# You, I and a Rabbi order a pizza?

You, I, and a rabbi order a 16” pizza. Being unreasonable, you order one of those awful stuffed

crust pizzas from pizza hut. Being equally unreasonable and hating the greasy cheese-wax they

put in the crust, I want as little of said crust as possible on my portion but insist on exactly one

third of the area. The rabbi decides the best way to deal with this debacle is to cut the pizza

with two vertical strokes. If you feel compelled to cut the pizza so that all three sections

are of equal area so as not to short me or the rabbi,

where (ie. at what distance from the center of the pizza) should you make the cuts?

### 2 risposte

- 9 anni faMigliore risposta
I see you ordered integral pizza....

You need to take the integral of the circumference

2 (1/2 r^2 arcsin (x/r) + x sqrt(r^2-x^2) )

I have doubled the formula putting a factor of 2 because we are computing the area, so you have to sum the two equal areas "above the x-axis" and "below the x-axis"

this you must integrate between -r and -r0 in order to give 1/3, and it will also be that integrating between -r0 and +r0 gives 1/3 (this is your slice), and of course the third slice is symmetric to the first.

r=8" as the diameter of pizza is 16"

Do you need that I compute r0 or can you finish it?

EDIT: I am going to try to explain in more detail.

Assuming that the Rabbi really wants to eat 1/3 of this nonkosher food, I don't think the best way is two vertical cuts. The best way is asking the Rabbi if he likes crust or if he doesn't: if he does, the best way is to make a circular cut centered on the center of the pizza, radius 1/(sqrt(3))*8", and give that to you with no crust, and the two people who like the crust cut the rest in half very easily. If he hates crust, the cut should be done at radius (sqrt(2)/sqrt(3))*8" and the central part divided in half among the two people who don't like crust.

However the text says the rabbi decides to use two vertical cuts. This is why I said he likes integral pizza. He would like that we compute an integral.

So we have a pizza which is circular

We know that the equation of a circle is

x^2+y^2=r^2

We want to deal with a function so we are going to take for example the upper branch that is

y=sqrt(r^2-x^2)

this is the upper half of the circle and is defined in the interval [-r, r] extremes included, and we are going to study this.

We can study the area which is included between such a function and the x-axis (y=0) and we will. Let us keep in mind the relationship with our original problem, we had a whole pizza so we will need to keep track of a factor of two, we cut it in three pieces with vertical cuts (the two cuts will be symmetrical because all pieces must have equal area, so the two opposite pieces at the sides will have indeed equal area, and the circle is of course symmetric). We will notice that the crust is not equally distributed between pieces, though.

And in your case, r is 8".

So now we want to ask ourselves: I have this plate in front of me, and I have a long bread knife, and I keep it oriented along the y-axis and move it to the left and to the right tracking the movement on the x-axis, and am looking for the best place to do the first cut. Where will that place be? It is going to be at the point that the area of the half-pizza on the external side (limited by knife, x-axis, and crust) is 1/6 of total area of the pizza, and the area of the quarter-pizza that is limited by: the knife, the x-axis, the y, axis, and the crust, is going to be 1/12 of the total area.

If you draw it, may be it is more easy to see and visualize.

OK so we need to compute the integral of the circle

integral(sqrt(r^2-x^2) dx)

let's renormalize to the unit circle

X=x/r

dX=dx/r

x varies in the interval [0, r] and X instead varies in the interval [0, 1]

(I am going to consider only a quarter of the pizza, see the previous email for the simmetry considerations)

= r^2 integral (sqrt(1-(X^2)) dX)

let's change variables - you can think about it a moment if you understand why, and if not, may be i try to explain why this "surprise"

X=sin(u)

dX=cos(u) du

u varies in the interval [0, pi/2]

notice that in this interval all the functions sin and cos are always positive so we may ignore the square and square root and just cancel them out

(i.e. sqrt( something^2) = |something| and this is equal to something if something >=0, but if it had been something<0, we would have that the result is -something. E.g. sqrt((-2)^2)=2 which is -(-2) of course.)

So, r^2 sqrt(1-(X^2))= r^2 sqrt(1-(sin(u))^2)=sqrt((cos(u))^2)=cos(u)

and dX is above

so we get r^2 int( (cos(u))^2) du)

u varies in the interval [0, pi/2]

Now there is a formula, look it up

(cos(u))^2 = 1/2 cos(2u) + 1/2

if you have never seen, you can check immediately, just take the formula for the cos(a+b) and use two identical angles as a and b

and this will make integration very easy

r^2 int( 1/2 cos(2u) + 1/2) = r^2 (1/2 sen (2u) *2 + 1/2 u) + integration constant = r^2 (1/2 sen(u) cos(u) + 1/2 u) + integration constant

So we have written the primitive for the circle function.

May be we want to get back to our original variables:

X=sin(u)

u=arcsin(X)

notice again that in the interval we took, the sin function is perfectly invertible

X=x/r

u=arcsin(x/r)

So the primitive becomes

r^2 ( 1/2 sin(arcsin (x/r)) cos(arcsin (x/r)) + 1/2 arcsin(x/r) )

notice that sin(arcsin(something))=something in this interval we have taken

Also in this interval cos(arcsin(something)) =sqrt (1- ((something)^2) )

You can check this formula: first square everything (it's all positive)

cos^2(arcsin(something))=1-((something)^2)

but cos^2(whatever)=1-(sin^2(whatever))

(i am using the convention (cos^2)(x)=(cos(x))^2 i hope this is not confusing)

so we substitute in the first half and get cos^2(arcsin(something))=1-sin^2(arcsin(something))=1-(something)^2 which is the second half

We can get back to the primitive and rewrite:

r^2 (1/2 (x/r) sqrt (1-(x^2/r^2)) + 1/2 arcsin (x/r) )

which, if I did not make mistakes, is what I wrote in the beginning as the primitive.

Next step is compute the area from our primitive

r^2 (1/2 (x/r) sqrt (1-(x^2/r^2)) + 1/2 arcsin (x/r) )

Now we have something unusual. What we usually do is to take the primitive of a function and compute it on a and b in order to figure out the (oriented) area on the [a,b] interval. But here we have a different problem which is called an integral equation.

We are supposed to compute this area between a fixed point (x=0 is the first one we are concerned about) and a variable point k, so that the integral from this fixed point and this movable point is equal to a fixed number, and determine the movable point accordingly.

Let's get back to the pizza. Area of the pizza is pi r^2 (in the end we will substitute r=8") area of a quarter of pizza is pi/4 r^2

We want to cut the pizza so that

integral on the interval [0, k] of the function = 1/3 area of the quarter = pi/12 r^2

This is because the person that does not like the crust and takes the center of the pizza will get a slice that is made of four such pieces, one in each quadrant of the cartesian plane (center is of course in the center of the pizza).

The value k=r0 that will solve the equation is going to tell is how to make the cuts, which will be vertical lines at x=-r0 and x=r0.

The variable k is on the interval (0, r)

We also demand that integral on the interval [k, r] of the function = 2/3 total area = pi/6 r^2 but this is verified automatically by the simmetry considerations. Each person will get, one of them the two pieces on the right (above and below the x-axis), and the other the two pieces on the left.

OK so let us try to solve the integral equation

[r^2 (1/2 (x/r) sqrt (1-(x^2/r^2)) + 1/2 arcsin (x/r) ) ]_(0, k)= pi/12 r^2

that is

(k/r) sqrt(1-k^2/r^2)) + arcsin (k/r) = pi/6

You can plot this thing on the left of the equation with your favorite graphic tool but you can see already that is monotonously growing on the interval (sum of monotone functions on the interval) and that it is zero at k=0 (of course!) and it has value pi/2 at k=r and so it will have one and only one solution.

You can also figure out numerically and it will be a little larger than r/4, something like about 0.26r, that is to say the cuts are approximately 2.12" from the center of the pizza, on either side.

Here it is with a graphic program:

http://fooplot.com/index.php?&type0=0&type1=0&type...

Let me know if it is ok or if there is something not clear!

And this is how the pizza looks like (because of the program I have cut this one horizontally, but it's the same; also I have put the radius equal to 1)

http://fooplot.com/index.php?&type0=1&type1=0&type...

you can see how the center slice has equal area but much less crust than the two side slices (I did not compute, I can compute if you want.)

- Anonimo9 anni fa
Cut it in six pieces so each gets two slices...if you don't want to eat the crust, that's *your* problem.