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.
Possible solution, maybe bogus?
The only way for AD and BC to be tangent to circle O and have AB pass through O is if and are both 90. But since ABCD is cyclic, the other angles must be 90 as well. Now call the point of tangency of CD E, and since AO=EO, AEOD is a square. Similarily, BCEO is a square, too, so DA=AO and CB=BO. Therefore, AD+BC=AB.
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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