Mathematics
An Olympiad Geometry Problem Solved Using Radical Axes
The following problem is a nice application of the use of the properties of radical axes.
This is an enjoyable Olympiad problem and I highly recommend trying to solve it yourself before reading the solution.
Solution
This problem is quite hard (or maybe even impossible) to solve using elementary Geometry techniques, such as angle chasing. Therefore, we present a solution that uses the properties of the radical axes and their intersection — the radical center. If you are unfamiliar with those terms, I recommend you reading my handout.
Our goal is to prove that four points lying on two circles are concyclic. This might immediately remind you of the theorem about the radical center. Here it is, in case you have forgotten about it:
If we use this theorem then it will be sufficient to prove that the intersection of MN and PQ lies on the radical axis of ω₁ and ω₂. We know that both lines MN and PQ are altitudes of the triangle ABC and therefore, they intersect at the orthocenter, call it H. Now, vertex A is clearly an intersection of ω₁ and ω₂ and thus we get that the radical axis passes through points A and H. But this is exactly the altitude of the triangle ABC from vertex A! Hence it is left to prove that the point of intersection of of ω₁ and ω₂ coincides with the foot of the altitude from A to BC (let it be called A₁). This is trivial to show, by using the technique of phantom points.
So here is the full solution
Let A’ be the point of intersection of ω₁ with BC. Since AB is the diameter of ω₁, the angle ∠AA’B = 90°. Hence we get that AA’ is an altitude and therefore A’ = A₁. Similarly, ω₂ also intersects with BC at A₁. Thus the altitude AA₁ is also a radical axis of ω₁ and ω₂. Now we get that MN and PQ intersect on the radical axis and therefore M, N, P, and Q are concyclic. □
