Can someone help me with this? Suppose you have a large pizza, Assume that any time you pick two points on the outside ednge of a pizza you can cut the pizza witha straight cut that connects the two points. If you repeat the process then depending on where you choose the two points the pizza is now cut into either 3 or 4 pieces. Now look at cutting the pizza 5 times, 8 times, 10 times or 20 times. Can you find a lower bound and an upper bound on the numbe of pieces? What are the best bounds you can find? Can you say what will happen if you make 'n' cuts? You can never repeat a cut that you made previously.
Pizza Math problem
- Thread starter Hilo
- Start date
These, *, **, and ***, are all questions for you to provide some work on first,
before you're to be given help on them.
12-year-old-poet
Junior Member
- Joined
- Jan 25, 2012
- Messages
- 78
that would be a good idea Denis. WHat kind of pizza shall it be?
12-year-old-poet
Junior Member
- Joined
- Jan 25, 2012
- Messages
- 78
Haha, I'm not a waitress, sorry. Ask someone else for that. I'd be happy to get money for that................ and a Dr.pepper. nothing better than a nice hit pizza and a cold drink. Yummy!!!!!!!!!
Pizza Heaven
Can someone help me with this? Suppose you have a large pizza, Assume that any time you pick two points on the outside ednge of a pizza you can cut the pizza witha straight cut that connects the two points. If you repeat the process then depending on where you choose the two points the pizza is now cut into either 3 or 4 pieces. Now look at cutting the pizza 5 times, 8 times, 10 times or 20 times. Can you find a lower bound and an upper bound on the numbe of pieces? What are the best bounds you can find? Can you say what will happen if you make 'n' cuts? You can never repeat a cut that you made previously.
Perhaps the following might open up some doors to your thinking.
Regions Created by Crossed Lines Within a Circle or The Pizza Saga
A popular problem in the field of recreational mathematics revolves around the ever popular pizza pie (or cake). It traditionally seeks the number of pieces the pie can be cut into by any number of straight slices across the pie.
The derivation of the number of regions created within different shaped figures comes in a variety of forms. Consider the following variations.
1--Irregular convex polygons divided into regions by connecting each vertex with every other vertex by straight lines.
2--Circles divided into regions by connecting "n" randomly located points on the circumference with each other by straight lines without any three lines passing through the same point. This is essentially joining all "'n" points in sequence and creating all the inner diagonals between the points.
3--Circles divided up by a series of lines that cross one another within the circle without any three lines passing through the same point. This is the same as the #2 variation with the exception that the points of intersection of the lines on the circumference are not joined to one another.
It is immediately obvious that the connection of "n" randomly located points by straight lines within the circle is simply the polygonal n-gon inscribed in a circle. Therefore, the number of regions created within the circle is the number of regions defined within the n-gon plus the "n" arc/chord regions created when inscribing the n-gon within the circle.
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
Another way of deriving this expression results from the analysis of the particular sequence of numbers that the data creates. Looking at the preliminary data again, we have
n = lines..........1.....2.....3.....4.....5
N = regions.....2.....4.....7....11...16
Taking the successive differences we get
n = lines..........1.....2.....3.....4.....5.....6.....7
N = regions.....2.....4.....7....11...16
1st Diff................2.....3.....4.....5
2nd Diff..................1......1.....1
With the 2nd differences being constant, we recognize that we have a finite difference sequence with 2nd differences equal to 1. Therefore, the general expression for N in terms of "n" is of the form N = an^2 + bn + c. Using the data, we can write
a(1^2) + b(1) + c = 2 or a + b + c = 2
a(2^2) + b(2) + c = 4 or 4a + 2b + c = 4
a(3(2) + b(3) + c = 7 or 9a + 3b + c = 7
Solving, a = 1/2, b = 1/2 and c = 1 leading to N = n^2/2 + n/2 + 1 = (n^2 + n + 2)/2, the same as we deduced earlier.
One cute across the pie has created two pieces. Two cuts have created a total of four pieces. Notice that the cuts do not have to go through the center of the pie, but they do have to be straight. Also, the pieces do not have to be the same size or shape. The next is three cuts and it produces seven pieces. You should be able to convince yourself that seven is the largest number of pieces that can be created by three cuts across the pie.
1) Find the largest number of pieces you can get from four cuts and from five cuts. Include those numbers in you table.
Number of Cuts 1 2 3 4 5 ... 10 n
Max. #of pieces 2 4 7
2)a. Try to find a pattern describing what is happening in the table.
b. Use your pattern to find the largest possible number of pieces from ten cuts.
c. Try to explain why this pattern is occuring (look back at your drawings).
3) Find a rule for the nth term in your table above.
Help! I don't know how to get a formula! Can you help? If so, please e-mail me back at Tiger8685@aol.com Thanks! >>
If memory serves me correctly, extending the table out looks something like
Number of Cuts 0 1 2 3 4 5 6 7 8 9 10 n
Max. # of pieces 1 2 4 7 11 16 22 29 37 46 56 (n^2 + n)/2 + 1
Difference 1 2 3 4 5 6 7 8 9 10 n
The differences are simply the sequence of positive integers from 1 on up. The maximum number of pieces, Pm, is the sum of the positive integers from 1 to n plus one or Pm(n) = (n^2 + n)/2 + 1. Thus, Pm(3) = (3^2 + 3)/2 + 1 = 7. Pm(4) = (4^2 + 4)/2 + 1 = 11. Pm(5) = (5^2 + 5)/2 + 1 = 16.
Can someone help me with this? Suppose you have a large pizza, Assume that any time you pick two points on the outside ednge of a pizza you can cut the pizza witha straight cut that connects the two points. If you repeat the process then depending on where you choose the two points the pizza is now cut into either 3 or 4 pieces. Now look at cutting the pizza 5 times, 8 times, 10 times or 20 times. Can you find a lower bound and an upper bound on the numbe of pieces? What are the best bounds you can find? Can you say what will happen if you make 'n' cuts? You can never repeat a cut that you made previously.
Perhaps the following might open up some doors to your thinking.
Regions Created by Crossed Lines Within a Circle or The Pizza Saga
A popular problem in the field of recreational mathematics revolves around the ever popular pizza pie (or cake). It traditionally seeks the number of pieces the pie can be cut into by any number of straight slices across the pie.
The derivation of the number of regions created within different shaped figures comes in a variety of forms. Consider the following variations.
1--Irregular convex polygons divided into regions by connecting each vertex with every other vertex by straight lines.
2--Circles divided into regions by connecting "n" randomly located points on the circumference with each other by straight lines without any three lines passing through the same point. This is essentially joining all "'n" points in sequence and creating all the inner diagonals between the points.
3--Circles divided up by a series of lines that cross one another within the circle without any three lines passing through the same point. This is the same as the #2 variation with the exception that the points of intersection of the lines on the circumference are not joined to one another.
It is immediately obvious that the connection of "n" randomly located points by straight lines within the circle is simply the polygonal n-gon inscribed in a circle. Therefore, the number of regions created within the circle is the number of regions defined within the n-gon plus the "n" arc/chord regions created when inscribing the n-gon within the circle.
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
Another way of deriving this expression results from the analysis of the particular sequence of numbers that the data creates. Looking at the preliminary data again, we have
n = lines..........1.....2.....3.....4.....5
N = regions.....2.....4.....7....11...16
Taking the successive differences we get
n = lines..........1.....2.....3.....4.....5.....6.....7
N = regions.....2.....4.....7....11...16
1st Diff................2.....3.....4.....5
2nd Diff..................1......1.....1
With the 2nd differences being constant, we recognize that we have a finite difference sequence with 2nd differences equal to 1. Therefore, the general expression for N in terms of "n" is of the form N = an^2 + bn + c. Using the data, we can write
a(1^2) + b(1) + c = 2 or a + b + c = 2
a(2^2) + b(2) + c = 4 or 4a + 2b + c = 4
a(3(2) + b(3) + c = 7 or 9a + 3b + c = 7
Solving, a = 1/2, b = 1/2 and c = 1 leading to N = n^2/2 + n/2 + 1 = (n^2 + n + 2)/2, the same as we deduced earlier.
One cute across the pie has created two pieces. Two cuts have created a total of four pieces. Notice that the cuts do not have to go through the center of the pie, but they do have to be straight. Also, the pieces do not have to be the same size or shape. The next is three cuts and it produces seven pieces. You should be able to convince yourself that seven is the largest number of pieces that can be created by three cuts across the pie.
1) Find the largest number of pieces you can get from four cuts and from five cuts. Include those numbers in you table.
Number of Cuts 1 2 3 4 5 ... 10 n
Max. #of pieces 2 4 7
2)a. Try to find a pattern describing what is happening in the table.
b. Use your pattern to find the largest possible number of pieces from ten cuts.
c. Try to explain why this pattern is occuring (look back at your drawings).
3) Find a rule for the nth term in your table above.
Help! I don't know how to get a formula! Can you help? If so, please e-mail me back at Tiger8685@aol.com Thanks! >>
If memory serves me correctly, extending the table out looks something like
Number of Cuts 0 1 2 3 4 5 6 7 8 9 10 n
Max. # of pieces 1 2 4 7 11 16 22 29 37 46 56 (n^2 + n)/2 + 1
Difference 1 2 3 4 5 6 7 8 9 10 n
The differences are simply the sequence of positive integers from 1 on up. The maximum number of pieces, Pm, is the sum of the positive integers from 1 to n plus one or Pm(n) = (n^2 + n)/2 + 1. Thus, Pm(3) = (3^2 + 3)/2 + 1 = 7. Pm(4) = (4^2 + 4)/2 + 1 = 11. Pm(5) = (5^2 + 5)/2 + 1 = 16.