Difference between revisions of "1985 IMO Problems/Problem 1"
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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. | 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. | ||
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=== Solution 2 === | === Solution 2 === | ||
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{{alternate solutions}} | {{alternate solutions}} | ||
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{{IMO box|year=1985|before=First question|num-a=2}} | {{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
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 |