Lesson 12-3: Properties of Size Changes (Day 123)

 Today on her Daily Epsilon on Math 2021, Rebecca Rapoport writes:

What is the minimum number of faces required to form an Archimedean solid?

Of course, until we know what an Archimedean solid is, we can't even start this problem. Many of us have heard of Platonic solids -- a Platonic solid is what the U of Chicago text calls a "regular polyhedron" in the Exploration section of Lesson 9-7:

"A regular polyhedron is a convex polyhedron in which all faces are congruent regular polygons and the same number of edges intersect at each of its vertices."

As it turns out, an Archimedean solid has almost the same definition, except for a few changes:

"An Archimedean solid is a convex polyhedron in which all faces are regular polygons that are not all congruent and the same number of edges intersect at each of its vertices..."

Notice that we can't simply drop the word "congruent" from the Platonic solid definition -- otherwise it would lead to an inclusive definition in which all Platonic solids are Archimedean. In particular, the tetrahedron would become Archimedean, and the answer to today's Rapoport question would be four. I will just say it now -- the answer isn't four. The definition of Archimedean solid is exclusive, so that no Platonic solid is Archimedean.

So how can the faces of a solid be all regular yet not all congruent? There are only two ways for regular polygons not to be congruent -- either they have different side lengths or different numbers of sides. But the faces of a solid share edges, thus forcing all side lengths to be equal. Thus in an Archimedean solid, the faces must have different numbers of sides -- for example, it might contain triangles and squares.

Oh, and there's another part of the definition of Archimedean solid that we must consider:

"...and is not a prism..."

It's easy to form a triangular prism with two equilateral triangles and three squares as faces. The definition of Archimedean solid excludes prisms, otherwise the answer to this problem is five. Here's a hint -- the correct answer isn't five.

One way to form an Archimedean solid is to take our simplest Platonic solid -- the tetrahedron -- and modify it so that it becomes Archimedean. For example, from each triangular face we might try cutting off a small triangle from each vertex.

If we try cutting halfway down each edge and removing the corners, what's left is a smaller triangle -- and so doing it to the whole tetrahedron, we're left with just a smaller tetrahedron.

Instead, we try cutting one-third of the way down each edge rather than halfway. Then voila -- what's left of each face is a regular hexagon. If we try doing to the entire tetrahedron, then the original four faces become regular hexagons, and the parts where we cut the four vertices become four new equilateral triangles.

Since we formed this new solid by cutting off or truncating the vertices of a tetrahedron, the resulting solid is called a truncated tetrahedron:

https://mathworld.wolfram.com/TruncatedTetrahedron.html

As it happens the truncated tetrahedron with its eight faces is the simplest Archimedean solid. Therefore the desired answer to today's problem is eight -- and of course, today's date is the eighth.

By the way, here's a link to the Wolfram page for all thirteen Archimedean solids. While eight is indeed the fewest number of faces one can have, the most is 92:

https://mathworld.wolfram.com/ArchimedeanSolid.html

According to this link, there's one part of the definition that we left out:

"...or an antiprism."

Antiprisms are tricky to define -- they're sort of like prisms, but not really. But as it turns out, the simplest antiprism also has eight faces, so we don't need to know what antiprisms are in order to get the Rapoport problem correct.

Also, Wolfram mentions another term, semiregular polyhedron. Since the U of Chicago text uses the term "regular polyhedron" for Platonic solid, we ought to use "semiregular polyhedron" instead of Archimedean solid. There's just one difference though -- a prism (or antiprism) counts as a semiregular polyhedron but not an Archimedean solid.

Archimedean solids are, of course, named for the Greek mathematician Archimedes, just as Platonic solids are named for Plato.

Lesson 12-3 of the U of Chicago text is called "Properties of Size Changes." In the modern Third Edition of the text, properties of size changes appear in Lesson 12-1.

This is what I wrote last year about today's lesson. Admittedly it isn't much.

Finally, here are the Geometry worksheets for today. They are based on Lesson 12-3, with an extra page for the proof of the Dilation Distance Theorem -- this proof comes directly from PARCC.

Meanwhile, the old first page makes an reference to FTS (an old Hung-Hsi Wu proof). Even though I no longer include Wu's FTS as part of the lesson, it's actually too much work to redo the entire page just to get rid of two little FTS mentions (especially when I'm already posting this late). So you'll just have to ignore FTS.