# Does the boundary of a convex body contain a regular planar pentagon?

How to prove or disprove that the boundary of any convex body in $\mathbb{R}^3$ includes 5 points which form a regular planar pentagon? The following consideration suggests the answer "yes": if we assume the boundary is described by the equation $f(x,y,z)=0$, then we have a system of $5+5+3=13$ equations in $15$ variables. In the general case an infinite solution set is expected. However, the convexity is not used here. The question was asked, but not answered, in MSE.

• The title is a bit misleading. (And it suggests that the answer is "yes"...) Maybe something like "Does the boundary of a convex body contain a regular planar pentagon?" – Joonas Ilmavirta Nov 13 '14 at 20:26
• @ Joonas Ilmavirta: Thank you. I have edited the title up to your suggestion. – user64494 Nov 13 '14 at 20:31
• Does a convex body need to have nonempty other interior? Otherwise the answer is no. – Dirk Nov 13 '14 at 21:42
• If so, perhaps this can be used together with some intermediate value argument. That is, take the quadrilateral made of 4 points of a regular pentagon, then it can be inscribed in any cross section. In some, the phantom fifth point falls outside the section and in some inside, but by IV it also sometimes falls on the boundary. – Yoav Kallus Nov 14 '14 at 0:34
• @Yoav Kallus: No non-square rhombus can be inscribed in a circle. Perhaps cyclic quadrilaterals have the property you need. – Douglas Zare Nov 14 '14 at 15:32

Corollary. Each convex body $K\subset \mathbb{R}^3$ is circumscribed about an affine-regular pentagon whose vertices lie on planes of support of $K$ parallel to one line.