**Spheres
in High Dimensions, Part 1**

The problem we'll discuss here is intended for a motivated high school student or an adult with a comparable understanding of mathematics. Yet I used it on my department's Ph.D. qualifying exam.

How can a high school student hope to understand a problem that stretched the minds of most doctoral candidates at one of the most prestigious research institutions in the world? By my providing many of the thought processes that I expected the doctoral candidates to furnish on their own.

Even with that help, the problem may seem to hopelessly confuse you at first, in which case I encourage you to give it a rest and come back to it when you're refreshed. Even better, get a small group of friends to work on it with you. I've often been amazed at how a group can quickly solve difficult problems that each individual student found impossible.

Turning back to my use of this problem on our Ph.D. qualifying exam, since different students will have taken different elective courses, I try not to ask questions that rely on knowledge of a specific fact. Rather, I seek questions that test a student's ability to do original research since that is the biggest hurdle in doctoral studies. So here's my favorite question of that nature:

I have a solid gold sphere of radius 10, that is worth one million dollars. The unit of length isn't inches or centimeters or anything else you're familiar with. Rather, it was chosen so that a solid gold sphere of radius 10 is worth exactly one million dollars.

This solid sphere is then broken into two pieces:

- an inner sphere of radius 9, and
- an outer shell of thickness 1.
You can have either the inner sphere of radius 9 or the outer shell of thickness 1. Which piece do you want?

Your choices are shown below in a cross-sectional figure drawn to scale. Note that the outer shell has inner radius 9 and outer radius 10, so that it fits around the inner sphere and together they form a solid sphere of radius 10.

There is one complication in all this: You are not living in a 3-dimensional world. Rather, you and the sphere are 100-dimensional creatures.

**Solid
Gold Sphere Divided Into**

**An
Inner Sphere of Radius 9 and **

**An
Outer Shell of Thickness 1**

What I like about this question is that, just as in original research, it throws the student into the chaos of the unknown. What the heck is 100-dimensional space, much less a 100-dimensional sphere? But, as you'll see, you don't need to have any idea about that to solve the problem.

The real question is, what does the student do when he finds himself in unknown territory with seemingly no signposts to guide him? Does he freeze up out of fear of taking a step in the "wrong direction"? Or does he have the courage to make mistakes in seeking the answer and learn from his missteps? In the latter event, the seeming missteps often provide clues as to the approach that is needed and aren't missteps at all.

Almost all the students taking my exam needed some help in getting started, in which case I would say "OK, you don't know how to work this problem. Is there any related problem you do know how to solve?" After which most students would say they knew how to solve it in three dimensions, which I would encourage them to do. One or two truly exceptional students would start without my help, by saying they had no idea how to work the problem in 100 dimensions so, to try and get some feel for the problem, they would back up and first work it in 3-dimensions. That kind of thinking, and the courage to follow it, is an excellent indicator of ability to do original research.

With or without my help, once they were on this track, their work would proceed something like this: The volume of a normal, 3-dimensional sphere is

V = (4/3) π r^{3}

so the solid inner sphere of radius 9 has volume

V(inner) = (4/3) π
9^{3}

Since the outer shell of thickness 1 consists of a sphere of radius 10 minus a sphere of radius 9 it has volume

V(shell) = [(4/3)
π 10^{3}] - [(4/3) π 9^{3}]

So the ratio of the volumes, and the ratio of the value of the gold, is

V(shell ) / V(inner)
= {[(4/3) π 10^{3}] - [(4/3) π 9^{3}]}
/ [(4/3) π 9^{3}]

The factors of [(4/3) π] can be cancelled, resulting in

V(shell ) / V(inner)
= (10^{3} - 9^{3}) / 9^{3}

= (1000 - 729) / 729

= 271 / 729 = 0.371742

to six decimal places. Since this ratio is less than 1, the outer shell has less volume, and gold, than the inner sphere, so we choose the inner sphere. A little more math would show that the inner sphere has $729,000 worth of gold, while the outer shell is worth $271,000.

In most cases our intuition, or looking at the above figure, would also have told us to take the "larger" inner sphere. After all, the shell is only of thickness 1, while the inner sphere is solid gold of radius (thickness) 9. But, let's see what happens when we try to extend our work in three dimensions to the original, 100-dimensional problem.

The students who got this far usually noticed that, in 3 dimensions, the ratio reduced to

(10^{3}
- 9^{3}) / 9^{3}

Where do the three numbers that appear (10,9 and 3) come from? 10 is the radius of the entire sphere, 9 is the radius of the inner sphere, and 3 is the dimensionality. So a natural guess in 100 dimensions would be

(10^{100}
- 9^{100}) / 9^{100}

With a little algebra, this ratio can be reduced to

(10/9)^{100}
- 1^{100} = (10/9)^{100} - 1

Evaluating this expression on a calculator, we find that in 100 dimensions the outer shell has almost 38,000 times as much gold as the inner sphere. Again, a little additional math can find the actual amounts in each piece. The inner sphere has $26.56 worth of gold, while the outer shell is worth $999,973.44.

While the miniscule fraction of gold in the inner sphere offends our intuition, that is not surprising. We live in 3 dimensions, not 100. When we work on paper, we get some experience in 2 dimensions, but we have no experience that even approximates what goes on in 100 dimensions. So, in hindsight, it is understandable that our intuition leads us astray.

This is a good stopping point, but for those who would like to take this problem a bit further, you can go on to Part 2 whenever you're ready