Difference between revisions of "1985 IMO Problems/Problem 1"
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== Problem == | == Problem == | ||
− | A [[circle]] has center on the side <math> | + | A [[circle]] has center on the side <math>AB</math> of the [[cyclic quadrilateral]] <math>ABCD</math>. The other three sides are [[tangent]] to the circle. Prove that <math>AD + BC = AB</math>. |
== Solutions == | == Solutions == | ||
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=== Solution 2 === | === Solution 2 === | ||
− | Let <math> | + | Let <math>O </math> be the center of the circle mentioned in the problem, and let <math>T</math> be the point on <math>AB </math> such that <math>AT = AD </math>. Then <math>\angle DTA = \frac{ \pi - \angle DAB}{2} = \angle DCO</math>, so <math>DCOT </math> is a cyclic quadrilateral and <math>T </math> is in fact the <math>T</math> of the previous solution. The conclusion follows. |
=== Solution 3 === | === Solution 3 === | ||
− | Let the circle have center <math> | + | Let the circle have center <math>O </math> and radius <math>r </math>, and let its points of tangency with <math>BC, CD, DA </math> be <math>E, F, G </math>, respectively. Since <math>OEFC </math> is clearly a cyclic quadrilateral, the angle <math>COE </math> is equal to half the angle <math>GAO </math>. Then |
<center> | <center> | ||
<math> | <math> | ||
\begin{matrix} {CE} & = & r \tan(COE) \ | \begin{matrix} {CE} & = & r \tan(COE) \ | ||
− | & = & | + | & = &r \left( \frac{1 - \cos (GAO)}{\sin(GAO)} \right) \ |
& = & AO - AG \ | & = & AO - AG \ | ||
\end{matrix} | \end{matrix} | ||
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</center> | </center> | ||
− | Likewise, <math> | + | Likewise, <math>DG = OB - EB</math>. It follows that |
<center> | <center> | ||
<math> | <math> | ||
− | + | {EB} + CE + DG + GA = AO + OB | |
</math>, | </math>, | ||
</center> | </center> | ||
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=== Solution 4 === | === Solution 4 === | ||
− | We use the notation of the previous solution. Let <math> | + | We use the notation of the previous solution. Let <math>X</math> be the point on the ray <math>AD</math> such that <math>AX = AO</math>. We note that <math>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>OFC, OGX</math> are congruent; hence <math>GX = FC = CE </math> and <math>AO = AG + GX = AG + CE</math>. Similarly, <math>OB = EB + GD </math>. Therefore <math>AO + OB = AG + GD + CE + EB </math>, Q.E.D. |
{{alternate solutions}} | {{alternate solutions}} | ||
+ | |||
+ | Observations | ||
+ | Observe by take <math>M</math>, <math>N</math> on <math>AD</math> extended and | ||
+ | <math>BC</math> | ||
+ | |||
Revision as of 11:14, 9 March 2018
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.
Observations Observe by take , on extended and
1985 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |