TITLE: How to cut a cube out of a tree stump, such that a pair of opposing vertices are in the center?
QUESTION [72 upvotes]: I saw this picture of a cube cut out of a tree stump.

I've been trying to craft the same thing out of a tree stump, but I found it hard to figure out how to do it.
One of the opposing vertices pair is on the center of the tree stump:

I've been struggling to find the numbers and angle needed to make the cuts.
Any kind of advice would be greatly appreciated. Thanks for taking your time to read this and I'm sorry for my bad english.

REPLY [28 votes]: The Method
Here's an approach with which you are only ever dealing with planar surfaces.

Cut from the stump a (regular) hexagonal prism of radius $r$ and height $\frac32r\sqrt2\approx2.1213 r$. Note: The face-to-opposite-face "width" of the prism should be $r\sqrt3\approx1.732r$. (Sanity check for regularity: The edges of the hexagonal top and bottom should also be $r$.)

The figure shows the top view and the six "unrolled" lateral faces.


Divide each lateral edge of the prism into thirds, and draw a zig-zag across the lateral faces, joining alternating $\frac13$ and $\frac23$ marks on those edges. Each segment of the zig-zag will be an edge of the final cube. (Its length should be $e\approx 1.2247r$.)

Each consecutive triad of points on the zig-zag (along with either the top-center or bottom-center point of the prism) defines a face of the cube, so "all you have to do" is remove excess wood until those faces are planar.


Done!

The Math
For a prism with radius $r=2$ (to avoid fractions), let the bottom-center of the prism be the origin $O=(0,0,0\sqrt2)$, and the top-center be $T=(0,0,3\sqrt2)$.
The coordinates of the zig-zag of points ($ABCDEFA$ in the figure) have the form
$$\left(2\cos k\,60^\circ, 2\sin k\,60^\circ, (1\;\text{or}\;2)\sqrt{2} \right)$$
Specifically,
$$A = (\phantom{-}2,0,2\sqrt{2}) \qquad B = (\phantom{-}1,\phantom{-}\sqrt{3},1\sqrt2) \qquad C = (-1,\phantom{-}\sqrt3,2\sqrt2)$$
$$D = (-2,0,1\sqrt2) \qquad E = (-1,-\sqrt3,2\sqrt2) \qquad F = (\phantom{-}1,-\sqrt3,1\sqrt2)$$
(where $A$, $C$, $E$ are closer to the top of the prism, and $B$, $D$, $F$ are closer to the bottom). From here, we can verify that
$$\begin{align}
T &= B+(A-B)+(C-B) &&\quad\to\quad T,A,B,C\;\text{coplanar} \\
&= D+(C-D)+(E-D)   &&\quad\to\quad T,C,D,E\;\text{coplanar}\\
&= F+(E-F)+(A-F)   &&\quad\to\quad T,E,F,A\;\text{coplanar}\\[4pt]
O &= A+(B-A)+(F-A) &&\quad\to\quad O,F,A,B\;\text{coplanar}\\
&=C+(B-C)+(D-C)    &&\quad\to\quad O,B,C,D\;\text{coplanar}\\
&=E+(D-E)+(F-E)    &&\quad\to\quad O,D,E,F\;\text{coplanar}
\end{align}$$
and
$$\begin{align}2\,\sqrt{\frac32}=\sqrt{6}\;&=\;|AB|=|BC|=|CD|=|DE|=|EF|=|FA| \\
&=\;|TA|=|TC|=|TE| \\
&=\;|OB|=|OD|=|OE|\end{align}$$
so that the faces are all quadrilaterals with equal edge-lengths, making them at least rhombuses. Also,
$$\begin{align}
0\;&=\;(A-T)\cdot(C-T)=(C-T)\cdot(E-T)=(E-T)\cdot(A-T) \\
&=\;(B-O)\cdot(D-O)=(D-O)\cdot(F-O)=(F-O)\cdot(B-O)
\end{align}$$
so that
$$90^\circ = \angle ATC = \angle CTE = \angle ETA =\angle BOD = \angle DOF = \angle FOB$$
That is, each rhombus face has at least one right angle, making it a square. $\square$