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 1 === | === Solution 1 === | ||
− | Let <math> | + | 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 === | === 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. |
− | == | + | ===Solution 5=== |
+ | |||
+ | This solution is incorrect. The fact that <math>BC</math> is tangent to the circle does not necessitate that <math>B</math> is its point of tangency. -Nitinjan06 | ||
− | + | From the fact that AD and BC are tangents to the circle mentioned in the problem, we have | |
− | + | <math>\angle{CBA}=90\deg</math> | |
+ | and | ||
+ | <math>\angle{DAB}=90\deg</math>. | ||
+ | Now, from the fact that ABCD is cyclic, we obtain that | ||
+ | <math>\angle{BCD}=90\deg</math> | ||
+ | and | ||
+ | <math>\angle{CDA}=90\deg</math>, | ||
+ | such that ABCD is a rectangle. | ||
+ | |||
+ | Now, let E be the point of tangency between the circle and CD. It follows, if O is the center of the circle, that | ||
+ | <math>\angle{OEC}=\angle{OED}=90\deg</math> | ||
+ | |||
+ | Since <math>AO=EO=BO</math>, we obtain two squares, <math>AOED</math> and <math>BOEC</math>. | ||
+ | From the properties of squares we now have | ||
+ | |||
+ | |||
+ | <math>AD+BC=AO+BO=AB</math> | ||
+ | |||
+ | |||
+ | as desired. | ||
+ | |||
+ | === Solution 6 === | ||
+ | Lemma. Let <math>I</math> be the in-center of <math>ABC</math> and points <math>P</math> and <math>Q</math> be on the lines <math>AB</math> and <math>BC</math> respectively. Then <math>BP + CQ = BC</math> if and only if <math>APIQ</math> is a cyclic quadrilateral. | ||
+ | |||
+ | Solution. Assume that rays <math>AD</math> and <math>BC</math> intersect at point <math>P</math>. Let <math>S</math> be the center od circle touching <math>AD</math>, <math>DC</math> and <math>CB</math>. Obviosuly <math>S</math> is a <math>P</math>-ex-center of <math>PDB</math>, hence <math>\angle DSI=\angle DSP = \frac{1}{2} \angle DCP=\frac{1}{2} \angle A=\angle DAI</math> so DASI is concyclic. | ||
+ | |||
+ | == Video Solution == | ||
+ | |||
+ | === Solution 1 === | ||
+ | https://www.youtube.com/watch?v=tM0WhXNCWGU | ||
+ | |||
+ | |||
+ | === Solution 2 === | ||
+ | https://youtu.be/Ormv0y4ZM1E | ||
+ | |||
+ | |||
+ | {{alternate solutions}} | ||
+ | |||
+ | {{IMO box|year=1985|before=First question|num-a=2}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 06:28, 3 October 2021
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 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.
Solution 5
This solution is incorrect. The fact that is tangent to the circle does not necessitate that is its point of tangency. -Nitinjan06
From the fact that AD and BC are tangents to the circle mentioned in the problem, we have and .
Now, from the fact that ABCD is cyclic, we obtain that and , such that ABCD is a rectangle.
Now, let E be the point of tangency between the circle and CD. It follows, if O is the center of the circle, that
Since , we obtain two squares, and . From the properties of squares we now have
as desired.
Solution 6
Lemma. Let be the in-center of and points and be on the lines and respectively. Then if and only if is a cyclic quadrilateral.
Solution. Assume that rays and intersect at point . Let be the center od circle touching , and . Obviosuly is a -ex-center of , hence so DASI is concyclic.
Video Solution
Solution 1
https://www.youtube.com/watch?v=tM0WhXNCWGU
Solution 2
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 |