1983 IMO Problems/Problem 2
Problem
Let be one of the two distinct points of intersection of two unequal coplanar circles and with centers and respectively. One of the common tangents to the circles touches at and at , while the other touches at and at . Let be the midpoint of and the midpoint of . Prove that .
Solution 1
Let be one of the two distinct points of intersection of two unequal coplanar circles and with centers and respectively. Let be such point on line so that tangents on touches it at and and tangents on touches it at and . Let be the midpoint of and the midpoint of . Prove that .
Proof: Since is image of under inversion wrt circle we have:Since is image of under inversion wrt circle we have:Image of is in both cases itself, since it lies on both circles. Since we have:Now:
This solution was posted and copyrighted by Number1. The original thread for this problem can be found here: [1]
Solution 2
Let and meet at . Let meet at . Now, it is well-known that , , and are concurrent at , the center of homothety between and . Now, it is well-known that bisects . Since , we have that meets at its midpoint, , and is perpendicular to . Similarly, passes through and is perpendicular to . Since , we have that , which implies that is cyclic. Yet, since and lie on and respectively and are collinear with , we see that the homothety that maps to about maps to . Also, is mapped to by this homothety, and since and are corresponding parts in these circles, is mapped to by this homothety, so , from which we conclude that .
This solution was posted and copyrighted by The QuattoMaster 6000. The original thread for this problem can be found here: [2]
Solution 3
Let be the other intersection point of these two circles. Let , , meet at . Let meet at . Clearly, and are perpendicular at ; and are perpendicular at . Since ,Since is on the radical axis, , so in the right trapezoid , is the midsegment. So we have and . Let be a point on such that (which means ve ). So, from , we getThis means, is the angle bisector of . So,
This solution was posted and copyrighted by matematikolimpiyati. The original thread for this problem can be found here: [3]
Solution 4
Let, concur at .Then a homothety with centre that sends to .Let .Under the homothety is the image of .So, and .so Hence are concyclic.so
This solution was posted and copyrighted by spikerboy. The original thread for this problem can be found here: [4]
Solution 5
It can be easily solved with this lemma: Let -symmedian of intersect its circumscribed circle at on satisfying . Let us define center of as . Then are concyclic. Now let be intersection of tangents to circles from the problem, intersects at . One can prove that is harmonic qudrilateral, so is symmedian of . By lemma we get that are concyclic, thus is equivalent to . By homothety we get .
This solution was posted and copyrighted by LeoLeon. The original thread for this problem can be found here: [5]
1983 IMO (Problems) • Resources | ||
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