ABC is a triangle. C_{1} is a circle through A and B. We can find circle C_{2} through B and C touching C_{1}, circle C_{3} through C and A touching C_{2}, circle C_{4} through A and B touching C_{3} and so on. Show that C_{7} is the same as C_{1}.

**Solution**

Let O_{i} be the center of C_{i}. Evidently O_{i} lies on the perpendicular bisector of the relevant side. Since C_{1} and C_{2} touch, O_{1}, B and O_{2} must be collinear. Let M be the midpoint of AB. Let ∠MO_{1}A = x_{1}. Define x_{i} similarly. Since O_{1}, B and O_{2} are collinear, we have (90^{o} - x_{1}) + B + (90^{o} - x_{2}) = 180^{o}. So B = x_{1} + x_{2}. Similarly, x_{2} + x_{3} = C, x_{3} + x_{4} + A, x_{4} + x_{5} = B, x_{5} + x_{6} = C, x_{6} + x_{7} = A. Hence (x_{1} + x_{2}) + (x_{3} + x_{4}) + (x_{5} + x_{6}) = A + B + C = (x_{2} + x_{3}) + (x_{4} + x_{5}) + (x_{6} + x_{7}). So x_{1} = x_{7}. Hence O_{1} = O_{7}.

© John Scholes

jscholes@kalva.demon.co.uk

11 May 2002