A Euclidean Construction of a Regular Triacontagon

Steps of this construction:

  1. Use the green circles and blue lines to construct the yellow pentagon, along with its green inscribed pentagram.
  2. Construct the equilateral triangle shown in gray. This is needed to obtain a twelve degree angle. The triangle is needed for its sixty degree angle, because 72 – 60 = 12. (The 72 degree angle is found inside the pentagon.)
  3. Identify the twelve degree angle shown in bold. A twelve degree angle is needed because 360 / 30 = 12.
  4. Use the red circles to complete the thirty sides of the regular triacontagon, which is shown with bold black segments, inscribed inside a large, bold, red circle.

The Regular Enneagon, and Three Regular Enneagrams

enneagon and enneagrams

The red figure above is a regular enneagon, or nine-sided polygon, and it has three regular enneagrams (or “star enneagons”) inside it. The light blue figure is called a {9,2} enneagram. The green figure can be viewed two ways: as a {9,3} enneagram, or as a compound of three equilateral triangles. Finally, the yellow figure is a {9,4} enneagram.

To see what these numbers in braces mean, just take a look at one of the yellow enneagram’s vertices, then follow one of the yellow segments to the next vertex it touches. Count the vertices which are skipped, and you’ll notice each yellow segment connects every fourth vertex, giving us the “4” in {9,4}. The “9” in {9,4} comes from the total number of vertices in this enneagram, as well as the total number of segments it has. The blue and green enneagrams are analogous to the yellow one. These pairs of numbers in braces are known as Schläfli symbols.

I should mention that some people call these figures “nonagons” and “nonagrams.” Both “ennea- and “nona-” refer to the number nine, but the latter prefix is derived from Latin, while the former is based on Greek. I prefer to use the Greek, since that is consistent with such Greek-derived words as “pentagon” and “hexagon.”

Finally, there is also an “enneagram of personality,” in popular culture, which some use for analyzing  people. Aside from this mention of it, that figure is not addressed here — nor is the nine-pointed star used as a symbol for the Bahá’í faith. However, it’s easy to find information on those things with Google-searches, for those who are interested.

Zonohedra, Zonish Polyhedra, and Another Puzzle

In a recent post, I showed many images of zonohedra, then challenged readers to figure out, from the images, what zonohedra are: polyhedra with only zonogons as faces. Zonogons, I then explained, are polygons with (A) even numbers of edges, and with opposite edges always (B) congruent and (C) parallel. Here is another collection of zonohedra. (Individual images may be enlarged with a click.)

The next set of polyhedra shown, below, are not true zonohedra (as all the ones above are), but merely “zonish polyhedra.” From examination of the pictures above and below, can you figure out the difference between zonohedra and zonish polyhedra?

When you are ready to see the solution to the puzzle, simply scroll down.











While zonohedra have only zonogons as faces, this restriction is “loosened” for zonish polyhedra. Such solids are formed by zonohedrifying non-zonohedral polyhedra, and letting at least some of the faces of the resulting polyhedra remain non-zonogonal. Zonish polyhedra  are called “zonish” because many (usually most) of their faces are zonogons, but not all of them — in each case, some non-zonogonal polygons (such as triangles and/or pentagons, with their odd numbers of edges) do appear. Non-zonogonal polygons are not required to have odd numbers of edges, of course: simply having opposite edges be parallel, but of different lengths, is enough to prevent a polygon (such as a hexagon, octagon, or decagon) from being a zonogon. 


Software credit: I used Stella 4d to make these images. This program may be tried for free at this website.

Octagons Can Tile a Plane II

In April 2014, I found a tessellation of the plane which uses two kinds of octagons — both types equilateral, but only one type regular.

Now, I have found two more ways to tessellate a plane with octagons, and these octagons are also equilateral. However, in these new tessellations, only one type of octagon is used. One of them appears below, twice (the second time is with reversed colors), and the other one appears, once, in the next post.


tessoct 2

The Second of Dave Smith’s “Bowtie” Polyhedral Discoveries, and Related Polyhedra

Dave Smith discovered the polyhedron in the last post here, shown below, with the faces hidden, to reveal how the edges appear on the back side of the figure, as it rotates. (Other views of it may be found here.)

Smith's puzzle

So far, all of Smith’s “bowtie” polyhedral discoveries have been convex, and have had only two types of face: regular polygons, plus isosceles trapezoids with three equal edge lengths — a length which is in the golden ratio with that of the fourth side, which is the shorter base.

Smiths golden trapezoid

He also found another solid: the second of Smith’s polyhedral discoveries in the class of bowtie symmetrohedra. In it, each of the four pentagonal faces of the original discovery is augmented by a pentagonal pyramid which uses equilateral triangles as its lateral faces. Here is Smith’s original model of this figure, in which the trapezoids are invisible. (My guess is that these first models, pictures of which Dave e-mailed to me, were built with Polydrons, or perhaps Jovotoys.)


With Stella 4d (available here), the program I use to make all the rotating geometrical pictures on this blog, I was able to create a version (by modifying the one created by via collaboration between five people, as described in the last post) of this interesting icositetrahedron which shows all four trapezoidal faces, as well as the twenty triangles.

Smith's Icositetrahedron

Here is another view: trapezoids rendered invisible again, and triangles in “rainbow color” mode.

Smith's Icositetrahedron H

It is difficult to find linkages between the tetrapentagonal octahedron Smith found, and other named polyhedra (meaning  I haven’t yet figured out how), but this is not the case with this interesting icositetrahedron Smith found. With some direct, Stella-aided polyhedron-manipulation, and a bit of research, I was able to find one of the Johnson solids which is isomorphic to Smith’s icositetrahedral discovery. In this figure (J90, the disphenocingulum), the trapezoids of this icositetrahedron are replaced by squares. In the pyramids, the triangles do retain regularity, but, to do so, the pentagonal base of each pyramid is forced to become noncoplanar. This can be difficult to see, however, for the now-skewed bases of these four pyramids are hidden inside the figure.

J90 disphenocingulum

Both of these solids Smith found, so far (I am confident that more await discovery, by him or by others) are also golden polyhedra, in the sense that they have two edge lengths, and these edge lengths are in the golden ratio. The first such polyhedron I found was the golden icosahedron, but there are many more — for example, there is more than one way to distort the edge lengths of a tetrahedron to make golden tetrahedra.

To my knowledge, no ones knows how many golden polyhedra exist, for they have not been enumerated, nor has it even been proven, nor disproven, that their number is finite. At this point, we simply do not know . . . and that is a good way to define areas in mathematics in which new work remains to be done. A related definition is one for a mathematician: a creature who cannot resist a good puzzle.

On the Possible Numbers of Vertices of Extended Convex Polygons

A few days ago, I came across the phrase, while surfing the Internet, “the extended quadrilateral has six vertices.” Not understanding this, I researched the matter, and found out what this phrase means. If the normal four vertices of a quadrilateral are placed such that the quadrilateral has no parallel sides, then two additional vertices are created when the four sides are extended as lines, as shown below. This gives a total of six vertices for this extended quadrilateral.


Of course, one need not position the vertices in this way. Trapezoids, by definition, have one pair of parallel sides. The non-parallel sides of an extended trapezoid still intersect outside this quadrilateral, creating a five-vertex situation.


The other option is the parallelogram, with two pairs of parallel sides. This eliminates additional vertices altogether, so there are only four.


This exhausts the possibilities, so the possible numbers of vertices for extended quadrilaterals are 4, 5, and 6.

Realizing this, of course, just raises other questions: what about other extended convex polygons? What are the numbers of possible intersections for such polygons with varying numbers of sides? Is there a pattern? To investigate this, I needed data, and started by taking a step back from quadrilaterals, to briefly consider extended triangles.


Extending the sides of any triangle creates no additional vertices, beyond the three which exist before the extension. This is a result of the fact that three is the maximum number of intersections created by three coplanar lines. Three is, therefore, for triangles, the only answer.

The next step: consider extended convex pentagons. I decided to start by maximizing the number of vertices, by having no parallel sides at all.


As you can see above, this produces ten vertices — five for the non-entended pentagon, plus five more formed by the entensions. To reduce this number, I simply moved vertices of the original, non-extended pentagon to eliminate external vertices, one at a time, by creating pairs of parallel sides.


With one pair of parallel sides, as shown above, the number of vertices is reduced by one, from ten to nine.


For pentagons, the maximum number of pairs of parallel sides is two, as shown above, which lowers the total number of vertices to eight. For pentagons, then, there are three solutions to this puzzle: 8, 9, and 10.

Next, I considered hexagons.


The extended hexagon above has no parallel sides, which gives it the maximum number of vertices:  fifteen. In the figure below, by contrast, there is one pair of parallel sides, reducing this number to fourteen.



With two pairs of parallel sides, as shown above, this number again decreases by one, to thirteen. Three pairs of parallel sides is the maximum for hexagons, and is shown below; for this extended hexagon, there are twelve vertices.


For hexagons, then, there are four solutions:  15, 14, 13, and 12.

Still needing more data, I next investigated extended convex heptagons. With no parallel sides at all, as shown below, the total number of vertices is twenty-one.



If only one pair of parallel sides exist, there is one less vertex than the previous answer gave, and this number is twenty, as shown above. It is also possible for there to be two pairs of parallel sides, as shown below, and this yields nineteen vertices.


For heptagons, there is only one other option: three pairs of parallel sides. (A fourth pair would require eight sides.) This situation is shown below, and in it, there are eighteen vertices.


For heptagons, then, the solutions are four in number:  18, 19, 20, and 21.

Next: octagons. With no sides parallel, to obtain the maximum number of vertices (as shown below), there are twenty-eight of them.


In the diagram below, the solution immediately above is reduced by one, to twenty-seven, by making one pair of sides parallel.


The next solution is shown below, and is twenty-six. This is accomplished by making two pair of sides parallel.


To obtain twenty-five vertices, three pairs of sides are made to be parallel, as shown below.


Finally, twenty-four vertices, the minimum for octagons, requires all four pairs of opposite sides to be parallel. This solution is shown below.


Octagons, then, have five solutions:  24 to 28, inclusive.

Here is the data gathered above, in the form of a table, along with additional data obtained by extrapolation, work shown toward a generalized solution, and then that generalized solution itself, in three parts: one for triangles, one where the number of sides is even, and one where n, the number of sides, is odd, with n > 3.