# Eight Kite-Rhombus Solids, Plus Five All-Kite Polyhedra — the Convex Hulls of the Thirteen Archimedean-Catalan Compounds

In a kite-rhombus solid, or KRS, all faces are either kites or rhombi, and there are at least some of both of these quadrilateral-types as faces. I have found eight such polyhedra, all of which are formed by creating the convex hull of different Archimedean-Catalan base-dual compounds. Not all Archimedean-Catalan compounds produce kite-rhombus solids, but one of the eight that does is derived from the truncated dodecahedron, as explained below. The next step is to create the compound of this solid and its dual, the triakis icosahedron. In the image below, this dual is the blue polyhedron. The convex hull of this compound, below, I’m simply calling “the KRS derived from the truncated dodecahedron,” until and unless someone invents a better name for it. The next KRS shown is derived, in the same manner, from the truncated tetrahedron. Here is the KRS derived from the truncated cube. The truncated icosahedron is the “seed” from which the next KRS shown is derived. This KRS is a “stretched” form of a zonohedron called the rhombic enneacontahedron. Another of these kite-rhombus solids, shown below, is based on the truncated octahedron. The next KRS shown is based on the rhombcuboctahedron. Two of the Archimedeans are chiral, and they both produce chiral kite-rhombus solids. This one is derived from the snub cube. Finally, to complete this set of eight, here is the KRS based on the snub dodecahedron. You may be wondering what happens when this same process is applied to the other five Archimedean solids. The answer is that all-kite polyhedra are produced; they have no rhombic faces. Two are “stretched” forms of Catalan solids, and are derived from the cuboctahedron and the icosidodecahedron:

If this procedure is applied to the rhombicosidodecahedron, the result is an all-kite polyhedron with two different face-types, as seen below. The two remaining Archimedean solids are the great rhombcuboctahedron and the great rhombicosidodecahedron, each of which produces a polyhedron with three different types of kites as faces.

The polyhedron-manipulation and image-production for this post was performed using Stella 4d: Polyhedron Navigator, which may be purchased or tried for free at http://www.software3d.com/Stella.php.

# A Tetrahedral Exploration of the Icosahedron

Mathematicians have discovered more than one set of rules for polyhedral stellation. The software I use for rapidly manipulating polyhedra (Stella 4d, available here, including as a free trial download) lets the user choose between different sets of stellation criteria, but I generally favor what are called the “fully supported” stellation rules.

For this exercise, I still used the fully supported stellation rules, but set Stella to view these polyhedra as having only tetrahedral symmetry, rather than icosidodecahedral (or “icosahedral”) symmetry. For the icosahedron, this tetrahedral symmetry can be seen in this coloring-pattern. The next image shows what the icosahedron looks like after a single stellation, when performed through the “lens” of tetrahedral symmetry. This stellation extends the red triangles as kites, and hides the yellow triangles from view in the process. The second such stellation produces this polyhedron — a pyritohedral dodecahedron — by further-extending the red faces, and obscuring the blue triangles in the process. The third tetrahedral stellation of the icosahedron produces another pyritohedral figure, which further demonstrates that pyritohedral symmetry is related to both icosidodecahedral and tetrahedral symmetry. The fourth such stellation produces a Platonic octahedron, but one where the coloring-scheme makes it plain that Stella is still viewing this figure as having tetrahedral symmetry. Given that the octahedron itself has cuboctahedral (or “octahedral”) symmetry, this is an increase in the number of polyhedral symmetry-types which have appeared, so far, in this brief survey. Next, I looked at the fifth tetrahedral stellation of the icosahedron, and was surprised at what I found. While I was curious about what would happen if I continued stellating this polyhedron, I also wanted to see this fifth stellation’s convex hull, since I could already tell it would have only hexagons and triangles as faces. Here is that convex hull: For the last step in this survey, I performed one more tetrahedral stellation, this time on the convex hull I had just produced. # Variations of the Snub Dodecahedron To make the first of these variations, above, I augmented each triangular face of a snub dodecahedron with an antiprism 2.618 times as tall as the triangles’ edge length, and then took the convex hull of the result. The other polyhedra shown, below, were obtained by various other manipulations of the snub dodecahedron, all performed using a program called Stella 4d: Polyhedron Navigator, which you can try right here. The variant above looked like it needed a name, so I called it an expanded snub truncated dodecahedron. As for the one below, it is one of many facetings of the snub dodecahedron. Finally, the last figure shown (stumbled upon during a “random walk” with Stella) is one of many possible figures which are non-convex relatives of the snub dodecahedron. # Standard and Faceted Versions, Side by Side, of Each of the Thirteen Archimedean Solids

These two polyhedra are the truncated tetrahedron on the left, plus at least one faceted version of that same Archimedean solid on the right. As you can see, in each case, the figures have the same set of vertices — but those vertices are connected in a different way in the two solids, giving the polyhedra different faces and edges.

(To see larger images of any picture in this post, simply click on it.)

The next three are the truncated cube, along with two different faceted truncated cubes on the right. The one at the top right was the first one I made — and then, after noticing its chirality, I made the other one, which is the compound of the first faceted truncated cube, plus its mirror-image. Some facetings of non-chiral polyhedra are themselves non-chiral, but, as you can see, chiral facetings of non-chiral polyhedra are also possible.

The next two images show a truncated octahedron, along with a faceted truncated octahedron. As these images show, sometimes faceted polyhedra are also interesting polyhedra compounds, such as this compound of three cuboids.

The next polyhedra shown are a truncated dodecahedron, and a faceted truncated dodecahedron. Although faceted polyhedra do not have to be absurdly complex, this pair demonstrates that they certainly can be.

Next are the truncated icosahedron, along with one of its many facetings — and with this one (below, on the right) considerably less complex than the faceted polyhedron shown immediately above.

The next two shown are the cuboctahedron, along with one of its facetings, each face of which is a congruent isosceles triangle. This faceted polyhedron is also a compound — of six irregular triangular pyramids, each of a different color.

The next pair are the standard version, and a faceted version, of the rhombcuboctahedron, also known as the rhombicuboctahedron.

The great rhombcuboctahedron, along with one of its numerous possible facetings, comes next. This polyhedron is also called the great rhombicuboctahedron, as well as the truncated cuboctahedron.

The next pair are the snub cube, one of two Archimedean solids which is chiral, and one of its facetings, which “inherited” its chirality from the original.

The icosidodecahedron, and one of its facetings, are next.

The next pair are the original, and one of the faceted versions, of the rhombicosidodecahedron.

The next two are the great rhombicosidodecahedron, and one of its facetings. This polyhedron is also called the truncated icosidodecahedron.

Finally, here are the snub dodecahedron (the second chiral Archimedean solid, and the only other one, other than the snub cube, which possesses chirality), along with one of the many facetings of that solid. This faceting is also chiral, as are all snub dodecahedron (and snub cube) facetings.

Each of these polyhedral images was created using Stella 4d: Polyhedron Navigator, software available at this website.

# On Polyhedral Augmentation and Excavation

I have made many posts here using polyhedral augmentation, but what I haven’t done — yet — is explain it. I have also neglected the reciprocal function of augmentation, which is called excavation. It is now time to fix both these problems.

Augmentation is the easier of the two to explain, especially with images. The figure below call be seen as a blue icosahedron augmented, on a single face, by a red-and-yellow icosidodecahedron. It can also be viewed, with equal validity, as the larger figure (the icosidodecahedron) augmented, on a single triangular face, with an icosahedron. When augmenting an icosidodecahedron with an icosahedron in this manner, one simply attaches the icosahedron to a triangular face of the icosidodecahedron. The reciprocal process, excavation, involves “digging out” one polyhedral shape from the other. Here is what an icosidodecahedron looks like, after having an icosahedron excavated from it, on a single triangular face. Excavating the smaller polyhedron from the larger one is easier to picture in advance, just as one can imagine what the Earth would look like, if a Moon-sized sphereoid were excavated from it, with a large, round hole making the excavation visible. (This is mathematics, not science, so we’re ignoring the fact that gravity would instantly cause the collapse of such a compound planetary object, with dire consequences for all inhabitants.) What’s more difficult is picturing what would result if this were turned around, and the Earth was used to excavate the Moon.

This “Earth-excavated Moon” idea is analogous to excavating the larger icosidodecahedron from the smaller icosahedron. If one thinks of subtracting the volume of one solid from that of the other, such a creature should have negative volume — except, of course, that this makes no sense, which is consistent with the fact that it would be impossible to do such a thing with physical objects: there isn’t enough matter in the Moon to remove an Earth’s worth of matter from the Moon. Also, moving back to polyhedra, with excavation only into a single face, it turns out that there is no change in appearance when the excavation-order is reversed: (Well, OK, there was a small change in appearance between the two images, but that’s only because I changed the viewing angle a bit, to give you a better view of the blue faces.)

Things get different — and the augmentation- and excavation-orders begin to matter a lot more — when these operations are performed on all available faces at once, which, in this case, means on all twenty of each polyhedron’s triangular faces. Here is the easiest case to visualize: an icosidodecahedron, augmented by twenty icosahedra. If you use the reciprocal function, excavation, but leave the order of polyhedra the same, you get a central icosidodecahedron, excavated by twenty smaller, intersecting icosahedra: It is, of course, possible to have other combinations. The ones I find most interesting, using these two polyhedra, are “global” augmentation and excavation of the smaller figure, the icosahedron, by twenty of the larger ones, the icosidodecahedra. Why? Simple: putting the icosidodecahedra on the outside allows for maximum visibility of both pentagons and triangles. On the other hand, the central icosahedron is completely hidden from view, whether augmentation or excavation is used. Here is the augmentation case, or what I have called a “cluster” polyhedron, many varieties of which can be seen elsewhere on this blog (just search for “cluster,” or “cluster polyhedron,” to find them): The global-excavation case which has the icosahedron hidden in the middle is similar to the cluster immediately above, in that all that can be seen are twenty intersecting icosidodecahedra. However, it also varies noticeably, because, with excavation, the icosidodecahedra are closer to the center of the entire cluster (the invisible, central icosahedron’s center) than was the case with augmentation. The last image here is of an invisible, central icosahedron, with an icosidodecahedron excavated from all twenty triangular faces. The larger polyhedra “punch through” the smaller one from all sides at once, trapping the central polyhedron — the blue icosahedron — from view. The remaining object looks, to me at least, more like a faceted icosidodecahedron than a cluster-polyhedron. I am of the opinion, but have not verified, that this resemblance to a faceting of the icosidodecahedron is illusory. [Image credits: all images in this post were made using Stella 4d: Polyhedron Navigator. This program may be purchased, or tried as a free trial download, at http://www.software3d.com/Stella.php.]

# The Cone Problem (The Easier Sequel to the Hemisphere Problem) That hemisphere problem (see previous two posts) was quite difficult. I’m going to unwind a  bit with the much easier cone version of the same problem: at what height x above the ground, expressed as a fraction of h, must a cone of height h and radius r be cut, in order for the two pieces produced by the cut to have equal volume? The fact that a path down the lateral surface of a cone is a straight line, not a curve, should make this much easier than the hemisphere problem.

Since the volume of a cone is (1/3)πr²h, and the smaller cone created above the cut would be half that volume, it follows that

(1/3)πr²h = (2/3)π(r′)²h′                [equation 1]

By cancellation of (1/3)π, this equation becomes

r²h = 2(r’)²h’               [equation 2]

Also, based on divisions of the cone’s altitude, we know that

h = h′ + x                [equation 3]

Furthermore, since the problem asks that the height x be expressed as a fraction of h, we can let that fraction (a decimal between zero and one) be represented by f, so that

x = fh               [equation 4]

Also, by using similar right triangles’ corresponding legs, we know that

r/h = r′/h′                [equation 5]

which rearranges to

rh′ = r′h                  [equation 6]

There is a proportionality constant in play here, p, defined as the fraction of the length of one part of the larger cone which equals the length of the corresponding part of the smaller cone. As equations, then,

r′ = pr         and          h′ = ph              [equations 7a and 7b]

Also, because p is the fraction of h which is h′, and f is the fraction of h which is x, and h = h′ + x, it follows that

p + f = 1                  [equation 8].

Next, by substituting equations 7a and 7b into equation 2 for r′ and h′, we know that

r²h = 2(pr)²ph               [equation 9]

Which reduces to

1 = 2p³               [equation 10]

When equation 10 is solved for p, it becomes

p = (1/2)^(1/3)                [equation 11]

And, since equation 8 states that p + f = 1, it follows that f = 1 – p, and f is the fraction we seek. By substituting equation 11 for p in f = 1 – p, the following value for f can be determined:

f = 1 – (1/2)^(1/3)               [equation 12]

This leads to the following cleaned-up solution to the problem, shown in standard exact form, and with a decimal approximation as well. The cut, therefore, should be made approximately 20.6% of the way from the bottom to the top of the full cone.

To check this answer, I need only find the volume of the smaller cone, times two, and show that it equals the value of the larger cone.

2(volume of smaller cone) = (2/3)π(r′)²h′ = (2/3)π(pr)²ph =

(2/3)πp³r²h = (2/3)π(cube root of ½)³r²h = (2/3)π(1/2)r²h = (1/3)πr²h,

which is the volume of the full cone, as it should be. The problem has now been solved, and the solution f (by way of p, which equals 1 – f, by a rearrangement of equation 8) has been checked.

# Working Towards a Solution of the Hemisphere Problem

The hemisphere problem referred to here was described in the previous post. To reword it somewhat, consider this hemisphere, half of a sphere of radius r. The orange cross-section is a circle parallel to the hemisphere’s yellow, circular base. We are to find the height of the yellow section with the orange circular top (which I shall call x), as a fraction of r, such that the yellow and red sections above have equal volumes.

Since the volume of a hemisphere is (2/3)πr³ (that’s half a sphere’s volume), each of these two sections must have half the hemisphere’s volume, or (1/3)πr³.

Moreover, the top (red) portion is a “spherical cap,” described here on Wikipedia, as was pointed out to me, by a friend, on Facebook. On that Wikipedia page, you can find this diagram, as well as the formula shown below for the volume of the purple spherical cap in the diagram.  Now, as our goal is to find x, as described at the top of this post, it important to remember that r = x + h, where r is the radius of the original sphere (and height of the hemisphere), h is the height of the spherical cap, and x is the height of the hemisphere, after the spherical cap is removed. We now have two expressions for the volume of the spherical cap: (1/3)πr³ (because it is a fourth of the volume of the original sphere), and (1/6)πh(3a² + h²) from the Wikipedia article on the spherical cap (so all of this assumes, then, accuracy in that Wikipedia article). Setting them equal to each other,

(1/3)πr³ = (1/6)πh(3a² + h²)

Next, I’ll clean this up by multiplying left and right by 6/π, to cancel fractions and π from both sides.

2r³ = h(3a² + h²)

A right triangle exists in the blue-and-purple figure above, and the unlabeled leg is x, the problem’s original goal. I’ll add an “x” to this diagram. Using this right triangle, and the Pythagorean Theorem, it can be seen that a² = r² – x². Also, since r = x + h, it follows that h = r – x. By substitution for a and h, then,

2r³ = h(3a² + h²)

becomes

2r³ = (r – x)[3(r² – x²) + (r – x)²], which then distributes to

2r³ = (r – x)(3r² – 3x² + r² – 2rx + x²), which expands as

2r³ = 3r³ – 3rx² + r³ – 2r²x + rx² -3r²x  + 3x³ – r²x + 2rx² – x³, which simplifies to

0 = 2r³ – 6r²x + 2x³, which becomes, by division:

0 = r³ – 3r²x + x³.

I’m trying to find x, as a fraction of r, meaning that x = kr, and I want k. On that basis, I’ll now substitute kr for each x in the last equation above.

0 = r³ – 3r²(kr) + (kr)³, which then becomes

0 = r³ – 3kr³ + k³r³, or

3kr³ = r³ + k³r³, and then dividing by r³ yields

3k = 1 + k³, which can be rearranged to

3k – k³ = 1, which factors,on the left side, to yield

(k)(3 – k²) = 1.

For k and (3 – k²) to have a product of one, they must be reciprocals. Therefore, 1/k = 3 – k². I can then graph y = 1/k, as well as y = 3 – k², and find the solution by seeing where the graphed functions intersect above the k-axis, with a “k” value between zero and one, since no value of k less than zero or greater than one would make sense, as a solution to the original problem. (I’ll be using k coordinates, and a k-axis, in place of the usual x coordinates and x-axis.) Here’s the initial graph: The only intersection in the specified range of zero to one is between point A and point B, so I brought them together, as closely as I could get Geometer’s Sketchpad to let me. With points A and B almost on top of each other, k = 0.34958 by one equation, and k = 0.34820 by the other. To two significant figures, then, I can conclude that the horizontal cut in the original problem should be made 35% of the way from the base of the hemisphere to the hemisphere’s top.

Using a graphing calculator, a more precise answer of 0.34729636 was obtained. I’d still like to have an exact answer, but this will do for now.

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Later addition: a helpful reader led me to a Wolfram Alpha site where I could get an exact answer, as well as a decimal approximation with a greater degree of precision. In the pic below, I have omitted the two solutions of the third-order polynomial which are not the one solution of interest. Here’s the one which is: Now, however, I have another mystery: how can an exact answer, with all those imaginary units in it, have a real-only approximation? To this question, at least for now, I don’t even have the beginnings of an answer.

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Even later post-script: I have been assured by friends on Facebook that the imaginary units in the above exact solution somehow cancel, although I must concede that I still do not see how, myself. I’ve also been shown another way to express the solution, for 2sin(π/18) is also ≈ 0.3472963553338606977034333. This surprised me, due the the lack of any explicit appearance by a π/18 (= 10°) angle in the original problem, and the fact that no trigonometric functions were used to solve it.