CONJECTURE: Global suprema for both area and perimeter of pentagonal cross-sections of the cube are the same as (presumably) quadrilateral, and given by the first entry listed below. List of currently known local pentagonal suprema, with planes a x + b y + c z + d = 0 denoted [d,a,b,c] , points (x, y, z) denoted [1,x,y,z] --- Quadrilateral perimeter & area suprema at plane = [0, 0, 1, 1] , perimeter = 4(1 + sqrt(2)) ~ 9.65685425 , area = 4 sqrt(2) ~ 5.65685425 , vertices = [1,-1,-1,+1], [1,+1,-1,+1], [1,+1,+1,-1], [1,-1,+1,-1] . Quadrilateral perimeter & area suprema at plane = [0, 1, 1, 2] , perimeter ~ 8.94427191, area ~ 4.89897949 , vertices = [1,-1,-1,+1], [1,+1,-1,0], [1,+1,+1,-1], [1,-1,+1,0] . Quadrilateral perimeter supremum at plane = [-2, 1, 1, 4] , perimeter ~ 8.24621125 , vertices = [1,-1,-1,+1], [1,+1,-1,+1/2], [1,+1,+1,0], [1,-1,+1,+1/2] . Pentagonal area maximum at plane = [-y, 1, 1+y, 2] , area ~ 4.99517500 , vertices = [1,-1,-2*y,+1], [1,2*y-1,-1,+1], [1,+1,-1,y], [1,+1,+1,-1], [1,-1,+1,0] ; where y = (1/4)*(sqrt(17) - 3) ~ 0.28077641 . It can be shown that all these quadrilaterals lie on the boundary of the phase-space region of (planes yielding) pentagons, which is convex when restricted to polygonal vertices along generically selected edges of the cube: in this computation vertices [1,-1,y,+1], [1,x,-1,+1], [1,+1,-1,z], [1,+1,+1,z'], [1,-1,+1,z"] . All solutions were found using Monte-Carlo search for numerical solutions, followed by exact-values constructed via Ansatz (polite word for guess-work!). There probably exist further local maxima in the interior of the phase-space hexahedron, but highly unlikely that they out-perform the best already known. In the event, conventional attempts to search systematically for formal or numerical solutions using Maple equation solvers quickly proved abortive. Fred Lunnon On 8/3/19, Fred Lunnon <fred.lunnon@gmail.com> wrote:

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