Two circles centers A and B lie outside each other and touch at X. A third circle center C encloses both and touches them at Y and Z respectively. The tangent to the first two circles at X forms a chord of the third circle with midpoint M. Prove that ∠YMZ = ∠ACB.
Solution
Solution by Nizameddin Ordulu and Ali Adalý:
The circles center A and C touch at Y, so CAY is a straight line. Similarly, CBZ is a straight line. So ∠ACB = ∠YCZ. So we have to show that C, M, Y and Z lie on a circle.
Suppose that the tangents to the large circle at Y and Z meet at K. Suppose the line KX meets the circle center A again at X' and the circle center B again at X". Then X' and X" must be on opposite sides of X. But KY2 = KX.KX', since KY is also a tangent to the circle center A, and KZ2 = KX·KX", since KZ is also a tangent to the circle center B. So KX' = KX". Hence X' = X = X", so K also lies on the tangent to the two small circles at X.
But now we can see that Y, Z and M all lie on the circle diameter CK (because ∠CYK = ∠CZK = ∠CMK = 90o - the last because M is the midpoint of the chord MK of the large circle).
© John Scholes
jscholes@kalva.demon.co.uk
12 Aug 2002