ABC is a triangle. Show that there is a unique point P such that PA2 + PB2 + AB2 = PB2 + PC2 + BC2 = PC2 + PA2 + CA2.
Solution
PA2 + PB2 + AB2 = PB2 + PC2 + BC2 implies PA2 - PC2 = BC2 - AB2. Let the perpendicular from P meet AC at K. Then PA2 - PC2 = (PK2 + AK2) - (PK2 + CK2), so AK2 - CK2 = constant. Thus P must lie on a fixed line perpendicular to AC. Let A, B, C be the midpoints of the sides of B'C', C'A', A'B'. Then B'A2 - B'C2 = BC2 - BA2 (since B'A = BC, B'C = BA). So the fixed line passes through B'. Similarly, P must lie on the line through A' perpendicular to BC. So P must be the orthocenter of A'B'C'.
© John Scholes
jscholes@kalva.demon.co.uk
20 Sep 2002