On the area of the polygon determined by the short diagonals of a convex polygon (Q2300114)

Given \(K = A_1A_2 \dots A_n\), a convex polygon with \(n \geq 5\) vertices, the authors draw the short diagonals \(A_1A_3, A_2A_4, \dots, A_{n-2}A_n, A_{n-1}A_1\), and \(A_nA_2\). A new convex \(n\)-gon \(K_1 = B_1B_2 \dots B_n\) is created inside \(K\), where \(B_i\) is the intersection point of \(A_{i-1}A_{i+1}\) and \(A_iA_{i+2}\) for all \(1 \leq i \leq n\). This construction was introduced by \textit{R. Schwartz} [Exp. Math. 1, No. 1, 71--81 (1992; Zbl 0765.52004)]. It is known as the pentagram map. Let \(K\) be a polygon and let \(\Delta(K)\) be its area. Then, for every integer \(n \geq 5\) and every \(\varepsilon > 0\), there exists a convex \(n\)-gon \(K\) such that \(\Delta(K_1)/\Delta(K) < \varepsilon\) and, for every integer \(n \geq 6\) and every \(\varepsilon > 0\), there exists a convex \(n\)-gon \(K\) such that \(\Delta(K_1)/\Delta(K) > 1 - \varepsilon\). When \(K\) is a pentagon it is not known how large the ratio \(\Delta(K_1)/\Delta(K)\) is. The authors show: Let \(K\) be a convex pentagon and let \(K_1\) be the pentagon formed by the diagonals of \(K\). Then \(\Delta(K_1)/ \Delta(K) \leq (7- 3\sqrt{5})/2\) with equality if and only if \(K\) is an affine regular pentagon. They give a description of prior attempts to solve the problem, they compute \(\Delta(K_1)\) and they make a generalization attempt.