# Solving a Math Competition Problem: Part 3

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.) AHH! Insight! Each plane must intersect the others because they all pass through the center. And two planes intersect in a line. And the line must intersect the sphere at two points. SO, we can count intersection points: There are 9 planes, and each plane will intersect the other 8, so there are 9 ∗ 8 = 72 intersection points IF we arrange the planes for maximum regions. More generally, if we have n planes arranged for max intersection points, we will have 𝑛(𝑛 − 1) intersection points.
Wait, let’s do this carefully. There are 9 planes, and they can each intersect 8 different planes; but that counts the intersections of plane A and plane B twice, so there are (9*8)/2 = 36 lines of intersection, but 36 ∗ 2 = 72 points of intersection with the sphere. So our problem just got narrower: Given 72 intersection points defining various regions on the sphere, how many regions do we get?
And that’s where the problem stands as of this writing. My preliminary conjecture is that each region will be a “triangle” (officially, spherical triangle) on the surface of the sphere, especially if we are maximizing regions. I need to prove that conjecture and then count triangles, which shouldn’t be too hard.

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