Report on maths prize question for 10 October

The question was about how many regions you get if you mark n points on the circumference of a circle and draw all the lines connecting them.

Three students, Hayden Leroux, Deniz Yukselir, and Hamse Adam, got good answers and earned prizes.

At first it looks as if you get 2n–1 regions. You get 1 region for 1 point, 2 regions for 2 points, 4 regions for 3 points, 8 regions for 4 points, 16 regions for 5 points.

Then you see that the pattern is more complicated: 31 regions for 6 points, 57 for 7 points, 99 for 8 points…

Hayden and Deniz saw correctly that the rule is: the number of regions = the total of the first five numbers in the (n–1) row of Pascal’s triangle.

Hamse worked the numbers from what he saw as a modified Pascal’s triangle. See for a discussion, which also includes an alternative proof.

Deniz tried to work the numbers out by classifying the regions into segments, triangles, quadrilaterals, pentagons, and so on. But his counting was not quite right: for six points you get 6 segments, 19 triangles, 3 quadrilaterals, and 3 pentagons.


Claim: for n points on the circumference, we get nC4+nC2+1 areas. This is also equal to the sum of the first 5 elements of the (n−1) row of Pascal’s triangle.

Proof: If necessary, tilt the circle microscopically so that every meeting-point of lines (or of a line and the circumference) is at a different height.

Then every region in the circle has exactly one meeting-point as its lowest vertex, and every meeting-point which is inside the circle (rather than on the circumference) is the lowest vertex of exactly one region.

There are nC4 meeting-points inside the circle, because every four points on the circumference define such a meeting point. Each is a lowest vertex of one region.
There are nC2 lines joining points. If we go round the circle from the lowest point, each such line defines a region (which we have not already counted) to its right at the lower of the two points which the line connects.

And then there is one more region to the left of the left-most line at the bottom point.

Therefore total regions = nC4+nC2+1

Since nC2=(n−1)C1+(n−1)C2 and nC4=(n−1)C3+(n−1)C4 and (n−1)C0=1, also:

total regions = sum from r=0 to r=4 of (n−1)Cr = sum of first 5 elements of (n−1) row of Pascal’s triangle. ◻

[[nCr is the number of ways of choosing r different items from a collection of n items. There’s an nCr key on your calculator (shift of the multiply key). See]]