Radical Axis
Such an impressive way of handling hard problems in geometry!
Definition
Given two circles ω1 and ω2 with distinct centres, the radical axis of the circles is the set of points P such that,
Powω1 (P) = Powω2 (P).
This might seem rather confusing definition as P seems to be a phantom point. However, in other words, we can define the radical axis as follows.
Let ω(1) and ω(2) be circles with distinct centres O(1) and O(2).
The radical axis of ω1 and ω2 is a straight line perpendicular to O(1)O(2).
The definition is clear now. But we need to show that both definitions are equal.
Now, it is a perfect time to use cartesian coordinates.
Let’s say that O1 = (a, 0) and O2 = (b, 0) in the coordinate plane and the circles have radii r1 and r2 respectively. Then for any point P = (x, y) we have
Powω1 (P) = O(1)P² − r(1)² = (x − a)² + y² − r(1)²Similarly,
Powω2 (P) = O(2)P² − r(2)²= (x − b)²+ y² − r(2)²
Equating the two, we find the radical axis of ω1 and ω2 is the set of points P =(x, y) satisfying
0 = Powω1 (P) − Powω2 (P)
=(x − a)² + y² − r(1)² − [(x − b)² + y² − r(2)²]
= (−2a + 2b)x +a² − b² + r(2)²− r(1)²
which is a straight line perpendicular to the x-axis (as −2a + 2b = 0).
Theorem
Three circles intersect as shown below, prove that their radical axes concur at one point.
Proof
Let d(12) be the radical line of the circles w1 and w2. Let d(23) be the radical line of the circles w2 and w3. Let d(31) be the radical line of the circles w3 and w1. Let d(12) and d(23) intersect at point P. We have to show that P is also on line d(31). Now, because P is on the lines d(12) and d(23) we can write,
Powω1 (P) = Powω2 (P) and Powω2 (P) = Powω3 (P)
Thus we have,
Powω1 (P) = Powω2 (P) = Powω3 (P).
It means P is also on line d(31). This completes our proof.
There are 2 cases when we are talking about radical axes of circles.
From Evan Chen,
1. Usually, the pairwise radical axes concur at a single point K. In that case, we call K the radical centre of the three circles.
2. Occasionally, the three radical axes will be pairwise parallel (or even the same line). Because the radical axis of two circles is perpendicular to the line joining its centres, this (annoying) case can only occur if O1, O2, O3 are collinear.
Let’s give an example for case 2.
Now let’s make the use of the radical axis.
Iran Team Selection Test 2011/1
In acute triangle ABC, ∠B is greater than ∠C. Let M be the midpoint of BC and let E and F be the feet of the altitudes from B and C, respectively. Let K and L be the midpoints of ME and MF respectively and let T be on line KL such that AT//BC. Prove AT = TM.
Solution
First, let’s draw a diagram.
Let H be the orthocentre of the triangle ABC, which is, in this case, the intersection of the lines BE and CF. We know for sure that A, F, H, E are concyclic. Also, as K and L are the midpoints of the sides FM and EM, FE is parallel to LK. Moreover, as BFC and BEC are right-angled triangles and M is the midpoint of the hypotenuse BC, FM = BM = CM = ME. Thus, MEF is an isosceles triangle. Now, let O be the centre of the circle passing through A, F, H, E. Because ∠AFH = 90, O is on line AH.
Now, let D be the foot of the altitude from M to FE in triangle MEF. As the triangle MEF is isosceles, FD = ED. If we draw the line OD then OD must be perpendicular to EF. Because O is the centre and FE is a chord of the circle and D is the midpoint of EF. Thus, ∠ODE = ∠MDE = 90. So M, D, O is colinear.
Let’s use the parallelity here. F, B, C, E are concyclic as ∠BFC = ∠BEC = 90. So ∠EAT = ∠CAT = ∠ACB = ∠ECB = ∠EFA. Hence, TA is tangent to the circle centered at O.
Now the biggest part is coming.
Think of a circle centred at M with radius 0 and call it U. Let Q be the circle passing through A, F, H, E. The line KL is the radical axis of circles U and Q. Because of the fact that OM ⊥ KL.
As T is on line KL,
PowU (T) = PowQ (T)
As AT is tangent to Q power of T over the circle Q is TA². And as the radius of the circle U is 0, the power of T over the circle U is MT²- r² = MT²- 0² = MT².
Thus if PowU (T) = PowQ (T) then AT² = MT² which leads to AT = MT as desired.
I hope this has shown the beautiful use of the radical axis and also taking a circle with radius 0.
Resources,
- Euclidian Geometry in Mathematical Olympiads, Evan Chen.
