As usual, \(ABCD\) is a square, and we divide each side into three equal parts, and mark the points \(E,F,G,H,I,J,K,L\) in a clockwise manner starting from the vertex \(D\). This time, we connect those points to the furthest
vertex that is opposite to them to form a not-necessarily-regular octagon as shown in the figure above, for example, points \(B\) and \(E\) will be connected in this case. What is the ratio of the area of the square to that of the octagon?
You may wish to try Part 1 and Part 2 of this trilogy.