Mathematics
Another “Radical” Geometry Problem from IMO
This problem was offered to the participants of the 1995 International Mathematical Olympiad. For those who don’t know, the International Mathematical Olympiad (or IMO for short) is one of the world's most prestigious and hardest mathematical competitions.
The problem
As always, try solving this problem for yourself, before reading the solution. Even though this is an IMO problem, you still might be able to solve it, don’t give up too easily, and remember that it is always recommended to work on such problems for at least a few hours. Even if you don’t manage to solve it, you still might learn something new and gain experience.
Solution
This problem can be surprisingly easily solved using the properties of the radical axes, namely, the theorem about the radical center. If are unfamiliar with those terms, I highly recommend reading my handout, or any other source that you can easily find on the Internet. Radical axes have a lot of applications in Olympiad Geometry, so make sure to study them thoroughly.
First, we notice that XY is the radical axis of those two circles. Thus, using our theorem, we find that it is sufficient to prove that points A, M, N, and D are concyclic (indeed, this is equivalent to proving that AM and DN intersect on XY, which is our goal).
Now, we notice that BN and CM intersect on the radical axis XY. This implies (again, from the theorem) that the quadrilateral BMNC is cyclic. Also, it is easy to see that ∠AMC = ∠BND = 90°, since AC and BD are both diameters.
Hence it is only left to do some trivial angle chasing.
∠DAM = ∠CAM = 90° − ∠ACM = 90° − ∠BCM = 90° − ∠BNM = 90° − ∠BNM − (∠BND − 90°) = 180° − (∠BNM + ∠BND) = 180° − ∠DNM
Therefore, the quadrilateral AMND is also cyclic, and we are done by our original observation. □
