Two circles C and C' lying outside each other touch at T. They lie inside a third circle and touch it at X and X' respectively. Their common tangent at T intersects the third circle at S. SX meets C again at P and XX' meets C again at Q. SX' meets C' again at U and XX' meets C' again at V. Prove that the lines ST, PQ and UV are concurrent.
Solution
The similarity center X takes P to S and Q to X', so PQ is parallel to SX'. Similarly, UV is parallel to SX. Extend PQ and UV to meet at K. Then SUKP is a parallelogram so KU = SP and KP = SU. Also KQV and SX'X are similar triangles. So KQ/KV = SX'/SX.
Since ST is tangent to C and C', we have SP.SX = ST2 = SU. SX'. Hence SX'/SX = SP/SU. Hence KQ/KV = SP/SU = KU/KP, so KP.KQ = KU.KV.
If K does not lie on ST, then ST must cut one of the segments KP and KU. Suppose it is KU. So let ST meet KU at A. Extend PK to meet ST at B. Then KQ.KP < BQ.BP = BT2 < AT2 = AT.AU < KV.KU. Contradiction. Similarly, we get a contradiction if ST cuts KP. Hence K must lie on ST.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002