Find the area!

Geometry Level 5

$S=\{(x,y,z) \in \mathbb{R}^3: x^2+y^2+z^2=1\}$

Let $S$ be the surface of a unit sphere centered at the origin.

Let $F$ be a subset of $S$ . Suppose that there exist three subsets $F_1,F_2,F_3$ of $S$ , each of which is obtained by rigidly rotating the set $F$ around the origin such that:

These three sets are pairwise disjoint, i.e., $F_i\cap F_j = \phi, i\neq j$
Their union spans the entire surface of the sphere, i.e.,
$F_1\cup F_2 \cup F_3=S$

Find the area of any such set $F$ satisfying the above conditions.

P.S. Rotation is defined in the usual way: $F_i=\{(x,y,z): \exists (a,b,c) \in F \text{ s.t. } (x,y,z)'=Q_i (a,b,c)'\}$ for some orthonormal matrix $Q_i,i=1,2,3$ .

4 \pi

\frac{4 \pi}{3}

No such

F

exists

F

exists but its area might not be defined

1 solution

Abhishek Sinha
Apr 18, 2016

Select the set $F$ as in Banach Tarski Paradox. It satisfies all the given conditions, yet does not have a well-defined area.

I think the wording "which is obtained by rotating $F$ around the origin in some way" is a bit weak. Does it mean that $F, {F}_{1}, {F}_{2}, {F}_{3}$ are all the same, through combinations of rigid-body motions on the surface of the sphere (loosely think of the skin of an orange sliced into $3$ identical wedges--except at the tips!), or does it mean they're surface areas swept by $F$ , so that $F$ is not necessarily the same as the others via such rigid-body motions on the surface? Furthermore, does "rotating" imply a single application of the rotation matrix, or can combinations be used?

This whole thing has has the smell of that Banach Tarski Paradox, so I picked my answer accordingly. But what about the others not familiar with it?

Michael Mendrin - 5 years, 1 month ago

Rotation is defined in the usual way: $F_i=\{(x,y,z): \exists (a,b,c) \in F \text{ s.t. } (x,y,z)'=Q_i (a,b,c)'\}$ for some orthonormal matrix $Q_i,i=1,2,3$ .

The point of the problem is to introduce the Banach Tarski Paradox to others who are not familiar with it yet.

Abhishek Sinha - 5 years, 1 month ago

Thanks, that clarifies the matter. I think that should be part of the question, for those not familiar with this.

Michael Mendrin - 5 years, 1 month ago

Can you show us how the sets in your problem, $F,F_1,F_2,F_3$ relate to the Proof Sketch in your link. How do you end up with three sets and not four? Also, don't we have to worry about all this working only "up to a countable set?" Finally, your definition of a "rotation" is a bit unusual as it seems to allow for a reflection.

Otto Bretscher - 5 years, 1 month ago

Explicit construction of these sets are intractable. Usual proofs invoke Axiom of Choice in some way.
Anything larger than $2$ is attainable, i.e., there exists a set $F$ such that the surface of the sphere can be tessellated simultaneously in $3,4,5 \ldots,$ copies of rotated and disjoint version of $F$ .

3 and 4. Does not matter to the answer of this problem.

Abhishek Sinha - 5 years, 1 month ago

Thanks! Your link does not relate well to your problem, both in notation and content. Do you have a better link or can you perhaps write your own solution? In your link, $S$ is partitioned into four subsets that are not obtained from each other by rotations. Also, things only work out "up to a countable set".

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – Well, I got introduced to it long ago in my probability class, where this is the first result we were shown so that we appreciate the need for learning measure theory seriously. There is a classic book of David Williams named "Probability with Martingales", where this result appears in its very first page.

Abhishek Sinha - 5 years, 1 month ago

@Abhishek Sinha – Interesting! Thanks for introducing us to this result! This sounds very different from the usual formulations of Banach-Tarski (the sets in the partition are not usually required to be congruent), more in the spirit of the Hausdorff Paradox (a precursor of Banach-Tarski). I will have to take a look at Stan Wagon's classic text on the subject.

Otto Bretscher - 5 years, 1 month ago

Your definition of a rotation is still not quite right:You want an orthogonal matrix with determinant 1.

Otto Bretscher - 5 years, 1 month ago

I don't get your comment. The absolute value of determinant of any orthonormal matrix is always 1.

Abhishek Sinha - 5 years, 1 month ago

But a rotation is defined as an orthogonal transformation with determinant 1.

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – Again it does not hurt if we allow for generalized rotations (such as reflections).

Abhishek Sinha - 5 years, 1 month ago

@Abhishek Sinha – I have never heard anybody call a reflection a "generalized rotation", but we are all entitled to our own terminology ;)

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – Apparently the wikipedia article uses the terminology that I used here. They call it improper rotation.

Abhishek Sinha - 5 years, 1 month ago

@Abhishek Sinha – Well, the terminology you are using seems improper indeed ;)

I'm worried that you are planting improper ideas in our young Comrades' minds: 99% of the linear algebra instructors around the world would mark your definition of a rotation as wrong.

Otto Bretscher - 5 years, 1 month ago

Find the area!

1 solution

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