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How to prove or disprove that the boundary of any convex body in $\mathbb{R}^3$ (treated as a surface) includes 5 points which form a regular planar pentagon?

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    $\begingroup$ I asked that in Mathoverflow and was answered. $\endgroup$
    – user64494
    Dec 19, 2014 at 21:15
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    $\begingroup$ And the answer appears to be that the answer to the question is unknown. $\endgroup$ Dec 25, 2014 at 19:47
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    $\begingroup$ It reminds my the Inscribed square problem: Consider a Jordan curve in the plane. Do they exist $4$ distinct points of the curve which are the vertices of a square? This is still an open problem but it is know to be true if the curve is convex. If I remember correctly the proof of the convex case is not so hard. I wonder if we could generalize it to answer your question $\endgroup$ Feb 10, 2015 at 19:06
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    $\begingroup$ More generally we could even consider the question "If $S$ is the boundary of a convex body in $\mathbb{R}^n$, do there exist always $n+2$ points on $S$ which form a regular planar polygon". $\endgroup$ Feb 10, 2015 at 19:07
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    $\begingroup$ Suppose the body is a pentagonal prism... Unless you meant 'the boundary of every convex body in $\Bbb R^3$'...? $\endgroup$
    – CiaPan
    Apr 27, 2015 at 22:14

1 Answer 1

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This is about a special case raised by Han de Bruijn: yes, you can inscribe a regular planar pentagon in a regular tetrahedron.

And you can do that in many ways, probably. Here is one of them.

Take a regular tetrahedron $ABCD$ with the edge length equal $1$. Let $E$ be a point on the $BC$ edge at distance $x$ from $B$. Similary $F$ on $DC$, $x$ apart from $D$.
The triangle $\triangle EFA$ is isosceles, degenerating continuously from the equilateral $\triangle BDA$ to the segment $CA$ as $x$ changes from $0$ to $1$.
That means $\epsilon = \angle EAF$ changes from $60^\circ$ to $0$. 'Somewhere' during the change it is $\epsilon = 36^\circ$. Where?

Let $y=AE=AF$ and $z=EF$. Due to the law of cosines in $\triangle ABE$ $$y^2 = 1^2 + x^2 - 2\cdot 1\cdot x\,\cos(\angle ABE) = x^2 + x + 1$$ and by the similarity $\triangle EFC \sim \triangle BDC$ $$EF = EC\times BD:BC$$ $$z = 1-x$$

In the isosceles $\triangle EFA$ $$\sin\frac\epsilon 2 = \frac z2:y$$ We want $\epsilon=36^\circ$: $$\left .\sin 18^\circ = \frac{1-x}{\sqrt{x^2+x+1}} \ \ \right\vert \big(\big)^2$$ $$\sin^2 18^\circ = \frac{x^2-2x+1}{x^2+x+1} = 1 - \frac{3x}{x^2+x+1}$$ $$\cos^2 18^\circ = \frac{3x}{x^2+x+1}$$ $$x^2+x+1 = \frac{3x}{\cos^2 18^\circ}$$ $$x^2+\left(1-\frac{3}{\cos^2 18^\circ}\right)x+1 = 0$$ The parenthesized expression is approx. $-2.317$ and one of the roots falls between $0$ and $1$: $$x\approx 0.573$$ Once we have $36^\circ$ at $A$ we can fit a regular pentagon $JKLMN$ in the $\triangle EFA$ so that $JK \subset FA$, $LM \subset AE$, $KL \parallel EF$. If it is too small or too big, scale it so that $N$ becomes the midpoint of $EF$.

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  • $\begingroup$ This is a special case (the regular case) of the special case (the tetrahedron) raised by Hans de Bruijn. $\endgroup$
    – TonyK
    Feb 19, 2016 at 16:12

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