A pentagon is such that each triangle formed by three adjacent vertices has area 1. Find its area, but show that there are infinitely many incongruent pentagons with this property.
Solution
Let the pentagon be ABCDE. Triangles BCD and ECD have the same area, so B and E are the same perpendicular distance from CD, so BE is parallel to CD. The same applies to the other diagonals (each is parallel to the side with which it has no endpoints in common). Let BD and CE meet at X. Then ABXE is a parallelogram, so area BXE = area EAB = 1. Also area CDX + area EDX = area CDX + area BCX = 1. Put area EDX = x. Then DX/XB = area EDX/area BXE = x/1 and also = area CDX/area BCX = (1-x)/x. So x2 + x - 1 = 0, x = √5 - 1)/2 (we know x < 0, so it cannot be the other root). Hence area ABCDE = 3 + x = (√5 + 5)/2.
Take any triangle XCD of area (3 - √5)/2 and extend DX to B, so that BCD has area 1, and extend CX to E so that CDE has area 1. Then take BA parallel to CE and EA parallel to BD. It is easy to check that the pentagon has the required property.
© John Scholes
jscholes@kalva.demon.co.uk
13 Jul 2002