L1, L2, L3, L4 are diameters of a circle C radius 1 and the angle between any two is either π/2 or π/4. If P is a point on the circle, show that the sum of the fourth powers of the distances from P to the four diameters is 3/2.
Solution
Take rectangular coordinates with origin at the center of the circle. Take two of the diameters as axes. Then the other two diameters are y = x and y = -x. A point (a, b) on the circle is a distance |a|, |b|, |a-b|/√2, |a+b|/√2 from the four diameters. Thus the sum of the fourth powers is a4 + b4 + (a-b)4/4 + (a+b)4/4 = 3a4/2 + 3b4/2 + 3a4b4 = (3/2)(a2 + b2)2 = 3/2.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002