17th Swedish 1977 Problem 2 There is a point inside an equilateral triangle side d whose distance from the vertices is 3, 4, 5. Find d.
√(25 + 12√3) Solution Take the triangle to be ABC and the point P. Let PA = 3, PB = 4, PC = 5. Rotate the triangle and P through 60 o about C. Let A go to A' and P to P'. Then CP = CP' and ∠PCP' = 60 o , so PCP' is equilateral with side 5. So PAP' is a 3,4,5 triangle and hence ∠PAP' = 90 o . Hence ∠PAB + ∠P'AA' = 120 o - 90 o = 30 o . Hence ∠APB = 150 o . Now applying the cosine formula to PAB we have d 2 = 3 2 + 4 2 - 24 cos 150 o = 25 + 12√3. Thanks to Suat Namli 17th Swedish 1977 © John Scholes jscholes@kalva.demon.co.uk 13 February 2004 Last corrected/updated 13 Feb 04