math problem

turbodog

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Ok boys and girls, let's see who can tackle this one.

This is one of those things that has bugged me for a while now. There's no practical application that I can think of, but I'd still like to figure it out.

Let's say you have a pizza you want to cut into equal 1/3 size pieces, except you want to cut vertically... like cutting the pizza into strips (no pie shaped pieces). Where to cut?

I've approached this from several angles (haha), but have come up short. Best I can think of is integrating the general equation of a circle.
 
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turbodog

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Funny how when you describe something to someone else it takes you in a new direction. I think I can do this with chords and circular segments.

Will see...
 

turbodog

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As a full circle has an angle total of 360° you´ll just have to make kind of a star shaped a jig with a three piece splitting system at 120° inside angle and that is it, will fit any pizza.

Cheers

RBR

As originally said, I want to cut vertical strips of 1/3 area each. Not pie shaped.
 

Kestrel

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An experimental solution instead ... ?

Cut out a circular piece of corrugated cardboard the same size as the pizza.
Weigh the cardboard and divide by three.
Cut vertical strips off of one side of the cardboard until the one-third weight is obtained.
Cut the pizza to model the remaining 2/3 of the pizza & repeat the same cut for the other side.

Edit: Bad idea to do this just before lunch, now I want to go out for pizza ...
 
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reppans

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^^ haha... good one.

Hope 2 of the 3 like crust more than cheese and tomato sauce :D
 

turbodog

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Re: An experimental solution instead ... ?

Cut out a circular piece of corrugated cardboard the same size as the pizza.
Weigh the cardboard and divide by three.
Cut vertical strips off of one side of the cardboard until the one-third weight is obtained.
Cut the pizza to model the remaining 2/3 of the pizza & repeat the same cut for the other side.

Edit: Bad idea to do this just before lunch, now I want to go out for pizza ...

Excellent 'analog' idea.
 

G. Scott H.

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You just had to go and burden me with this question, didn't you? Sheesh! I've been trying to figure out how many licks it takes to get to the center of a Tootsie Pop for 40 years! Now I have to worry about this as well? What kind of monster are you? :D
 

turbodog

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You just had to go and burden me with this question, didn't you? Sheesh! I've been trying to figure out how many licks it takes to get to the center of a Tootsie Pop for 40 years! Now I have to worry about this as well? What kind of monster are you? :D

Someone figured out the tootsie pop thing not that long ago. Takes about 1000 licks per inch of diameter.
 

TEEJ

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Ok boys and girls, let's see who can tackle this one.

This is one of those things that has bugged me for a while now. There's no practical application that I can think of, but I'd still like to figure it out.

Let's say you have a pizza you want to cut into equal 1/3 size pieces, except you want to cut vertically... like cutting the pizza into strips (no pie shaped pieces). Where to cut?

I've approached this from several angles (haha), but have come up short. Best I can think of is integrating the general equation of a circle.

If this is a sicilian pizza, it is, potentially, a very easy problem. You did not specify the type of pizza though....or whether it was perfectly square, perfectly round or rectangular, etc. For example, no real pizza is dimensionally symmetrical....so, the "round" ones tend to be somewhat out of round, and so forth.

Pizza also have texture, so, if concerned with "area" this typically means surface area, and the surface area will vary if its got chunky sauce or if its a plain white pie vs a Hawaiian or pepperoni and sausage, and so forth. IE A rough textured topping morphology can have over twice the surface area per pizza of the same perimeter dimensions.

:D
 

orbital

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+

cup+o+pizza.jpg
 
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bestellen

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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.
 
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