The Pizza Theorem: A proof using... calculus?
by greenturtle3141, Nov 9, 2021, 12:28 AM
Reading Difficulty: 3/5
Minimum Prerequisites: Know what an integral is.
You're having a great time at a party with friends. The doorbell rings. It's the pizza delivery! Y'all ordered a big pie.
It's time to cut the pizza. There are
people at the party. Since you're the math guy/girl, they let you divide the pizza. It can't be that hard though, right? Just make 4 equally spaced cuts through the center and you win... So you go ahead and make the cuts.
It's a success! Everyone enjoys the pizza. But then a nervous thought sets in...
Welp, maybe you did. But for next time, here's a totally different way to do it that will ease all your nervesassuming that you can make cuts at perfect angles!
THEOREM: The red and orange areas here are the same!
Also, the red, orange, and yellow areas here are the same too!
In general, to feed
people, follow this procedure:
A proof of this involves the somewhat surprising intersection of two totally different areas of math: Euclidean geometry and calculus.
Let us first step into geometry...
LEMMA 1: Let
be a cyclic, quadrilateral. Then:
![[asy]
import olympiad;
pair A,B,C,D;
A = dir(110);
C = dir(-110);
B = dir(170);
D = dir(10);
draw(A--B--C--D--cycle);
draw(A--C);
draw(B--D);
draw(circle((0,0),1));
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$D$",D,dir(D));
pair X = extension(A,C,B,D);
dot("$X$",X,NE);
[/asy]](//latex.artofproblemsolving.com/5/3/d/53dd56d8294f53675eccf6a4b678ab16ed2d06b4.png)
Proof. Let
be the intersection of 
and 
. Getting the first claim is , whereas the second claim is more specific and requires another insight.
Proving the second claim uses a cute trick that I like to call "chord swapping". What happens if we "swap" the chords
and 
?
![[asy]
import olympiad;
pair A,B,C,D;
A = dir(110);
C = dir(-110);
B = dir(170);
D = dir(10);
draw(B--C--D--cycle);
draw(A--C,dashed);
draw(A--B,dashed);
draw(A--D,dashed);
draw(circle((0,0),1));
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$D$",D,dir(D));
pair A2 = dir(70);
draw(B--A2--D);
dot("$A'$",A2,dir(A2));
[/asy]](//latex.artofproblemsolving.com/2/b/e/2be688b51fd8e8c72237ad588635ccdadd9e4655.png)
By the first claim, we have that
. Huh, could 
be a right angle, so that 
is actually a diameter? Indeed, observe that:
So 
.
Wow! So has to be a diameter. Particularly:
Substituting back 
and 
gives the second claim. 
As a corollary, note that:
This is the specific result that we'll need.
Now let us step into calculus...
THEOREM: The Pizza Theorem holds.
Proof. Fix some arbitrary right-angled "cross".
![[asy]
import olympiad;
pair A,B,C,D;
A = dir(110);
C = dir(-110);
B = dir(170);
D = dir(10);
draw(A--C);
draw(B--D);
draw(circle((0,0),1));
dot("$A_0$",A,dir(A));
dot("$B_0$",B,dir(B));
dot("$C_0$",C,dir(C));
dot("$D_0$",D,dir(D));
pair X = extension(A,C,B,D);
dot("$X$",X,NE);
[/asy]](//latex.artofproblemsolving.com/e/b/e/ebe473e9333ff2ca9a02a94daa9d5e157b042540.png)
If we "rotate" the cross by some angle
, then a certain amount of area is displaced.
![[asy]
import olympiad;
pair A,B,C,D;
A = dir(110);
C = dir(-110);
B = dir(170);
D = dir(10);
pair X = extension(A,C,B,D);
pair A1,B1,C1,D1;
real t = 20;
A1 = intersectionpoints(X--((A-X)*dir(t)*3+X), circle((0,0),1))[0];
B1 = intersectionpoints(X--((B-X)*dir(t)*3+X), circle((0,0),1))[0];
C1 = intersectionpoints(X--((C-X)*dir(t)*3+X), circle((0,0),1))[0];
D1 = intersectionpoints(X--((D-X)*dir(t)*3+X), circle((0,0),1))[0];
fill(A1--X--A..arc((0,0),1,degrees(A),degrees(A1),CCW)--cycle,gray(0.7));
fill(B1--X--B..arc((0,0),1,degrees(B),degrees(B1),CCW)--cycle,gray(0.7));
fill(C1--X--C..arc((0,0),1,degrees(C),degrees(C1),CCW)--cycle,gray(0.7));
fill(D1--X--D..arc((0,0),1,degrees(D),degrees(D1),CCW)--cycle,gray(0.7));
draw(A1--X,dashed);
draw(B1--X,dashed);
draw(C1--X,dashed);
draw(D1--X,dashed);
draw(A--C);
draw(B--D);
draw(circle((0,0),1));
dot("$A(\theta)$",A1,dir(A1));
dot("$B(\theta)$",B1,dir(B1));
dot("$C(\theta)$",C1,dir(C1));
dot("$D(\theta)$",D1,dir(D1));
dot("$A_0$",A,dir(A));
dot("$B_0$",B,dir(B));
dot("$C_0$",C,dir(C));
dot("$D_0$",D,dir(D));
dot("$X$",X,NE);
[/asy]](//latex.artofproblemsolving.com/3/e/e/3ee2d6564a4b69badaec29ad3298029b8a993556.png)
If we can prove that the shaded area depends only on the radius
and the angle , then the theorem is proven. (Exercise: Why?)
We find the area by integration. Let
be the point on the arc between 
and 
such that 
. So, 
and, well, 
... Define 
, 
, and 
similarly.
The area of the "
"-shaded region is given by the following integral:
Where we identify 
as length. This comes from the formula for finding . Hence the total shaded area is just:
But the segments 
are equally spaced, at angles of 
. So by Lemma 1, we have that:
Therefore the area is just 
. This proves the Pizza Theorem.
Addendum
The Pizza Theorem usually refers to this result, which I will refer to as the "romantic" version because it only involves two:
Theorem (Romantic Pizza Theorem): Pick any point on the pizza and make
equally-spaced full cuts through some random point, which makes 
slices of pizza. Then alternate slices form exactly half the pizza.
EXERCISE: Why does the previous version of the Pizza Theorem NOT imply this one?
In order to hide the fact that I have no proof of this to give at the moment, I'll leave it as an exercise.
Further reading:
http://www.cs.umd.edu/~gasarch/BLOGPAPERS/pizza.pdf
Minimum Prerequisites: Know what an integral is.
You're having a great time at a party with friends. The doorbell rings. It's the pizza delivery! Y'all ordered a big pie.
It's time to cut the pizza. There are
people at the party. Since you're the math guy/girl, they let you divide the pizza. It can't be that hard though, right? Just make 4 equally spaced cuts through the center and you win... So you go ahead and make the cuts.It's a success! Everyone enjoys the pizza. But then a nervous thought sets in...
Oh no, what if I didn't make the cuts through the exact center??? Did I mess up???
Welp, maybe you did. But for next time, here's a totally different way to do it that will ease all your nerves
THEOREM: The red and orange areas here are the same!
Also, the red, orange, and yellow areas here are the same too!
In general, to feed
people, follow this procedure:- Pick a random point on the pizza. Let be the number of people you need to feed.
- Make
full cuts through the point, spaced out at equal angles. This makes 
pieces. - Line up the party members and have them take the slices in some counter-clockwise order. Repeat until the pizza is empty.
- ???
- Profit!
full cuts through the point, spaced out at equal angles. This makes 
pieces.A proof of this involves the somewhat surprising intersection of two totally different areas of math: Euclidean geometry and calculus.
Let us first step into geometry...
LEMMA 1: Let
be a cyclic, quadrilateral. Then:
- We have that
. - If is the circumradius, then
.
.
.![[asy]
import olympiad;
pair A,B,C,D;
A = dir(110);
C = dir(-110);
B = dir(170);
D = dir(10);
draw(A--B--C--D--cycle);
draw(A--C);
draw(B--D);
draw(circle((0,0),1));
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$D$",D,dir(D));
pair X = extension(A,C,B,D);
dot("$X$",X,NE);
[/asy]](http://latex.artofproblemsolving.com/5/3/d/53dd56d8294f53675eccf6a4b678ab16ed2d06b4.png)
Proof. Let
be the intersection of 
and 
. Getting the first claim is , whereas the second claim is more specific and requires another insight.
Now we cleverly add the first equation with the third, and the second equation with the fourth.
This proves that 
,Proving the second claim uses a cute trick that I like to call "chord swapping". What happens if we "swap" the chords
and 
?![[asy]
import olympiad;
pair A,B,C,D;
A = dir(110);
C = dir(-110);
B = dir(170);
D = dir(10);
draw(B--C--D--cycle);
draw(A--C,dashed);
draw(A--B,dashed);
draw(A--D,dashed);
draw(circle((0,0),1));
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$D$",D,dir(D));
pair A2 = dir(70);
draw(B--A2--D);
dot("$A'$",A2,dir(A2));
[/asy]](http://latex.artofproblemsolving.com/2/b/e/2be688b51fd8e8c72237ad588635ccdadd9e4655.png)
By the first claim, we have that
. Huh, could 
be a right angle, so that 
is actually a diameter? Indeed, observe that:
So 
.Wow! So
has to be a diameter. Particularly:
Substituting back 
and 
gives the second claim. 
As a corollary, note that:
This is the specific result that we'll need.Now let us step into calculus...
THEOREM: The Pizza Theorem holds.
Proof. Fix some arbitrary right-angled "cross".
![[asy]
import olympiad;
pair A,B,C,D;
A = dir(110);
C = dir(-110);
B = dir(170);
D = dir(10);
draw(A--C);
draw(B--D);
draw(circle((0,0),1));
dot("$A_0$",A,dir(A));
dot("$B_0$",B,dir(B));
dot("$C_0$",C,dir(C));
dot("$D_0$",D,dir(D));
pair X = extension(A,C,B,D);
dot("$X$",X,NE);
[/asy]](http://latex.artofproblemsolving.com/e/b/e/ebe473e9333ff2ca9a02a94daa9d5e157b042540.png)
If we "rotate" the cross by some angle
, then a certain amount of area is displaced.![[asy]
import olympiad;
pair A,B,C,D;
A = dir(110);
C = dir(-110);
B = dir(170);
D = dir(10);
pair X = extension(A,C,B,D);
pair A1,B1,C1,D1;
real t = 20;
A1 = intersectionpoints(X--((A-X)*dir(t)*3+X), circle((0,0),1))[0];
B1 = intersectionpoints(X--((B-X)*dir(t)*3+X), circle((0,0),1))[0];
C1 = intersectionpoints(X--((C-X)*dir(t)*3+X), circle((0,0),1))[0];
D1 = intersectionpoints(X--((D-X)*dir(t)*3+X), circle((0,0),1))[0];
fill(A1--X--A..arc((0,0),1,degrees(A),degrees(A1),CCW)--cycle,gray(0.7));
fill(B1--X--B..arc((0,0),1,degrees(B),degrees(B1),CCW)--cycle,gray(0.7));
fill(C1--X--C..arc((0,0),1,degrees(C),degrees(C1),CCW)--cycle,gray(0.7));
fill(D1--X--D..arc((0,0),1,degrees(D),degrees(D1),CCW)--cycle,gray(0.7));
draw(A1--X,dashed);
draw(B1--X,dashed);
draw(C1--X,dashed);
draw(D1--X,dashed);
draw(A--C);
draw(B--D);
draw(circle((0,0),1));
dot("$A(\theta)$",A1,dir(A1));
dot("$B(\theta)$",B1,dir(B1));
dot("$C(\theta)$",C1,dir(C1));
dot("$D(\theta)$",D1,dir(D1));
dot("$A_0$",A,dir(A));
dot("$B_0$",B,dir(B));
dot("$C_0$",C,dir(C));
dot("$D_0$",D,dir(D));
dot("$X$",X,NE);
[/asy]](http://latex.artofproblemsolving.com/3/e/e/3ee2d6564a4b69badaec29ad3298029b8a993556.png)
If we can prove that the shaded area depends only on the radius
and the angle , then the theorem is proven. (Exercise: Why?)We find the area by integration. Let
be the point on the arc between 
and 
such that 
. So, 
and, well, 
... Define 
, 
, and 
similarly.The area of the "
"-shaded region is given by the following integral:
Where we identify 
as length. This comes from the formula for finding . Hence the total shaded area is just:
the area between 
and 
is just 
.
.
by the sine area formula. Then there's some small part that should become really negligible as we let the number of triangles go to infinity.
is really small, 
.
.
.
But the segments 
are equally spaced, at angles of 
. So by Lemma 1, we have that:
Therefore the area is just 
. This proves the Pizza Theorem. Addendum
The Pizza Theorem usually refers to this result, which I will refer to as the "romantic" version because it only involves two:
Theorem (Romantic Pizza Theorem): Pick any point on the pizza and make
equally-spaced full cuts through some random point, which makes 
slices of pizza. Then alternate slices form exactly half the pizza.EXERCISE: Why does the previous version of the Pizza Theorem NOT imply this one?
In order to hide the fact that I have no proof of this to give at the moment, I'll leave it as an exercise.
Further reading:
http://www.cs.umd.edu/~gasarch/BLOGPAPERS/pizza.pdf
This post has been edited 2 times. Last edited by greenturtle3141, Nov 9, 2021, 3:48 PM
March 2025
November 2024