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In Famous Puzzles of Great Mathematicians by Miodrag S. Petković there is this problem:

A large avalanche in the Alps traps an unhappy mole. When the avalanche stops, it turns out that the poor mole has been buried somewhere inside a snowball with an ellipsoidal shape with a volume of 500 cubic meters. The mole can dig a hole through the snow advancing at one meter per minute, but he only has the strenght and breath for 24 minutes. Can the mole reach the surface of the snowball and save his life?

In the provided answer, Petković states that the mole should dig a tunnel through the points $A$, $B$, $C$, $D$ in the picture below, which are vertices of a cube with sides 8 meters. Then he says: «Indeed, if all four points $A$, $B$, $C$, $D$ would lie inside the snowball, then all interior points of the cube constructed from the perpendicular segments $AB$, $BC$ and $CD$ would belong to the interior of the snowball.» (After this, the result follows immediately, of course)

Trouble is, I don't know how to prove that this must be the case. Of course if all eight vertices lied in the snowball this would be true, but here I may only say that the convex hull of the four point are within the snowball. Am I forgetting something obvious?

escaping path

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  • $\begingroup$ If the cube edges are aligned with the axes of the ellipsoid, then I think the symmetry of the ellipsoid makes the statement true. However, if for example AD were parallel with the long axis of the ellipsoid, but near the surface with B,C inside the ellipsoid, it seems false, or at least non-obvious. $\endgroup$ Mar 14, 2017 at 14:58
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    $\begingroup$ I thought of this, but I don't think that the problem let us use this hypothesis. After all, the poor mole does not have any idea of the orientation of the snowball. $\endgroup$
    – mau
    Mar 14, 2017 at 15:12
  • $\begingroup$ In general, when considering three vertices of a cube inside an ellipsoid, one can not infer that the whole cube is insode the same volume. So I think the truth of the above statement follows from the mere length of the sides compared to the axes of the ellipsoid. But I also think that the conclusion is far from trivial. $\endgroup$
    – M. Winter
    Mar 14, 2017 at 15:38
  • $\begingroup$ Actually, none of the other four cube vertices are going to belong to the smallest volume ellipsoid around $A$, $B$, $C$, $D$. However, the smallest volume is indeed larger that that of the cube. $\endgroup$
    – A.Γ.
    Mar 14, 2017 at 19:05

1 Answer 1

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Let us pick the coordinates for the points as $$ A:(1,0,0),\quad B:(0,0,0),\quad C:(0,1,0),\quad D:(0,1,1). $$ It is easy to verify that the tetrahedron with the vertices $$ A':(1,0,0),\quad B':(0,0,0),\quad C':\left(\frac12,\frac{\sqrt{3}}{2},0\right),\quad D':\left(\frac12,\frac{1}{2\sqrt{3}},\frac{\sqrt{2}}{\sqrt{3}}\right) $$ is regular. We can map the original tetrahedron ABCD to the regular one $A'B'C'D'$ by a linear map: $$ BA\mapsto B'A',\quad BC\mapsto B'C',\quad BD\mapsto B'D'. $$ The standard constructions gives the matrix of the map as $$ M=\begin{bmatrix}B'A' & B'C' & B'D'-B'C'\end{bmatrix}= \begin{bmatrix}1 & \frac12 & 0\\0 & \frac{\sqrt{3}}{2} & -\frac{1}{\sqrt{3}}\\ 0 & 0 & \frac{\sqrt{2}}{\sqrt{3}}\end{bmatrix} $$ with $\det(M)=\frac{1}{\sqrt{2}}$. It is known that the determinant is the volume scaling under linear mapping. It is also known (and clear from the symmetry) that the minimal volume ellipsoid around the regular tetrahedron is the circumscribed ball. The ball radius can be found as $R=\frac{\sqrt{3}}{2\sqrt{2}}$, so the ball volume is $$ V'=\frac{4\pi}{3}R^3=\frac{\pi\sqrt{3}}{4\sqrt{2}}. $$ Thus the original minimum volume ellipsoid has the volume $$ V=\det(M^{-1})V'=\sqrt{2}\frac{\pi\sqrt{3}}{4\sqrt{2}}=\frac{\pi\sqrt{3}}{4}\approx 1.3603 $$ which is larger than the volume of the unit cube. Hence, an ellipsoid of a volume that is smaller than that of the cube cannot include all the points $A,B,C,D$.

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