# Solving a Math Competition Problem: Part 5

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.

For a while, I toyed with the situation where we have

• Plane 1 β equator (this always happens: Just make plane 1 the equator) π1(0π, 0πΈ).
• Plane 2 β Prime Meridian π2(90π, 0πΈ)
• Plane 3 β Intl Date Line π3(90π, 90πΈ)
• Plane 4 β at an angle to all of those π4(45π, 45πΈ)

Here is our mapping with P1, P2, and P3 on it:

Now, how to represent P4? Aha! The inside of the unit circle is the southern hemisphere, and the outside is the northern. P4 must hit the equator a two points 180 degrees apart, go inside the southern hemisphere, and then outside to the northern. Thus:

The white region is a NONtriangular region created by the intersection of four planes. These are strange-looking regions, and I spent a long time β several days β vainly trying to count max regions created when I added P5, P6 etc. But one thing was clear: not all of the regions are triangular, nor can they be. For if a plane (say P4) cuts through a triangular region, it will create a new triangular region and a non-triangular βquadrilateralβ, as in the figure below. So counting triangles from points is NOT the solution here!

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