## Response from Prof. B.

The Java applet looks really good on my SUN. Is it possible to get the code that you wrote to obtain this example? One way for me to try to understand what you have done is for me to try to modify the program to get another example, and I'm looking for
ward to that.

By the way, whatever program you used to write your Week 10 response will not let me make a modification to insert my standard "Prof. Banchoff's Response" line. Do you know how that happened?

It seems to me that your analysis of the slicing along the long diagonal is correct and I'm not sure what remaining problems there are. The lowest point will always be the origin, with all coordinates 0, and the highest will be at level n, where a
ll coordinates are 1. At level 1, you will have all the points with one coordinate 1 and all the rest 0, and there will be n of them. The centroid of this collection of points is just their sum divided by n, so we get the point with all coordinates equa
l to (1/n). The distance from this point to the origin is the square root of the sum of the squares, i.e. the square root of n times (1/n)^2, so sqrt(1/n). In the case of the square, the centroid is (1/2,1/2) at distance sqrt(1/2) for example. For the
hypercube, you get something nicer since the centroid is (1/4,1/4,1/4,1/4) and the distance from this to the origin is sqrt(4 * (1/16)) = sqrt(1/4) = 1/2. So between 1/2 and 1 along the long diagonal, the slices will be truncated tetrahedra, then at 1 we
get the octahedron, with six coordinates, each with two 1's and two 0's. These are precisely the vertices where the sum of the coordinates equals 2, and the centroid of these will be the point with all coordinates 1/2 (since each coordinate appears in t
hree of the six points). This centroid has distance 1 from the origin, as predicted.

So, is that what you were expecting? I look forward to seeing this slicing on a nice applet soon!

**Additional Comment**I certainly wish that I had read
Jeremy's contribution before I answered
yours. I would have saved me some time! So now there are two versions of the same an
swer to your good question.