Suppose you have a large pizza (or a large circle). Assume that any time you pic

Question

Suppose you have a large pizza (or a large circle). Assume that any time you pick two points on the outside edge of the pizza you can "cut" the pizza with a straight cut that connects the two points (this cut is called a chord). Obviously, if you make such a cut. the pizza is cut into two pieces. If you repeat this process (i.e. make another cut), then depending upon where you choose the two points, your pizza is now cut into either 3 pieces (lower bound) or 4 pieces (upper bound). Your task is to learn more about how many pieces you might have if you cut the pizza 5 times, 8 times, or 10 times (see the table below). Can you find the "lower bound" and the "upper bound" on the number of pieces of pizza? Can you generalize this to what will happen if you make n cuts? How do you make the cuts to find the lower and upper bounds? [It should be a rule that you never repeat a cut - i.e. cut the same line - that you have made previously.] Here's a table to start the process: make sure you complete the table. Remember: Solutions must be typed together by each group of 2-3 members with careful attention paid to explaining your reasoning.

Explanation / Answer

So first we go for the lower bound. To have the minimum number of pieces the lines must not cross each other.

So,

Number of lines Number of pieces

1 2

2 3

3 4

4 5

5 6

6 7

7 8

8 9

9 10

10 11

n n+1

For upper bound

Eliminating the lines joining the adjacent points on the circumference, the chords, creates the familiar pizza puzzle where you are asked to determine the number of pieces a round pizza can be cut into with "n" straight line cuts across the pizza, without any three lines passing through the same point. In other words, how many regions are created with the increasing number of "n" lines? Lets explore.

Draw a circle and draw the diameter. Obviously, the circle is divided into two regions, or slices.
Draw another line crossing over the first line, Clearly, we can see that we now have four regions.
Draw a third line crossing over both of the previous two lines. Once again, we can readily see that we now have 7 regions.
Draw a fourth line crossing over the previous three lines. It is still fairly easy to see that we now have 11 regions
Draw a fifth line crossing over the previous four lines. With a little care, we can see that we now have 16 regions.
Lets step back and see what we have so far.

Lines.......Regions....Difference
...0...............1................-
...1...............2...............1
...2...............4...............2
...3...............7...............3
...4..............11..............4
...5..............16..............5

What can we learn from this information? It is immediately obvious that with each increase in the number of lines, we gain an identical number of regions, i.e., we add a fifth line and we add 5 regions. Another careful observation tells us that each number of regions is simply the sum of 1 plus the sum of the lines used up to that point. 1 = 1 + 0, 2 = 1 + 1, 4 = 1 +3, 7 = 1 + 6, 11 = 1 + 10, and 16 = 1 + 15. You might now recognize that the sums added to the 1 each time are simply the triangular numbers, the sums of the integers from 1 to n, being defined by Tn = n(n + 1)/2. Therefore, if we call the number of lines "n", the "n" lines will divide the circle up into Rn = 1 + n(n + 1)/2 = [n^2 + n + 2]/2. We can now extend our tabular data to

Lines.......Regions....Difference
...0...............1................-
...1...............2...............1
...2...............4...............2
...3...............7...............3
...4..............11..............4
...5..............16..............5
...6..............22..............6
...7..............29..............7
...8..............37..............8
...9..............46..............9
..10.............56.............10
...n......[n^2+n+2]/2........n

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