# Cutting a pie with a fork

You baked a pie in the shape of a disc, with some cherries spread unevenly on its top. You want to give each of your two children a piece of cake such that:

1. The pieces are congruent - have the same shape and size;
2. Each piece has the same amount of cherries (cherries are considered divisible so each piece might contain fractions of cherries).

One way to do this is the following: hold a knife over the diameter of the pie and rotate it slowly, changing its angle from $0$ towards $\pi$. Let $l(t)$ be the number of cherries to the left of the knife at angle $t$ and $r(t)$ the number of cherries to its right. At angle $\pi$, right becomes left and left becomes right, $l(0)=r(\pi)$ and $r(0)=l(\pi)$, so if (for example) $l(0)>r(0)$, then $l(\pi)<r(\pi)$. Both functions are continuous, so by the intermediate value theorem, there must be a $T\in[0,\pi]$ in which $r(T)=l(T)$. Cutting at that angle gives two semi-discs with the same amount of cherries.

MY QUESTION IS: What happens if, instead of a straight knife, you have a fork, which leaves a small piece of cake undivided, like this? There, the green 3-part line represents the fork. It divides the brown pie to 3 parts, two of which (the larger ones) are given to your children, and the small triangular reminder is discarded.

The argument from the semi-disc case doesn't work, because it is no longer true that $r(0)=l(\pi)$. However, by playing with the GeoGebra simulation it seems to be always possible to find an angle in which $r(T)=l(T)$, so I am trying to prove this.

I thought of the following 'proof' which I am trying to complete:

Because the two pieces are symmetric, their integral is the same, i.e.:

$$\int_{t=0}^{2\pi} l(t)dt = \int_{t=0}^{2\pi} r(t)dt$$

So:

$$\int_{t=0}^{2\pi} (l(t)-r(t))dt = 0$$

So there must be a $T\in[0,2\pi]$ in which $l(T)-r(T) = 0$.

Is this proof correct?

• Now I'm hungry. – user132181 Dec 14 '14 at 11:57
• I don't get what the fork is.. – user21820 Dec 14 '14 at 12:10
• @user21820 I've edited the picture, I hope it is clearer now. The fork is the green 3-part line. Each child receives one of the large pieces at the right/left of the fork. The small triangular piece at the bottom is not divided (e.g. you eat it yourself). – Erel Segal-Halevi Dec 14 '14 at 14:26
• I don't understand your reasoning for saying the two integrals are the same. They are reflection symmetric but not rotation symmetric. I believe your hypothesis is in fact correct but for a non-trivial reason. – user21820 Dec 14 '14 at 15:38
• Does my answer make sense? It is indeed along the lines of what you were trying but I think the justification is non-trivial.. Note that symmetry is not enough. It must cover the same 'amount' of each annulus around the centre of the pie. – user21820 Dec 15 '14 at 6:58

True theorem if I didn't make a mistake, but I don't think there is a very simple proof even if the cherry density is Riemann integrable. Please tell me if you see any error in my solution!

Let $v(p)$ be the non-negative finite integrable cherry density at point $p$ within the circular pie $P$.

Let $r(p,x)$ be the rotation of point $p$ by angle $x$ about the centre of $P$.

Let $A(S,x) = \int_P v(p) \mathbf{1}_S(r(p,x))\ dp$ for any measurable $S \subseteq P$.

Then $(x \mapsto A(S,x))$ is continuous for any measurable $S \subseteq P$ because $|S|$ is finite and $v$ is finite.

Take any measurable $S,T \subseteq P$ such that $S,T$ are reflections about the diameter of $P$ that is at angle 0.

$\int_{[0,2\pi]} A(S,x)\ dx$

$= \int_{[0,2\pi]} \int_P v(p) \mathbf{1}_S(r(p,x))\ dp\ dx$

$= \int_P \int_{[0,2\pi]} v(p) \mathbf{1}_S(r(p,x))\ dx\ dp$ [because $v,\mathbf{1}_S$ are non-negative]

$= \int_P v(p) \int_{[0,2\pi]} \mathbf{1}_S(r(p,x))\ dx\ dp$

$= \int_P v(p) \int_{[0,2\pi]} \mathbf{1}_T(r(p,-x))\ dx\ dp$ [because S,T are reflections about angle 0]

$= \int_P v(p) \int_{[-2\pi,0]} \mathbf{1}_T(r(p,x))\ dx\ dp$

$= \int_P v(p) \int_{[0,2\pi]} \mathbf{1}_T(r(p,x))\ dx\ dp$ [because rotation by $2\pi$ is identity]

$= \int_{[0,2\pi]} \int_P v(p) \mathbf{1}_T(r(p,x))\ dp\ dx$ [because $v,\mathbf{1}_T$ are non-negative]

$= \int_{[0,2\pi]} A(T,x)\ dx$

If $A(S,x) \ne A(T,x)$ for any $x \in [0,2\pi]$:

Assume $A(S,0) < A(T,0)$ by swapping $S,T$ if necessary.

Then $A(S,x) < A(T,x)$ for any $x \in [0,2\pi]$. (*)

Thus $\int_{[0,2\pi]} A(S,x)\ dx < \int_{[0,2\pi]} A(T,x)\ dx$. (**)

Therefore $A(S,x) = A(T,x)$ for some $x \in [0,2\pi]$.
(*) by intermediate value theorem on the continuous function $(x \mapsto A(S,x)-A(T,x))$.
(**) because $(x \mapsto A(S,x)),(x \mapsto A(T,x))$ are continuous.
• By $A(S,x)$ you mean the value contained inside the shape $S$ when it is rotated in angle $x$ around the center? – Erel Segal-Halevi Dec 15 '14 at 6:59
• @ErelSegalHalevi: Yes that is what it is intended to be. Note that the way I defined it means that it is rotated in the backward direction with respect to the direction that $r$ rotates. – user21820 Dec 15 '14 at 7:01
• $A(S,x)$ and $A(T,x)$ have the same integral. So, if they are not the same function, there must be an $x_S$ such that $A(S,x_S)>A(T,x_S)$ and an $x_T$ such that $A(S,x_T)<A(T,x_T)$. By the IVT, there is an $x_1\in[x_S,x_T]$ such that $A(S,x_1)=A(T,x_1)$. This is what you proved. I add that, because of periodicity, there is also an $x_2\in[x_1,x_0+2\pi]$ such that $A(S,x_2)=A(T,x_2)$. So there are at least two different equalizing points. – Erel Segal-Halevi Dec 16 '14 at 20:05