Problem ==
 
== Problem ==
  
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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>.
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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>.
  
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== Solution ==
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== Solutions ==
  
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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
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=== 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.
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=== 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.
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=== 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>
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Q.E.D.
 
Q.E.D.
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=== Solution 4 ===
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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.
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== Resources ==
 +
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* [[1985 IMO Problems]]
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* [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

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

Solution 2

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.

Resources

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