Mathcad Community Challenge November 2024 - Frustums and Pyramids

[DaveMartin]

This month’s challenge is based around pyramids and frustums (a pyramid that’s had its top removed by a plane parallel to the base). Choose any or all of the following and create a Mathcad worksheet:

Moderate Challenge
Calculate the volume and surface area of the frustum in the image below. A Creo 7.0 part model is attached to this challenge for verification.

Note that the original pyramid is a right pentahedron. (A line joining the center of the base and the vertex of the pyramid is at a right angle / perpendicular to the base. See the CAD model and image for additional clarification.) The base and top surface of the frustum are both square.

Optional 1: Create a function that calculates the volume given the 3 dimensions above: edge length of the base, side edge length, and top face edge length.

Optional 2 (for Mathcad Prime 10 users): Use the slider advanced input control to change either the height of the slicing plane or length of the side edge, and recalculate the volume.

Hard Challenge: The Pyramid of Least Volume
“Of all the planes tangent to the ellipsoid

 one of them cuts the pyramid of least possible volume from the first octant x ≥ 0, y ≥ 0, z ≥ 0. Show that the point of tangency of that plane is the centroid of the face ABC.”

The pyramid in this situation has four sides. One of the corners is at the origin (0,0,0). The centroid is this case means the intersection of its medians.
(Source: “The Mathematical Mechanic” by Mark Levi, section 3.5.)

Documentation Challenge
Create a Mathcad worksheet that uses text and image tools to explain the derivation of the formula for the volume of a pyramid.

Find the Mathcad Community Challenge Guidelines here!

[LucMeekes]

"Note that this is a right pentahedron":

I count 6 surface faces, I'd call it a hexahedron....

Luc

[DaveMartin]

You are correct. "Right pentahedron" described the original pyramid before the top was cut off to form the frustum. I have edited the original post to reflect that distinction.

[Werner_E]

I am not sure if "right pentahedron" really describes the object fully.
I would have called it a "straight regular square four-sided pyramid" and I guess we could remove the "regular"  and "four-sided" when we have 'straight' and "square". But then, I am all but an English native speaker ...

BTW, I just looked up the book of Mark Levi and the general non-math proof in 3.6 is a real nice one!

[DaveMartin]

I took the phrase "right pentahedron" from another source (a video, I believe). The following source explains the term "right polyhedron:"

https://www.alloprof.qc.ca/en/students/vl/mathematics/polyhedrons-m1229

(Original page may be in French.) I will add verbiage if people are having trouble understanding the geometry I am describing.

However, note that the original pyramid is not a regular polyhedron, as all the faces are not the same size and shape. The base has 4 edges, and the other faces have 3 edges.

[Werner_E]

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However, note that the original pyramid is not a regular polyhedron, as all the faces are not the same size and shape. The base has 4 edges, and the other faces have 3 edges.

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Sure, the only pyramid which also is a regular polyhedron is a regular tetrahedron, isn't it?

Given that 'regular polyhedron' means all faces being equal.
As with any pyramid the side faces that are connected to the top of the pyramid are always triangles. For a pyramid to be a regular polyhedron this means that the base must be triangular as well.

[StuartBruff]

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@Werner_E wrote:

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However, note that the original pyramid is not a regular polyhedron, as all the faces are not the same size and shape. The base has 4 edges, and the other faces have 3 edges.

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Sure, the only pyramid which also is a regular polyhedron is a regular tetrahedron, isn't it?

Given that 'regular polyhedron' means all faces being equal.
As with any pyramid the side faces that are connected to the top of the pyramid are always triangles. For a pyramid to be a regular polyhedron this means that the base must be triangular as well.

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I take it we're limiting this discussion to Euclidean space?

Stuart

@DaveMartin I intuitively understood the term "right pentahedron" to mean a pyramid with a square base and four equal-length triangular sides, the frustrum being the hexahedron resulting from lopping off a similar smaller pentahedron from the top of the pyramid.

There are other forms of pentahedron which meet the criterion of a "right" pentahedron being one where "A line joining the center of the base and the vertex of the pyramid is at a right angle / perpendicular to the base", but I did indeed see the CAD model and image for additional clarification.  

Perhaps, a further stipulation is that the right pentahedron under discussion is congruent under right rotations?

[Werner_E]

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@StuartBruff wrote:

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@Werner_E wrote:

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However, note that the original pyramid is not a regular polyhedron, as all the faces are not the same size and shape. The base has 4 edges, and the other faces have 3 edges.

---

Sure, the only pyramid which also is a regular polyhedron is a regular tetrahedron, isn't it?

Given that 'regular polyhedron' means all faces being equal.
As with any pyramid the side faces that are connected to the top of the pyramid are always triangles. For a pyramid to be a regular polyhedron this means that the base must be triangular as well.

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I take it we're limiting this discussion to Euclidean space?

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Yes. The term ‘regular’ would first have to be defined in non-Euclidean spaces ...

[StuartBruff]

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@Werner_E wrote:

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@StuartBruff wrote:

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I take it we're limiting this discussion to Euclidean space?

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Yes. The term ‘regular’ would first have to be defined in non-Euclidean spaces ...

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Hmph! Spoilsport.  👿

I was just wondering, "What if the top of the pyramid was sliced off by being within the Schwarzschild radius of a black hole formed from a newly collapsed star with the mass of the remnant exceeding the Tolman–Oppenheimer–Volkoff limit?".  And people wonder why nobody wants to talk to me at parties.  And why I'm so easily distracted ...

Stuart

[AlanStevens]

Ok, here's my entry with Prime 10 Express.

Alan

[ppal]

Volume:

[DJNewman]

Oh, is this why you made that thread asking about the slider and units?

[DJNewman]

It's come to my attention that November is only 30 days long, so...  as we are over two-thirds through November now, time is running out for entries to this challenge! Don't miss the deadlines...!

We'd love to see as many entries as we can. It's very good seeing different perspectives on how to tackle problems.