### 35th IMO 1994 shortlist

**Problem G3**

A circle C has two parallel tangents L' and L". A circle C' touches L' at A and C at X. A circle C" touches L" at B, C at Y and C' at Z. The lines AY and BX meet at Q. Show that Q is the circumcenter of XYZ.

**Solution**

The first step is to show that AY is tangent to C and C" at Y. Let L' touch C at H. Let the other tangent from A to C touch at Y'. Let the lines HY' and L" meet at B'. Let O, O', O" be the centers of C, C', C" respectively. Let the line through B' perpendicular to L" meet the line OY' at S. Since SB' is parallel to OH, ∠SB'Y' = ∠OHY' = ∠OY'H (OY' = OH) = ∠SY'B' (opposite angles). So SB' = SY'. Let SB' = r".

Let L" touch C at K. Let HA = x, KB' = y. Let r, r' be the radius of C, C'. OA is perpendicular to HY', so ∠OAH = 90^{o} - ∠AOH = ∠OHB' = ∠B'HK. Hence triangles OHA and B'KH are similar. So OH/HA = KB'/HK or xy = 2r^{2}. But OO' = r + r'. So x^{2} = (r + r')^{2} - (r - r')^{2} = 4rr'. Similarly, y^{2} = 4rr". Hence O'O"^{2} = (x - y)^{2} + (2r - r' - r")^{2} = (r' + r")^{2}. So the circle center S radius r" touches L", C and C'. So S = O", Y' = Y, B' = B. Hence AY is tangent to C and C".

Similarly BX is tangent to C and C'. So Q X = QY. It remains to show that QZ is tangent to C' and C". If it is not tangent, then it cuts C' again at Z' and C" again at Z". Z will lie between Z' and Z", so QZ' is not equal to QZ". But QZ'·QZ = QX^{2} = QY^{2} = QZ"·QZ. Contradiction.

35th IMO shortlist 1994

© John Scholes

jscholes@kalva.demon.co.uk

22 Sep 2002