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

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=== Solution 2 ===
 
=== Solution 2 ===
  
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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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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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{{alternate solutions}}
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== Resources ==
 
== Resources ==

Revision as of 14:32, 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 $\displaystyle O$ 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.


Resources

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