1st Iberoamerican 1985 Problem A2 P is a point inside the equilateral triangle ABC such that PA = 5, PB = 7, PC = 8. Find AB. Solution
: √129. Let the side length be x. Using the cosine formula, we have cos APB = (74 - x 2 )/70, cos APC = (89 - x 2 )/80, cos BPC = (113 - x 2 )/112. But cos BPC = cos APC cos BPC - sin APC sin BPC, so (113 - x 2 )/112 = (74 - x 2 )/79 (89 - x 2 )/80 - √( (1 - (74 - x 2 ) 2 /70 2 ) (1 - (89 - x 2 ) 2 /80 2 ) ). We isolate the square root term, then square. We multiply through by 25.256.49 and, after some simplification, we get x 6 - 138x 4 + 1161x 2 = 0. Hence x = 0, ±3, ±√129. We discard the zero and negative solutions. x = 3 corresponds to a point P outside the triangle. So the unique solution for a point P inside the triangle is x = √129. Alternative solution by Johannes Tang Rotate the triangle about C through 60 o . Let P go to P'. We have AP' = 7, CP' = 8 and angle PCP' = 60 o , so PP'C is equilateral. Hence angle CPP' = 60 o . Also PP' = 8. Using the cosine formula on triangle APP' we find angle APP' = 60 o . Hence angle APC = 120 o . Now applying cosine formula to triangle APC, we get result. 1st Ibero 1985 © John Scholes jscholes@kalva.demon.co.uk 1 July 2002 Last corrected/updated 24 Dec 2002