Pentagon ABCDE satisfies AB=BC=CD=DE=2. Find the maximum possible area of ABCDE.
My first thought was that all five vertices must lay on a common circle. Then the simplest resolution is for the fifth side to be a diameter of the circle.
Let R be the radius of the circle. Then the pentagon is partitioned into four isosceles triangles with base 2, sides R and vertex angle 45 deg.
Then by law of cosines we have 4 = 2R^2 - 2R^2*cos45, which after simplifying makes R^2 = 4+2*sqrt(2)
Then the area of the pentagon is four times the area of one of the triangles, which by the sine angle formula is (1/2)*R^2*sin45.
So then multiplying by 4 substituting the expression for R^2, we get an area of 4*((1/2)*(4+2*sqrt(2))*1/sqrt(2)) = 4*sqrt(2) + 4 ~= 9.657.
I have not actually proved this is the maximum, but it at least agrees with Larry's solution.
Thoughts
