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Suppose we have a tetrahedron defined by points $(0,0,0),(1,1,0),(0,1,1),(1,0,1)$. Now define surface by $(a,b,a + b - 2ab)$ for $a,b$ between $0$ and $1$. Let $E_1$ be the set of points inside the tetrahedron and $E_2$ be the set of points on that surface, $A$ the area of that surface and $d$ be Euclidian distance

Is it true that for all surfaces $E_3$ with area not greater than $A$?

$$\max_{p \in E1} (\min_{q \in E2} (d(p,q))) \le \max_{p \in E1} (\min_{q \in E3} (d(p,q)))$$

g1 = Graphics3D[{Blue, Opacity[.1], 
    GraphicsComplex[{{0, 0, 0}, {1, 1, 0}, {0, 1, 1}, {1, 0, 1}}, 
     Polygon[{{2, 3, 4}, {3, 2, 1}, {4, 1, 2}, {1, 4, 3}}]]}];
g2 = Plot3D[a + b - 2*a*b, {a, 0, 1}, {b, 0, 1}, 
   PlotStyle -> Opacity[.5]];
Show[g1, g2]
  • 1
    Your surface $f(a,b)=a-2ab+b$ does not seem to be a minimal surface; the expression I'm getting for the mean curvature is not 0 but $\frac{2(2a-1)(2b-1)}{\sqrt{\left(4a(a-1)+4b(b-1)+3\right)^3}}$2010-08-27
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    oops, I meant to say "saddle surface", not "minimal surface"2010-08-28

1 Answers 1

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No. Your defined objective is the Hausdorff distance $d_H$ between a surface and the tetrahedron -- at least, it is as long as the surface is inside the tetrahedron. For the chosen surface, the distance is $d_H(E_1,E_2) = \frac{1}{2}$, attained at $p = \left(\frac{1}{2},\frac{1}{2},0\right), q = \left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$. You can have $d_H(E_1,E_3) \le \frac{1}{2}$ with the area of $E_3$ arbitrarily small by letting $E_3$ be a small sphere at $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$ with lots of very thin "spikes" reaching the boundary of the tetrahedron.