Solving a Math Competition Problem: Part 8

This series of posts concerns solving the following problem from the 2016 University of Maryland High School Mathematics Competition.

A sphere is divided into regions by 9 planes that are passing through its center. What is the largest possible number of regions that are created on its surface?

a. 2^8

b. 2^9

c. 81

d. 76

e. 74

This series was actually written by my friend Jeff Cagle, department head for mathematics at Chapelgate Christian Academy, as he tried technique after technique to solve this problem. I thought that his resolution to the problem was an excellent example of the process of mathematical problem-solving, and (with his permission) I am posting the process of his solution here. (For the record, I have no doubt that I would not have been able to solve this problem.)

OK, so I wanted to prove that each region would be a triangle. So I decided to project the sphere onto a plane.

The projection of four planes:

Conjecture: The max number of regions is the number of intersection points plus 2.
Proof (by induction)
If we have 1 plane, we have no intersection points and 2 regions. Suppose we have n planes with 𝑛(𝑛 − 1) intersection points and 𝑛(𝑛 − 1) + 2 regions. Now we add the next plane to our figure. The plane creates a circle on the sphere. To maximize the number of regions, we angle the plane so that our circle does not intersect any already-existing intersection points. So the circle goes through a number of segments. Each time it does, it cuts the region bounded by that segment into two. So for each new intersection point, we lose one region and gain two, for a net gain of one region. That is, however many intersection points are added, that will be the number of regions added as well. And since 𝑛 + 1 planes have (𝑛 + 1)(𝑛) intersection points, we will
have (𝑛 + 1)(𝑛) + 2 max regions. DONE.
For the original competition problem, we have 9 planes and hence 9*8 + 2 = 74 regions, answer e.

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