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Is it possible to inscribe at least one regular tetrahedron in every convex body?

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Is the 2d case known? I mean, is it possible to inscribe at least one regular triangle in every convex body in the plane? –  lhf Dec 13 '11 at 0:43
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I suppose you haven't seen this, or this, or this... –  J. M. Dec 13 '11 at 0:45
    
No, I haven't. Thanks for the links! –  Vladimir Reshetnikov Dec 13 '11 at 19:05

3 Answers 3

up vote 7 down vote accepted

The first theorem of Section 4 of this paper, mentioned by J.M. in the comments, gives an affirmative answer, citing V.V Makeev, Inscribed simplices of a convex body (in Russian), Ukr. Geom. Sb. 35 (1992), 47-49 = J. Math. Sci. 72 (1994) (4), 3189-3190, MR 95d:52006:

Theorem. Let $K\subset\mathbb{R}^n$ be a convex body. Then $K$ admits an inscribed similar copy of any prescribed simplex.

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Suppose this is in $\mathbb{R}^3$, and of course it is necessary for the convex body to affinely span the whole $3$-dimensional space (since a regular tetrahedon within the body would space that whole space). (Is that part of the definition of "body"?) Then there are four affinely independent points $a,b,c,d$ within the convex body. Consider the point $o=(a+b+c+d)/4$. Let $p,q,r,s$ be the vertices of a regular tetrahedron centered at $o$, and the consider the points $o+\varepsilon(p-o)$, $o+\varepsilon(q-o)$, $o+\varepsilon(r-o)$, $o+\varepsilon(s-o)$, where $\varepsilon>0$. These are vertices of a regular tetrahedron. For $\varepsilon$ small enough, the weights $w,x,y,z$ such that $o+\varepsilon(a-o)=wa+xb+yc+zd$ should be close to $1/4$, hence all positive, and similarly for $b, c, d$. So those vertices should be in the convex hull of $\{a,b,c,d\}$, hence within the original convex body.

However, I wonder if I'm missing something?

Later edit: My suspicion is confirmed below: I missed one of the definitions.

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Inscribed means that the vertices of the tetrahedron lie on the boundary of the convex body. –  Brian M. Scott Dec 13 '11 at 8:07
    
Thanks. I suspected, from the way things were phrased, that there was some definition involved that I didn't know about. That's what "inscribed" means when you're talking about conics, etc., so I'm not surprised. –  Michael Hardy Dec 13 '11 at 18:39

Here is a simple proof.

Fix a coordinate system. Suppose that the body sits between $z=z_0$ and $z=z_1$.

For each $z_0 < z < z_1$, cut the body with a plane parallel to $xy$-plane. Inscribe an equilateral triangle in this intersection (to show it is possible, repeat this proof for two dimensions).

Let $G_z$ be the circumcircle of this triangle and $l_z$ be the length of the edge of the triangle, lets call the triangle $A_zB_zC_z$.

Draw a semi-line parallel to the $z$ axis, going up, and starting at $G_z$. Let $L_z$ be the intersection of this semiline with the body (it's unique by convexity)$.

Let $f(z)=\frac{G_zL_z}{l_z}$. This ratio is continuous as a function in $z$.

It is easy to prove that

$$\lim_{z \to z_0^+} f(z)= \infty \,,\, \lim_{z \to z_1^-} f(z)=0 \,.$$

Then, by the intermediate value theorem, there exists some $z_3$ so that $f(z_3)$ is exactly the ratio of the height of the regular tetrahedron/edge of the regular tetrahedron.

Now it is trivial to prove that $A_{z_3}B_{z_3}C_{z_3}L_{z_3}$ is a tetrahedron which satisfies the requirements.

Moreover, you can change the proof to show that you can inscribe such a tetrahedron so that its "basis" is parallel to any given plane....

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I think the "easy to prove" property only works for a body that has a smooth (positive) curvature at the $z_0$ $z_1$ extreme points. Nice proof, though. –  leonbloy Dec 13 '11 at 19:19
    
Yea missed that... Anyhow this construction can be done with the plane perpendicular on a given direction, so I guess one could start by taking a point on the body of positive curvature, and take the normal direction to the tangent plane... This should work, unless the "trivial" case when the curvature is always zero is not really trivial... –  N. S. Dec 19 '11 at 0:29

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