Category Archives: Mathematics

The dominant topic of my blog. You figured that out from the blog’s name, though, right?

Polyhedron Featuring Decagons and Triangles

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Polyhedron Featuring Decagons and Triangles

This was created using Stella 4d, which you may try for free at http://www.software3d.com/stella.php.

A Survey of Right Interior Angles in Hexagons

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A Survey of Right Interior Angles in Hexagons

A regular hexagon, of course, has no right angles, but irregular, convex hexagons can have one, two, or three right angles.

With one right angle, there is only one basic configuration, but, with two right angles, there are three: the right angles may be consecutive, have one non-right angle between them, or be opposite angles.

There are also three possible configurations with three right angles: the three angles can be consecutive, or two can be consecutive with one non-right angle separating the other right angle from the consecutive pair, or every other angle can be a right angle.

Four right angles cannot exist in a convex hexagon, nor can five, nor, of course, six. Four right interior angles are possible, however, for non-convex hexagons, and, again, there are three possible configurations. In the first, the four right angles are consecutive. In the second, three are consecutive, then a non-right angle separates the fourth right angle from the other three. In the third, there are two pairs of consecutive right angles, with single non-right angles separating the pairs on opposite sides of the hexagon.

A Survey of Right Angles in Convex Pentagons

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A Survey of Right Angles in Pentagons

A regular pentagon, of course, has no right angles, but irregular pentagons can have one, two, or three (but not four, nor five). There are two varieties for both two and three right angles in pentagons — the right angles can be consecutive, or non-consecutive.

Compound of the Rhombic Triacontahedron and a Truncated Icosahedron

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Compound of the Rhombic Triacontahedron and a Truncated Icosahedron

I stumbled across this while manipulating polyhedra with Stella 4d, which you can try for free at http://www.software3d.com/stella.php.

The title describes the blue and yellow figure as “a” truncated icosahedron, rather than “the” truncated icosahedron, because of the slight irregularity of the hexagonal faces, a result of the truncation-planes being slightly closer to the center than is the case for the true Archimedean solid. It should be possible to fix this, but that may be beyond my abilities.

Sixty Irregular Pentagons in a Regular Pattern

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Sixty Irregular Pentagon in a Regular Pattern

Another accidental discovery made while manipulating polyhedra with Stella 4d, which you can try for free at http://www.software3d.com/stella.php.

Polyhedron Featuring Nonagons and Rhombi

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Polyhedron Featuring Nonagons and Rhombi

I stumbled across this while manipulating polyhedra with Stella 4d, which you can try for free at http://www.software3d.com/stella.php.

I don’t think the nonagons are quite regular, and the yellow figures may be near-rhomboidal kites, rather than true rhombi. Nevertheless, I find it an interesting figure, and am posting it here so I can find it for further investigations later.

Icosahedral Shell

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Icosahedral Shell

This was created with Stella 4d, software you can try for free at http://www.software3d.com/stella.php.

A “Bowtie” Variant of the Truncated Icosahedron

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This has the twelve regular pentagons and twenty regular hexagons of a truncated icosahedron (most familiar to the world as the shape upon which most soccer balls are based), but also has pairs of trapezoids in “bowtie” configurations. I discovered this polyhedron using Stella 4d, which you can try for free at http://www.software3d.com/stella.php.

The 27-63-90 Triangle

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The 27-63-90 Triangle

I’m not as pleased about finding this special right triangle as I was with the previous ones. For one thing, I didn’t use a regular polygon to derive it, but instead used the 36-54-90 triangle I had previously found, and applied a half-angle trigonometric identity to find the tangent of 27 degrees, as the tangent of half of 54 degrees. According to this identity, tan(27⁰) = (1 – cos(54⁰))/sin(54⁰). Once this gave me the leg lengths, I simply used the Pythagorean Theorem to determine the length of the hypotenuse. Finally, I checked all of this using decimal approximations.

The triple-nested radical in the expression for the hypotenuse is no cause for celebration, either. If anyone knows of a way to put this in simpler exact terms, please let me know.

The 15-75-90 Triangle

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The 15-75-90 Triangle

As the 30-60-90 triangle is based on an equilateral triangle, the 45-45-90 triangle is based on a square, the 18-72-90 and 36-54-90 triangles are based on the regular pentagon (see https://robertlovespi.wordpress.com/2013/03/12/the-18-72-90-and-36-54-90-triangles/), and the 22.5-67.5-90 triangle is based on the regular octagon (see previous post), so the 15-75-90 triangle is based on the regular dodecagon, shown here with three radii (red) and a single diagonal (purple). The 15-75-90 triangle is shown in yellow. An argument from symmetry is sufficient to show that angle EFC is the right triangle in this triangle, and the larger of its two acute angles (angle FCE) is one-half of an interior angle of this dodecagon. The interior angle of a regular decagon measures 150 degrees (the proof of this is trivial), and so angle FCE must measure half that amount, or 75 degrees. This leaves 15 degrees for angle CEF, via the triangle sum theorem.

What about the side lengths of the 15-75-90 triangle, though? First, consider the red diagonals shown, and let them each have a length of 2. Angles DAF and FAE each measure 30 degrees, since 360/12 = 30, and they are central angles between adjacent radii. This makes angle DAE 60 degrees by angle addition, and triangle DAE is known to be isosceles, since the two red sides are radii of the same regular dodecagon, and therefore are congruent. By the isosceles triangle theorem and triangle sum theorem, then, angles ADE and AED each also measure (180-60)/2 = 60 degrees, so triangle ADE is therefore equilateral, with the purple side, DE, also having a length of two. Symmetry is sufficient to see that DE is bisected by radius AC, which leads to the conclusion that EF, the long leg of the 15-75-90 triangle, has a length of 1.

Segment AF is a median, and therefore also an altitude, of equilateral triangle ADE, and splits it into two 30-60-90 triangles, one of which is triangle AEF. Its hypotenuse, AE, is already known to have a length of 2, while its short leg, EF, is already known to have a length of 1. Segment AF is therefore the long leg of this 30-60-90 triangle, with a length of √3.

AF, length √3, and FC, the short leg of the 15-75-90 triangle, together form dodecagon radius AC, already set at length 2. By length subtraction, then, FC, the 15-75-90 triangle’s short leg, has a length of 2 – √3. A test is prudent at this point, by taking the tangent of the 15 degree angle FEC in the yellow triangle. Tan(15 degrees) is equal to 0.26794919…, which is also the decimal approximation for FC/EF, or (2 – √3)/1.

All that remains to know the length ratios for the sides of the 15-75-90 triangle is to determine the length of EC, its hypotenuse, via the Pythagorean Theorem. The square of length EC must equal the square of 1 plus the square of (2 – √3), so EC, squared, equals 1 + 4 – 4√3 + 3, or 8 – 4√3. The hypotenuse (EC) must therefore be the square root of 8 – 4√3, which is √(8-4√3)) = 2√(2-√3)).

The short leg:long leg:hypotenuse ratio in a 15-75-90 triangle is, therefore, (2-√3):1:2√(2-√3)).