2nd Mexico 1988 Problem A3 Two circles of different radius R and r touch externally. The three common tangents form a triangle. Find the area of the triangle in terms of R and r.
Let the triangles have centers O, O'. Let the common tangents meet at A, B, C as shown. Take the radii to be r, R as shown. Let the circle touch at Z. Note that BZA, OXA and O'YA are all similar. So area ABC = 2 area BZA = 2 AZ 2 /AX 2 area OXA = AZ 2 r/AX. By the similarity of AXO, AYO', r/AO = R/(AO+R+r), so AO = r(R+r)/(R-r). Hence AZ = AO + r = 2Rr/(R- r). By Pythagoras, AX 2 = AO 2 - r 2 = (r 2 /(R-r) 2 )((R+r) 2 - (R-r) 2 ) = 4Rr 3 /(R-r) 2 . Hence area ABC = 2(Rr) 3/2 /(R-r). 2nd Mexican 1988 © John Scholes jscholes@kalva.demon.co.uk 21 February 2004 Last corrected/updated 21 Feb 04