29th USAMO 2000

Problem B2

ABC is a triangle. C1 is a circle through A and B. We can find circle C2 through B and C touching C1, circle C3 through C and A touching C2, circle C4 through A and B touching C3 and so on. Show that C7 is the same as C1.



Let Oi be the center of Ci. Evidently Oi lies on the perpendicular bisector of the relevant side. Since C1 and C2 touch, O1, B and O2 must be collinear. Let M be the midpoint of AB. Let ∠MO1A = x1. Define xi similarly. Since O1, B and O2 are collinear, we have (90o - x1) + B + (90o - x2) = 180o. So B = x1 + x2. Similarly, x2 + x3 = C, x3 + x4 + A, x4 + x5 = B, x5 + x6 = C, x6 + x7 = A. Hence (x1 + x2) + (x3 + x4) + (x5 + x6) = A + B + C = (x2 + x3) + (x4 + x5) + (x6 + x7). So x1 = x7. Hence O1 = O7.



29th USAMO 2000

© John Scholes
11 May 2002