Keith F. Lynch <kfl@keithlynch.net> asked << I have a related question, in support of testing my program that finds the average perimeter and area of random intersections of cubes and planes. I know what the maximum perimeter and maximum area are when the intersection is a triangle, a quadrilateral, or a hexagon. (Or at least I think I do.) But I don't know what it is when the intersection is a pentagon. Does anyone know? Thanks. >> In principle such questions can definitively be answered by the method of Lagrange multipliers, followed by inspection of the (lower-dimensional) semi-algebraic boundary; but in practice I prefer to start by trying to get a provisional feel for the shape of solutions via a (quick-and-dirty) numerical Monte-Carlo attack. [ Maybe not so quick but certainly dirty, since convergence to any critical point is necessarily ill-conditioned! ] The cube will have vertices with components in {+1,-1} . For this problem we get off to a good start by observing that, modulo symmetries of the cube, the interior of the phase space of potential plane sections is connected: for example, plane x + y + 3 z = 2 is continuously connected to every possible pentagonal cross-section modulo congruence. Our luck quickly runs out when it transpires that any solutions have at least one pentagon corner with a zero component, causing large cancellation errors in the generated approximate solutions. But, after a number of runs with different settings of convergence parameters and initial planes, a hazy picture starts to emerge. Apparently all solution pentagons are degenerate: no proper maximal pentagons exist, only quadrilaterals yielding perimeter suprema. We find the following provisional solutions (there may be others): A local interior maximum perimeter 8.24621125 at quadrilateral on plane x + y + 4 z = 2 ; A global(?) boundary maximum perimeter 4 sqrt(5) ~ 8.94427191 at quadrilateral on plane x + y + 2 z = 0 . Notice that the latter is NOT the globally maximal quadrilateral with perimeter 4(1 + rt2) ~ 9.65685425 , which it seems cannot tilt to a pentagon while maintaining planarity. The area problem remains pending. Further details on request. Fred Lunnon