# Octahedron Pyramid

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

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

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

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

and

$$S_1 =(13 | 1 | 9)$$

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 $$0 , the plane $$E_a$$ intersects a pyramid of the octahedron with the peak $$S_1$$:

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

$$P_a=(13-4a|1-2a|9-4a)$$

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.

Thx

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 $${\rm vol}(Y_a)=144a^3\ .$$