Equal slicing of my spherical cake

Recently I baked a spherical cake ($3$ cm radius) and invited over a few friends, $6$ of them, for dinner. When done with main course, I thought of serving this spherical cake and to avoid uninvited disagreements over the size of the shares, I took my egg slicer with parallel wedges(and designed to cut $6$ slices at a go; my slicer has $5$ wedges) and placed my spherical cake right in the exact middle of it before pressing uniformly upon the slicer.

What should the relative placement of my wedges of the slicer be like so that I am able to equally distribute the cake among my $6$ friends?The set-up of the egg slicer looks something like this :
enter image description here

Solutions Collecting From Web of "Equal slicing of my spherical cake"

Let $R$ be the radius of your spherical cake. We will consider the case when we need to divide the cake into $n$ equally sized pieces.

Each piece will then have volume $\frac{4}{3}\frac{\pi R^3}{n}.$

We can calculate the volume of each slice by setting up a series of integrals, where we integrate over an infinitesimally thin cylinder with volume $dV=\pi r^2 dy,$ where $r\leq R$ is the radius of the cylinder (like this, but where $r$ here is the radius at height $h=R-y$). We have (draw this for yourself to see it) $r^2+y^2=R^2,$ which gives

$$\frac{4}{3}\frac{\pi R^3}{n}=\pi\int_{a_i}^{b_i}(R^2-y^2)dy=\pi\left(R^2(b_i-a_i) +\frac{a_i^3-b_i^3}{3} \right),$$
where $a_i$ is the value of $y$ where the $i$th slice starts and $b_i$ is the value where it ends, with $a_1=-R$ and $b_n=R$. Note that we have $a_{i+1}=b_i.$

We thus have $n$ equations with $n-1$ unknowns:

\begin{align}
\frac{4}{3}\frac{ R^3}{n}&=R^2(b_1-a_1) +\frac{a_1^3-b_1^3}{3}=R^2(b_1-(-R)) +\frac{(-R)^3-b_1^3}{3}\\
&=R^2(b_2-a_2) +\frac{a_2^3-b_2^3}{3}=R^2(b_2-b_1) +\frac{b_1^3-b_2^3}{3} \\
&\quad\quad \quad\;\;\;\;\quad \quad\quad\quad\;\quad\quad \quad\vdots \\
&= R^2(b_n-a_n) +\frac{a_n^3-b_n^3}{3}=R^2(R-b_{n-1}) +\frac{b_{n-1}^3-R^3}{3}.
\end{align}

As this is non-linear, it probably most fruitful to solve it numerically.

For $n=6$, in units of cake-radii (so $R=1$, but we can always scale), we have

$$b_1\approx-0.4817, \quad b_2\approx-0.2261,$$

which is all we need because of the symmetry ($b_3=0$). So your cake/egg-slicer should have distance

$$\Delta b_{01}=0-b_1\approx 0.4817,\Delta b_{12}=b_2-b_1\approx 0.2556,\Delta b_{23}=0-b_2 \approx 0.2261.$$

So your egg-slicer should look a little something like this:

$\quad\quad\quad\;\;\;$enter image description here

Here is the same as above, but for $n\in[1,20]$:

enter image description here

I must say that I’m surprised how broad the outermost pieces should be in order for them to have a volume equal to the rest, but this is probably my brain not being able to grasp exponentiation (here taking the third power).

Here is the (probably very ugly, but working) Mathematica code:

ClearAll["Global`*"];
M = 20;
For[
 n = 1, n < M, n++,
 ans = b /. 
   FullSimplify[
    Assuming[Element[n, Integers], 
     Solve[4/n == 3 (b - a) + a^3 - b^3, b]]];
 B[a_] = N[Re[ans[[3]]]];
 tabs[n] = {RecurrenceTable[{h[k + 1] == B[h[k]], h[1] == -1}, 
    h, {k, 1, n + 1}]};
 ]
NumberLinePlot[Table[tabs[k], {k, 1, n}], {x, -1, 1}]

As shown in this answer, we can find the position of the first blade to be (in the case of $R=1$)

$$ x(n)=2\cos \left( {\frac{{\alpha – 2\pi }}
{3}} \right), \quad\quad \text{with } {\,\alpha = \arctan \left( {\frac{{2\sqrt {\left( {n – 1} \right)} }}
{{n – 2}}} \right)}, \text{ for } n>2.$$

But feeding this answer into the next cubic equation and trying to solve just seems masochistic, hence the above call for numerical methods. Note that as $\tan (0)=0$ we have $\lim_{n\rightarrow \infty}x(n)=-1$ as expected.

Let me recast Lovsovs‘ answer in another perspective.
Consider at first to split an hemi-sphere into $n$ parts of equal volume.
Let designate as $h(k,n)$ the relative position of the $k$-th cut along the radius of the emisphere starting from the base circle
(corresponding to $k=0$).
$$
0 \leqslant h(k,n) \leqslant 1\quad \left| {\;0 \leqslant k \leqslant n} \right.
$$
Thus we shall have
$$
\frac{2}
{3}\pi R^{\,3} \frac{k}
{n} = \pi \int_{y\; = \;0}^{h\,R} {\left( {R^{\,2} – y^{\,2} } \right)dy} = \pi \left( {R^{\,2} R\,h – \frac{1}
{3}R^{\,3} h^{\,3} } \right)
$$
that means
$$
h^{\,3} – 3h + 2\frac{k}{n} = 0
$$
We will proceed and solve such depressed cubic equation
according to the method indicated in this work by Alessandra Cauli, refer also to the answer to this post.
Putting apart the case $k=0$, for $\;1 \leqslant k \leqslant n$ we define
$$
u = \sqrt[{3\,}]{{ – \frac{q}
{2} + \sqrt {\frac{{q^{\,2} }}
{4} + \frac{{p^{\,3} }}
{{27}}} }}
$$
where $p$ and $q$ are respectively the coefficients of $h^1$ and $h^0$.
The 2nd radical is
$$
\frac{{q^{\,2} }}
{4} + \frac{{p^{\,3} }}
{{27}} = \left( {\frac{k}
{n}} \right)^{\,2} – 1 = – \frac{{n^{\,2} – k^{\,2} }}
{{n^{\,2} }} \leqslant 0
$$
which being non-positive tells us that there are three real solutions.
Completing the calculation for $u$
$$
\begin{gathered}
u = \sqrt[{3\,}]{{ – \frac{k}
{n} + i\frac{1}
{n}\sqrt {n^{\,2} – k^{\,2} } }} = \frac{1}
{{\sqrt[{3\,}]{n}}}\sqrt[{3\,}]{{n\,e^{\,i\,\alpha } }} = \hfill \\
= e^{\,\,i\,\alpha \,/\,3} \quad \left| \begin{gathered}
\;1 \leqslant k \leqslant n \hfill \\
\;\alpha = \arctan _{\text{4Q}} \left( { – k,\sqrt {n^{\,2} – k^{\,2} } } \right) = \pi – \arctan \left( {\sqrt {\left( {\frac{n}
{k}} \right)^{\,2} – 1} } \right) = \hfill \\
\;\;\;\; = \pi – \beta \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered}
$$
We take then
$$
v = – \frac{p}
{{3\,u}} = \frac{1}
{u}\quad \quad \omega = e^{\,i\,\frac{{2\pi }}
{3}} = 1/2\left( { – 1 + i\sqrt 3 } \right)
$$
and arrive to obtain the three solutions as
$$
\begin{gathered}
1 \leqslant k \leqslant n\quad 0 \leqslant \beta = \arctan \left( {\sqrt {\left( {\frac{n}
{k}} \right)^{\,2} – 1} } \right) < \frac{\pi }
{2} \hfill \\
\left\{ \begin{gathered}
h_{\,1} = e^{\,i\,\alpha /3} + e^{\, – \,i\,\alpha /3} = 2\cos \left( {\frac{\pi }
{3} – \frac{\beta }
{3}} \right) \hfill \\
h_{\,2} = e^{\,i\,\alpha /3 + 2\pi /3} + e^{\, – \,i\,\alpha /3 – 2\pi /3} = – 2\cos \left( {\frac{\beta }
{3}} \right) \hfill \\
h_{\,3} = e^{\,i\,\alpha /3 – 2\pi /3} + e^{\, – \,i\,\alpha /3 + 2\pi /3} = 2\cos \left( {\frac{\pi }
{3} + \frac{\beta }
{3}} \right) \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered}
$$
of which we can easily determine that only $h_3$
respect the physical conditions of our problem.

Passing now to split the entire sphere, imagine it consisting of the two halves, placed base-to-base, one on the positive and one on the negative $h$ axis.
So, when $n$ is even we will place the cuts at $k = 0,\; \pm 2, \cdots ,\; \pm \left( {n – 2} \right)$,
while for $n$ odd at $k = \pm 1,\; \pm 3, \cdots ,\; \pm \left( {n – 2} \right)$.
In conclusion, always as a ratio to the radius, the cutting blades shall be positioned at:
$$
\left\{ \begin{gathered}
2 \leqslant n\quad 0 \leqslant k \leqslant \left\lfloor {\frac{n}
{2}} \right\rfloor – 1 \hfill \\
\beta \left( {k,n} \right) = \arctan \left( {\sqrt {\left( {\frac{n}
{k}} \right)^{\,2} – 1} } \right) \hfill \\
h_{\,T} (k,n) = \pm 2\cos \left( {\frac{\pi }
{3} + \frac{{\beta (2k + \bmod (n,2),\,n)}}
{3}} \right) \hfill \\
\end{gathered} \right.
$$
(if you can accept that$\pm 0=0$).