Given a circle, find the inscribed polygon with the largest sum of the squares of its sides.
Solution
If there are more than 3 vertices, then the mean angle (between two adjacent sides) is ≥ 90o, so there must be a vertex with angle ≥ 90o. But now we can remove that vertex without increasing the sum of the squares.
So let us consider triangles.
Let M be the midpoint of BC. Applying the cosine formula to AMB and AMC and adding, we get AB2 + AC2 = 2AM2 + 2BM2. So if we keep B and C fixed and allow A to vary on the circle, then we maximize AB2 + AC2 by maximizing AM2. But that is clearly achieved by taking AM to contain O the center of the circle (for AM2 = AO2 + OM2 - 2·AO·OM cos AOM, which is maximized by taking ∠AOM = 180o). Thus the triangle ABC is not optimal unless each pair of sides is equal, so the optimal triangle must be equilateral. Note that none of its sides are diameters, so the triangles we derive from any 4-gon must have smaller sum of squares, so no 4-gons are optimal. Since all n-gons are worse than 4-gons (since they have a vertex with angle > 90o), no n-gon is optimal for n > 4. Thus the unique optimal polygons are the equilateral triangles.
© John Scholes
jscholes@kalva.demon.co.uk
1 Nov 2003
Last corrected/updated 1 Nov 03