Q:

Consensus division of a cake

There is a pie and two people with different tastes. The goal is to cut a piece, using two radial cuts like this:

such that both people agree that the piece has a value of exactly a fraction $p$ of the total, where $p\in[0,1]$ is a given constant.
Formally, the pie is described as the interval $[0,1]$ whose two endpoints are identified (- a topological circle). There are two non-negative value measures over the interval, $V_A$ and $V_B$. They are absolutely continuous with respect to length (this means that they are non-atomic). Both measures assign the same value to the whole interval: $V_A([0,1])=V_B([0,1])=1$. The goal is to find an interval $[x,y]$ such that $V_A([x,y])=V_B([x,y])=p$.
I found a solution for the special case in which the measure $V_A$ is equal to the length measure. Here it is. Hold two knives over the pie, such that the distance between them is exactly $p$. Move the knives a whole round around the pie, always keeping the distance between them at $p$. I claim that there is a point in which $V_B$ of the piece between the knives is exactly $p$.
PROOF: At each time $t\in[0,1]$, the piece between the knives is $[t,t+p]$. Mark by $S$, the sum of the values of $V_B$ when the knives make a whole round:
$$ S = \int_{t=0}^1 V_B([t,t+p]) dt $$
Since $V_B$ is continuous, it has a derivative $v_b$, such that $V_B([t,t+p])=\int_{x=t}^{t+p} v_b(x)dx$. So:
$$ S = \int_{t=0}^1 \int_{x=t}^{t+p} v_b(x)dx dt = \int_{t=0}^1 \int_{x=0}^{p} v_b(t+x)dx dt $$
Substitute the order of integration and get:
$$ S = \int_{x=0}^{p} \int_{t=0}^1 v_b(t+x) dt dx = \int_{x=0}^{p} 1 dx = p$$
The integral of $V_B([t,t+p])$ equals $p$ and the functions are continuous, so by the Mean Value Theorem there must be a  $t$ in which $V_B([t,t+p])=p$. $\square$
Is it possible to generalize this argument to arbitrary continuous value measures?

EDIT: I just thought of an informal idea.
Let $N$ be a large integer. Normalize the value measures such that the value of the entire cake for each person is $N^2$. 
Divide the cake to $N$ arcs such that, in each arc, $V_A+V_B=2N$. Additionally, round the values of $V_A$ and $V_B$ in each arc to the nearest integers.
Now, we have a discrete problem. We can imagine that in each arc, there are $V_A$ Azure balls and $V_B$ Black balls.
Move two knives around the cake in the following way. In each step $t$, knife #1 is just before Azure ball $t$, and knife #2 is just before Azure ball $t + p N^2$. Hence the number of Azure balls between the knives is always $p N^2$. Move the knives until the knives return to their original location.
At each step, count the number of Black balls between the knives, and sum over the entire round-trip.  Each black ball is counted whenever there are between 1 and $p N^2$ azure balls before it. Hence, it is counted exactly $p N^2$ times. Hence, the sum of the counts over the entire trip is $p N^4$. This is a fraction $p$ of the product of the total number of balls.
An interactive illustration is available here: http://tube.geogebra.org/m/1355529 
There, $p N^2 = 5$. The green arc is the arc between the knives. You can increase $t$ and see that the green arc always covers exactly 5 azure balls, and each black ball is covered exactly 5 times.
When $N$ is sufficiently large, the sum becomes an integral, and by continuity, the integral of the $V_B$ is $p$.
Hence, by the Mean Value Theorem, there is a point in the trip in which $V_B=p$.
Can this idea be made formal?

A:

The cases $p = 0$ and $p = 1$ are either trivial or, if an empty sector/the whole pie are disallowed, in general not possible to achieve, so let's assume $0 < p < 1$.
Then let's lift the value functions to $\mathbb{R}$, so we have two functions $V_\ast \colon \mathbb{R}\to \mathbb{R}$ with $V_\ast(0) = 0$ and $\bigl(\forall x\in \mathbb{R}\bigr)\bigl(V_\ast(x+1) = V_\ast(x) + 1\bigr)$, where $V_\ast(x)$ is the value that person $\ast$ ascribes to the piece of the pie between angle $0$ [arbitrarily determined] and angle $2\pi x$ for $0 \leqslant x < 1$, and the function is extended by setting $V_\ast(x) = \lfloor x\rfloor + V_\ast(x-\lfloor x\rfloor)$ for arbitrary $x$.
By assumption, the two functions $V_A$ and $V_B$ are continuous and monotonically increasing (but not necessarily strictly). Since $x \mapsto V_\ast(x) - x$ is continuous and has period $1$, $V_\ast$ is uniformly continuous.
First, let us assume that - without loss of generality - $V_A$ is strictly increasing. Then $V_A$ has a continuous inverse $\varphi$, and since $x\mapsto \varphi(x) - x$ also has period $1$, $\varphi$ is also uniformly continuous. Define $s(x) = \varphi(V_A(x) + p)$. Then $s$ is uniformly continuous, $s(x+1) = s(x) + 1$ for all $x\in \mathbb{R}$ and $V_A(s(x)) - V_A(x) = p$ for all $x\in \mathbb{R}$. We then look at
$$\psi \colon x \mapsto V_B(s(x)) - V_B(x).$$
We need to show that there is an $x_0\in\mathbb{R}$ with $\psi(x_0) = p$. Since $\psi$ is continuous and has period $1$, if there were no such $x_0$, there would be an $\varepsilon > 0$ such that either $\psi(x) \leqslant p-\varepsilon$ for all $x\in \mathbb{R}$ or $\psi(x) \geqslant p+\varepsilon$ for all $x\in \mathbb{R}$. Suppose $\psi(x) \neq p$ for all $x$.
There is a $\delta > 0$ such that $\lvert x-y\rvert \leqslant \delta \implies \lvert V_\ast(x) - V_\ast(y)\rvert \leqslant \varepsilon/2$ by the uniform continuity of $V_\ast$ (we can choose the same $\delta$ for $V_A$ and $V_B$). The uniform continuity of $s$ gives us a $0 <\eta < \varepsilon/2$ with $\lvert x-y\rvert \leqslant \eta \implies \lvert s(x) - s(y)\rvert \leqslant \delta$.
Further, there are $k,m \in \mathbb{N}\setminus \{0\}$ such that $m - \eta \leqslant k\cdot p \leqslant m$. For rational $p$ we can achieve equality ($kp = m$), and for irrational $p$ the sequence $k p - \lfloor kp\rfloor$ is dense in $[0,1]$.
Now, letting $s^j$ denote the $j$-fold iterate of $s$, we have $[0,s^k(0)[ = \bigcup\limits_{j=1}^k [s^{j-1}(0), s^j(0)[$ where the union is disjoint, and $V_A(s^j(0)) - V_A(s^{j-1}(0)) = p$ for all $j$, so $V_A(s^k(0)) = kp \in [m-\eta,m]$. Then $s^k(0) = \varphi(V_A(s^k(0)) \in [m-\delta,m]$ since $\varphi(m) = m$, and therefore $V_B(s^k(0)) \in [m-\varepsilon/2,m]$.
But on the other hand we have
$$V_B(s^k(0)) = \sum_{j=1}^k V_B(s^j(0)) - V_B(s^{j-1}(0)) = \sum_{j=1}^k \psi(s^{j-1}(0)) \begin{cases} \leqslant k(p-\varepsilon) \leqslant kp-\varepsilon < m-\varepsilon/2\\ \geqslant k(p+\varepsilon) \geqslant kp+\varepsilon > m.\end{cases}$$
This contradiction shows that $\psi$ attains the value $p$. By periodicity, it attains the value $p$ in $[0,1[$.

If neither of $V_A$ and $V_B$ is strictly increasing, we can approximate the value functions by strictly increasing ones. For $n \geqslant 1$, consider the modified value functions $W_\ast^{(n)}$ given by
$$W_\ast^{(n)}(x) = \frac{n}{n+1}\biggl(V_\ast(x) + \frac{x}{n}\biggr)$$
on $[0,1[$, and extended by the relation $W_\ast^{(n)}(x+1) = W_\ast^{(n)}(x) + 1$ to all of $\mathbb{R}$. These are continuous and strictly increasing, so by the above, we have $x_n \in [0,1[$ and $y_n > x_n$ with
$$W_A^{(n)}(y_n) - W_A^{(n)}(x_n) = p = W_B^{(n)}(y_n) - W_B^{(n)}(x_n).\tag{1}$$
Pick a subsequence such that $x_{n_k}$ and $y_{n_k}$ converge, say to $x_0$ resp. $y_0$. Since $W_\ast^{(n)}$ converges uniformly to $V_\ast$, we have $W_\ast^{(n_k)}(x_{n_k}) \to V_\ast(x)$ and $W_\ast^{(n_k)}(y_{n_k}) \to V_\ast(y)$ and therefore obtain
$$V_A(y) - V_A(x) = \lim_{k\to\infty} W_A^{(n_k)}(y_{n_k}) - W_A^{(n_k)}(x_{n_k}) = p = \lim_{k\to\infty} W_B^{(n_k)}(y_{n_k}) - W_B^{(n_k)}(x_{n_k}) = V_B(y) - V_B(x)$$
from $(1)$.

