So, each octahedron can be inscribed in a cube, so that the corner points of the octahedron are in the midpoints of the side areas of the cube, am I right?

From the octahedron $ABCDS_1S_2$, shown in the image, the vertices

$A =(13 | -5 | 3)$

, $B =(11 | 3 | 1)$

, $C =(5 | 3 | 7)$


$S_1 =(13 | 1 | 9)$

are given. This octahedron is inscribed in the illustrated cube with the corners $P_1$ to $P_8$.

Now let

$E_0: 2x_1 + x_2 + 2x_3 + 9 * (2a-5) = 0, a ∈ ℝ$

be a set of planes $aEa$; Let $h$ be the line passing through the points $S_1$ and $S_2 =(5 | -3 | 1).$

Now the task: For $0 <a ≤ 1$, the plane $E_a$ intersects a pyramid of the octahedron with the peak $S_1$:

Octahedron with Pyramid

I have to find the point of intersection $P_a$ of the plane $E_a$ with the line $h$ and then the volume $V_a$ of the truncated pyramid.

This is what I've done: I have determined


but how can I find the volume? If anyone needs the math for determining $P_a$, please say so and I'll add my calculations.



For heaven's sake write points data as $A=(13,-5,9)$, and so on!$\quad$ [Note that $A=(13,-5,9)$ is a proposition, while $A(13,-5,9)$ is a function value.]

One finds that the center of the octahedron $O$ is at $M=(9,-1,5)$, so that $\vec{MS_1}=(4,2,4)$, which is orthogonal to the planes $2x_1+x_2+2x_3={\rm const.}$ It follows that all these planes intersect the axis $S_1\vee S_2$ of the octahedron orthogonally.

You have obtained $P_a=(13-4a,1-2a,9-4a)$ [I have not checked this], so that $P_0=S_1$ and $P_1=M$. This allows to conclude that for $0< a<1$ the plane $E_a$ cuts off a small square pyramid $Y_a$ from $O$ with apex at $S_1$. The volume of this pyramid can be computed using elementary geometry. Note that the edge length $s$ of $O$ satisfies $s^2=|AB|^2=72$, and the height of the "upper half" $Y_1$ of $O$ is given by $h=|MS_1|=6$. It follows that ${\rm vol}(Y_1)={1\over3}s^2 h=144$. Since each $Y_a$ is similar to $Y_1$ with a linear factor $a$ we finally obtain $${\rm vol}(Y_a)=144a^3\ .$$


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