P is a point inside the equilateral triangle ABC such that PA = 5, PB = 7, PC = 8. Find AB.
Solution
Answer: √129.
Let the side length be x. Using the cosine formula, we have cos APB = (74 - x2)/70, cos APC = (89 - x2)/80, cos BPC = (113 - x2)/112. But cos BPC = cos APC cos BPC - sin APC sin BPC, so (113 - x2)/112 = (74 - x2)/79 (89 - x2)/80 - √( (1 - (74 - x2)2/702) (1 - (89 - x2)2/802) ).
We isolate the square root term, then square. We multiply through by 25.256.49 and, after some simplification, we get x6 - 138x4 + 1161x2 = 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 60o. Let P go to P'. We have AP' = 7, CP' = 8 and angle PCP' = 60o, so PP'C is equilateral. Hence angle CPP' = 60o. Also PP' = 8. Using the cosine formula on triangle APP' we find angle APP' = 60o. Hence angle APC = 120o. Now applying cosine formula to triangle APC, we get result.
© John Scholes
jscholes@kalva.demon.co.uk
1 July 2002
Last corrected/updated 24 Dec 2002