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.

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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?

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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.

[Werner_E]

OK, here are my 2 cents

Only the second part of the challenge, the proof.

Prime 10 was able to handle this almost entirely on its own, only a little help from legacy Mathcad was needed in one place.

Maybe an incentive to improve Primes Symbolics a little further in this direction... 😉

[AlanStevens]

Nice work!

It's also easy enough to get a symbolic solution to:

by hand, without resorting to another CAS program:

Alan

[Werner_E]

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It's also easy enough to get a symbolic solution to:

Agreed on - especially because both equations simplify to equations in just one variable each.
It was surprising that Prime could not deal with that situation.

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by hand, without resorting to another CAS program:

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"by hand", that's the point - my goal was not just using Prime as an equations editor (MathType would do a much better job if that was all it was about) but to let the program do the (not so hard) work on its own (and of course I couldn't resist to show that Mathcad can do it 🙂 ).

BTW, even Prime is able to solve each of the equations on its own if we help it by telling it which variable to solve for.

Forcing the solution in the needed range from 0 to pi/2 seems to confuse Prime - the solution of the second equation is correct but written in a very unusual way:

adding "simplify" or even "rewrite, asin" did not help.

[AlanStevens]

What happens if you try to solve the first equation (on its own) for u, and the second equation (on its own) for v?

I can't try it as I only have Prime Express.

Alan 

[Werner_E]

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@AlanStevens wrote:

What happens if you try to solve the first equation (on its own) for u, and the second equation (on its own) for v?

I can't try it as I only have Prime Express.

Alan 

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Had the same idea and edited my previous reply.

It works but IMHO this would be too much help for Prime as normally you can't expect that the system falls apart in this convenient manner.

[AlanStevens]

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@Werner_E wrote:

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... as normally you can't expect that the system falls apart in this convenient manner.

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Very true, However, no CAS system is perfect, so I think it is always worth seeing if one can provide a "helping hand" when it struggles!

Alan 

[Werner_E]

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@AlanStevens wrote:

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@Werner_E wrote:

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... as normally you can't expect that the system falls apart in this convenient manner.

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Very true, However, no CAS system is perfect, so I think it is always worth seeing if one can provide a "helping hand" when it struggles!

Alan 

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Fully agreed on when it comes to everyday work. One has to know the pros and cons of ones tools and their limits. And so we have to act accordingly. And of course we would choose the tool which best fits the task to be done. When it comes to more elaborate symbolic calcs then the tool to choose at the time probably would not be Prime. OK, it possibly would not be Mathcad either unlesss we have access to MC11.

But I guess the goal of these challenges is to show off how great Prime is, not how good we are in doing math by hand 😉
Its a Prime only challenge, so Prime must show what its capable of out of the box and without having to resort to complicated tricky workarounds. After all that's not so bad in essence, but there is room for improvement in many areas.