## Step 1: Transform $P_1$ to match the rotated base of $P_2$

Let

$$
\phi=\frac{1+\sqrt5}{2}.
$$

Define the first prism as

$$
P_1=\left(\frac{15}{2\phi},\ \frac{8}{\sqrt3\,\phi},\ \sqrt3\,\phi\right),
\qquad
P_2=\left(\frac{5\sqrt3}{\phi^2},\ 2,\ \frac{2}{\sqrt3}\right).
$$

Here $P_2$ is the prism currently produced by the geometry pipeline, measured from the actual assembled oblate-prism vertices, up to axis order.

Write

$$
a=\frac{15}{2\phi},\qquad
b=\frac{8}{\sqrt3\,\phi},\qquad
c=\sqrt3\,\phi.
$$

So initially,

$$
P_1=[0,a]\times[0,b]\times[0,c].
$$

For Step 2, rotate $P_2$ so that its base is

$$
\frac{5\sqrt3}{\phi^2}\times 2
$$

and its stacking height is

$$
\frac{2}{\sqrt3}.
$$

So we want to convert $P_1$ into

$$
P_1'=\left(\frac{5\sqrt3}{\phi^2},\ 2,\ h\right).
$$

The height $h$ is determined by volume preservation:

$$
\operatorname{vol}(P_1)=\frac{15}{2\phi}\cdot\frac{8}{\sqrt3\,\phi}\cdot\sqrt3\phi=\frac{60}{\phi},
$$

while the new base area is

$$
\frac{5\sqrt3}{\phi^2}\cdot 2=\frac{10\sqrt3}{\phi^2}.
$$

Hence

$$
h=\frac{60/\phi}{10\sqrt3/\phi^2}=\frac{6\phi}{\sqrt3}=2\sqrt3\,\phi.
$$

So the Step 1 target becomes

$$
P_1'=\left(\frac{5\sqrt3}{\phi^2},\ 2,\ 2\sqrt3\,\phi\right).
$$

### Step 1A: Reshape the $xy$-face so the first dimension becomes $\frac{5\sqrt3}{\phi^2}$

Keep the $z$-depth fixed at $c$, and replace the base rectangle

$$
a\times b
$$

with an equal-area rectangle

$$
\frac{5\sqrt3}{\phi^2}\times d,
$$

where

$$
d=\frac{ab}{5\sqrt3/\phi^2}.
$$

Compute $ab$:

$$
ab=\frac{15}{2\phi}\cdot\frac{8}{\sqrt3\phi}
=\frac{60}{\sqrt3\,\phi^2}
=\frac{20\sqrt3}{\phi^2}.
$$

So

$$
d=\frac{20\sqrt3/\phi^2}{5\sqrt3/\phi^2}=4.
$$

After Step 1A, the target block is

$$
B_1=\left(\frac{5\sqrt3}{\phi^2},\ 4,\ \sqrt3\phi\right).
$$

Geometrically:

1. Take the face rectangle
   $$
   R_{xy}=[0,a]\times[0,b].
   $$
2. Dissect it into finitely many polygons.
3. Rearrange those polygons into
   $$
   R'_{xy}=
   \left[0,\frac{5\sqrt3}{\phi^2}\right]\times[0,4].
   $$
4. Extrude each polygon through the full interval $z\in[0,c]$.

This gives a 3D dissection of $P_1$ into prisms that reassemble into $B_1$.

### Step 1B: Reshape the $yz$-face to get the final desired dimensions

Now start from

$$
B_1=\left(\frac{5\sqrt3}{\phi^2},\ 4,\ \sqrt3\phi\right).
$$

Keep the $x$-length fixed at $\frac{5\sqrt3}{\phi^2}$, and reshape the $yz$-face.

Current $yz$-face:

$$
4\times \sqrt3\phi.
$$

Target $yz$-face:

$$
2\times 2\sqrt3\phi.
$$

These have the same area, because

$$
4\cdot \sqrt3\phi
=
4\sqrt3\phi,
$$

and

$$
2\cdot 2\sqrt3\phi
=
4\sqrt3\phi.
$$

Because this is the special case

$$
4\times h \longrightarrow 2\times 2h
\qquad\text{with}\qquad h=\sqrt3\phi,
$$

Step 1B can be done with a single straight cut.

After this step, we obtain

$$
P_1'=\left(\frac{5\sqrt3}{\phi^2},\ 2,\ 2\sqrt3\phi\right).
$$

Geometrically:

1. Take the $yz$-rectangle
   $$
   R_{yz}=
   [0,4]
   \times
   [0,\sqrt3\phi].
   $$
2. Cut it by the single line
   $$
   y=2.
   $$
   This splits the face into two congruent rectangles
   $$
   [0,2]\times[0,\sqrt3\phi]
   \qquad\text{and}\qquad
   [2,4]\times[0,\sqrt3\phi].
   $$
3. Keep the lower rectangle fixed.
4. Translate the upper rectangle by
   $$
   (0,-2,\sqrt3\phi)
   $$
   in $(y,z)$-coordinates, so it lands on
   $$
   [0,2]\times[\sqrt3\phi,2\sqrt3\phi].
   $$
5. Extrude both rectangles through the full $x$-interval
   $$
   \left[0,\frac{5\sqrt3}{\phi^2}\right].
   $$

At the end of Step 1, the first prism has been converted into

$$
P_1'=\left(\frac{5\sqrt3}{\phi^2},\ 2,\ 2\sqrt3\phi\right).
$$

## Step 2: Stack $P_1'$ with $P_2$

Now we have

$$
P_1'=\left(\frac{5\sqrt3}{\phi^2},\ 2,\ 2\sqrt3\phi\right)
$$

and

$$
P_2=\left(\frac{5\sqrt3}{\phi^2},\ 2,\ \frac{2}{\sqrt3}\right).
$$

These already share the same first two dimensions, so Step 2 is just stacking along the third axis.

Place $P_1'$ at

$$
\left[0,\frac{5\sqrt3}{\phi^2}\right]\times[0,2]\times[0,2\sqrt3\phi]
$$

and place $P_2$ at

$$
\left[0,\frac{5\sqrt3}{\phi^2}\right]\times[0,2]\times
\left[2\sqrt3\phi,\ 2\sqrt3\phi+\frac{2}{\sqrt3}\right].
$$

Together they form the rectangular block

$$
B=\left(\frac{5\sqrt3}{\phi^2},\ 2,\ 2\sqrt3\phi+\frac{2}{\sqrt3}\right).
$$

No additional cuts are needed in Step 2; it is just a placement of the two rectangular prisms along the third axis.