The proof is short, but frankly I can't read it (as I am not sure what all the products are sisj ; I presume they are line segments and the + is union/area or

The proof is short, but frankly I can't read it (as I am not sure what all the products are sisj ; I presume they are line segments and the + is union/area or some such). Yes, I know it because I didn't study classic geometry long enough. If we take the brutal step of resorting to trigonometry and calculus we can prove the statement using general techniques. It may seem long and technical- but I think it demonstrates re-usable methods that give some intuition (after you use them a few times!). To prove: the regular n-gon has the maximal area of any n-gon inscribed inside the unit circle. Proof sketch: We will prove only the lemma that if theta1, and theta2 are angles summing to C < pi then sum of the areas of the area of the figure formed by the convex hull of (0,0), (sin(theta1),cos(theta1)), (0,1), and (-sin(theta2),cos(theta2)) is maximized for a given C when theta1=theta2. That is: the area formed by subdividing a segment of the circle of angle theta1+theta2 is maximized by dividing in the middle. This would let us prove the theorem as it implies any non-regular division can be improved by re-centering one of the sub-divisions. Proof of lemma: The area of triangle formed by the convex hull of the center of a unit circle and the points (sin(theta/2),cos(theta/2)), (-sin(theta/2),cos(theta/2)) is (1/2) base height = (1/2) (2 sin(theta/2)) cos(theta/2) = sin(theta/2) cos(theta/2). So if f(theta1,theta2) is the area of the two inscribed triangles of angles theta1 and theta2 we have: f(theta1,theta2) = sin(theta1/2) cos(theta1/2) + sin(theta2/2) cos(theta2/2) If we add the constraint theta1+theta2=C then we expect the maximum to occur at one of theta1=0,theta2=C; theta1=C,theta2=0 or an point where the Lagrange auxiliary function L(theta1,theta2,lambda) = f(theta1,theta2) + lamba (theta1+theta2-C) has a zero gradient (gradient with respect to theta1, theta2)) Gradient((theta1,theta2),L) = (1/2) (cos(theta1/2)^2 -sin(theta1/2)^2, cos(theta2/2)^2 -sin(theta2/2)^2) + lambda (1,1) For this vector quantity to be equal to zero we would have to have the coordinates equal (so the lambda (1,1) can stomp them out): cos(theta1/2)^2 -sin(theta1/2)^2 = cos(theta2/2)^2 - sin(theta2/2)^2 But cos(x)^2 -sin(x)^2 = 1 - 2 sin(x)^2 which is a 1 to 1 function in the range [0,pi) so to have cos(theta1/2)^2 -sin(theta1/2)^2 = cos(theta2/2)^2 -sin(theta2/2)^2 we must have theta1=theta2. So we check that f(C/2,C/2) >= max(f(C,0),f(0,C)) and we are done. Again I know that is long and technical (and bringing in overly big hammers)- but the steps are all standard re-usable ideas. (actually longer than I even intended)