A Big Ball of Pyramids


A Big Ball of Pyramids

To create this, a central rhombicosidodecahedron had each of its faces augmented by pyramids, each of which uses only equilateral triangles for lateral faces.

Software credit: The program used to make this image, Stella 4d, may be tried for free at http://www.software3d.com/stella.php.

A Variant of the Great Icosahedron


Faceted Dual

The great icosahedron has 20 faces which are interpenetrating equilateral triangles, most of which are hidden in the interior of that polyhedron. The non-hidden, and therefore visible, parts are called “facelets” — and there are 180 of them: 120 scalene, and 60 isosceles.

In this variant of the great icosahedron, the sixty isosceles facelets are simply missing, which changes the shape of the remaining 120, still-scalene facelets. The color scheme is one which gives each facelet a different color — except for coplanar or parallel facelets, which are the same color, making them easier to spot.

Software credit: see http://www.software3d.com/stella.php — with a free trial download available there.

The 18-72-90 and 36-54-90 Triangles

It is well-known that an altitude splits an equilateral triangle into two 30-60-90 triangles, and that a diagonal splits a square into two 45-45-90 triangles. The properties of these “special right triangles,” as they are often called, are well-understood, and shall not be described here.

What happens if other polygons are split by diagonals, altitudes, or pieces thereof? Can more triangles be found which can allow, for example, exact determination of certain trigonometric ratios?

Yes, and the logical place to start looking is in the regular pentagon.


In this diagram, the yellow triangle is the 18-72-90 triangle. Its hypotenuse is a diagonal of the pentagon, and its short leg is a half-side of the pentagon. Since sides and diagonals of regular pentagons are in the Golden Ratio, (1 + √5)/2, these two sides must be in twice that ratio. Let their lengths, then, be 1 (short leg) and 1 + √5 (hypotenuse), for those are simple, and in the specified ratio. The Pythagorean Theorem may then be applied to find the length of the long leg; the result is sqrt((2√5) + 5). Yes, nested radicals appear at this point, and they resist efforts to make them go away. No one promised this would be simple!

The blue triangle is the 36-54-90 triangle. Its long leg is a half-diagonal of the pentagon, while its hypotenuse is a full side of the pentagon. These triangle sides must, therefore, be in half the Golden Ratio, so the simplest lengths for those sides (which work) are 1 + √5 for the long leg, and 4 for the hypotenuse. Applying the Pythagorean Theorem to find the length of the short leg, nested radicals appear again in the solution:  sqrt(10 – 2√5).

How I Found the Nagel Line While Playing with Triangles

Several days recently swirled down the drain in a depression-spiral. Needing a way out, I spent my Saturday morning playing with triangles, after first getting plenty of sleep. It worked. This technique, however, probably would not transfer to those who are not geometry obsessives. Perhaps any favorite activity would work? I leave that to others to explore.

Here’s what I did that worked for me:

ImageThe original triangle is ABC, and is in bold black. The bold blue line is its Euler Line, and contains the orthocenter (M), circumcenter (G), nine-point center (K), and centroid (point W). It does not, however, contain the incenter (S).

It struck me as odd that the incenter would be different in this way, so I investigated it further. It is the point of concurrence of the three angle bisectors of a triangle. On a lark, I constructed the midsegments of triangle ABC, forming a new, smaller triangle, shown in red. When I then found the incenter of this smaller triangle (Z), it appeared to be collinear with S and W. I checked; it was, and this line is shown in bold yellow. Moreover, the process could be continued with even smaller midsegment-triangle incenters, and they were also on this yellow line.

I wondered if I had discovered something new, and started to check. It didn’t take long to find out that Nagel had beaten me to it. The Nagel line is the official name of this yellow line I stumbled upon, and here is my source:  http://mathworld.wolfram.com/NagelLine.html — but, as far as I know, I did discover that these midsegment-derived points also lie on the Nagel line.

Someone else may have known this before, of coruse. I don’t know, and it doesn’t matter to me, for I had my fun morning playing with triangles, and now feel better than I have in days.

[Side note:  this is my 100th post, and I’d like to thank all my readers and followers, and also thank, especially, those who encouraged me to try WordPress to get a fresh start after Tumblr-burnout. It worked!]