Cake cutting conjecture 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. 
 A: 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.
A: If $M$ is rotationally symmetric around a line (for example, the union of the unit balls with centers at $(0,0,0)$ and $(0,0,1)$), then there is a ham sandwich cut through every point of $M$, and both parts are always connected. So the convexity is not a necessary condition.
