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:


*

*The pieces are congruent - have the same shape and size;

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