1985 IMO Problems/Problem 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.
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.
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
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.
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
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.
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
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