The following 'volume splitting property' occurred to me for convex euclidean manifolds $M \subset \mathbb{R}^3$ homeomorphic to the sphere:


$\forall p \in \partial M,$ there exists a
 hyperplane $H$ such that $p \in H$ and $M$ can be decomposed into path connected components $M\_1,M \cap H, M\_2$ so that


$$
\begin{cases}
M=M\_1 \cup \{M \cap H\} \cup M\_2\\
Vol(M\_1)=Vol(M\_2)=Vol(M)/2\\
\end{cases} \tag{\*}
$$


The challenge is to show that $(\*)$ holds for compact $M$ if and only if $M$ is convex. I managed to prove that $(\*)$ holds for any convex $M$ but I'm having difficulties showing that $(\*)$ doesn't hold if $M$ isn't convex. 


**Note 1:** This problem occurred to me while thinking of how to fairly divide a cake among an even number of people without leaving any crumbs. 


**Note 2:** Due to the path-connectedness constraint, this conjecture doesn't follow directly from the Ham sandwich theorem. 



Cake cutting conjecture
The result is true even if $M$ is not convex.


Fix $n\geq 2$ and suppose that $M$ is a bounded domain in $\newcommand{\bR}{\mathbb{R}}$ $\bR^n$. We denote by $S^{n-1}$ the unit *sphere* in $\bR^{n}$ centered at the origin, i.e., the set of unit vectors in $\bR^n$.


For any $\newcommand{\bnu}{\boldsymbol{\nu}}$ $\newcommand{\bp}{\boldsymbol{p}}$ $(\bnu, \bp)\in S^{n-1}\times \bR^n$ we denote by $H\_{\bnu,p}$ the hyperplane $\newcommand{\bx}{\boldsymbol{x}}$
$$H\_{\bnu,\bp}=\big\{\; \bx\in \bR^n;\;\;(\bx-\bp,\bnu)=0\,\big\}, $$
where $(-,-)$ denotes the usual inner product in $\bR^n$.


We set


$$H\_{\bnu,\bp}^\pm:=\big\{\; \bx\in \bR^n;\;\;\pm (\bx-\bp,\bnu)\geq 0\,\big\}, $$


Fix $\newcommand{\pa}{\partial}$ $\bp\in \bR^n$ and define $\DeclareMathOperator{\vol}{vol}$


$$ D^M\_{\bp}: S^{n-1}\to \bR,\;\;D\_{\bp}^M(\bnu)=\vol( M\cap H^+\_{\bnu,\bp})- \vol( M\cap H^-\_{\bnu,\bp}).\tag{\*} $$


Note that the function $D\_{\bp}^M$ is continuous. More precisely, if $\DeclareMathOperator{\diam}{diam}$ $R> \diam(M)$ and $\omega\_n$ denotes the volume of the unit balls then
$$
\big\vert\;\vol(M\cap H^\pm\_{\bp,\bnu\_0})-\vol(M\cap H^\pm\_{\bp,\bnu\_1})\;\big\vert\leq \frac{\arccos\, (\bnu\_0,\bnu\_1)}{\pi}\omega\_nR^n.
$$
Moreover $D^M\_{\bp}$ is odd, i.e., and $D^M\_{\bp}(-\bnu)=D^M\_{\bp}(\bnu)$, for any $\bnu\in S^{n-1}$. 


Fix $\bnu\_0\in S^{n-1}$. For any meridian that connects $\bnu\_0$ to $\bnu$ there exists $\nu$ on that meridian such that $D^M\_{\bp}(\bnu)=0$. Thus, if $n\geq 3$, for any $\bp\in\pa M$ there exists infinitely many ways of cutting $M$ into two parts of equal volumes by a hyperplane through a given point $\bp$ of the boundary.


**Note** As pointed out by Aidan Rocke, the sides $M\cap H^\pm\_{\bp,\nu}$ may not be connected. Let me explain how to produce a nonconvex region in $\bR^3$ that can be cut into connected pieces by a hyperplane containing an arbitrary point in $\bR^3$.


Consider now a region $R$ in $\bR^3$ defined as the union of a cube $C\_1$ and of size $1$ sitting on top of a cube $C\_2$ of size $2$, so that the center of the bottom face of $C\_1$ coincides with the center of the top face of $C\_2$. Denote by $\newcommand{\bc}{\boldsymbol{c}}$ $\bc\_1$ the center of $C\_1$ and by $\bc\_2$ the center of $C\_2$. By construction $\bc\_1$ and $\bc\_2$ lie on a vertical line (parallel to the $z$-axis).


Fix $\bp\in \bR^3$. Any hyperplane containing the points $\bp,\bc\_1,\bc\_2$ cuts both $C\_1$ and $C\_2$ into convex parts of equal volumes. Such a hyperplane contains the vertical line determined by $\bc\_1,\bc\_2$. The bottom vertices of $C\_1$ that lie on one side of the hyperplane are also points in $C\_2$ showing that such a hyperplane cuts $R$ into connected parts of equal volumes.


