# Tag Archives: triangle

# A Golden Tessellation of Quadrilaterals

In this tessellation, golden rectangles are shown in yellow. The orange darts are each made of two golden gnomons, joined at a leg — while the blue rhombi are each made of two golden triangles, sharing a base.

# Thirteen Equilateral Triangles

## Triangles and Circles

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# Tessellation of Blue Triangles and Yellow Concave Pentagons

Alternately, this can be seen as a tessellation of blue diconcave hexagons and yellow triconcave enneagons. Which do you see?

## Spectral Golden Spiral II

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# Polygons Related to the Golden Ratio, and Associated Figures in Geometry, Part 1: Triangles

There are two isosceles triangles which are related to the golden ratio, [1 + sqrt(5)]/2, and I used to refer to them as the “golden acute isosceles triangle” and the “golden obtuse isosceles triangle,” before I found out these triangles already have other names –the ones shown above. The golden triangle, especially, shows up in some well-known polyhedra, such as both the great and small stellated dodecahedron. The triangles which form the “points” or “arms” of regular star pentagons (also known as pentagrams) are also golden triangles.

These triangles have sides which are in the golden ratio. For the golden triangle, it is the leg:base ratio which is golden, as shown above. For the golden gnomon, this ratio is reversed: the base:leg ratio is φ, or ~1.61803 — the irrational number known as the golden ratio.

The angle ratios of each of these triangles are also unique. The golden triangle’s angles are in a 1:2:2 ratio, while the angles of the golden gnomon are in a ratio of 1:1:3.

Another interesting fact about these two triangles is that each one can be subdivided into one of each type of triangle. The golden triangle can be split into a golden gnomon, and a smaller golden triangle, while the golden gnomon can be split into a golden triangle, and a smaller golden gnomon, as seen below.

This process can be repeated indefinitely, in each case, creating ever-smaller triangles of each type.

Polyhedra which use these triangles, as either faces or “facelets” (the visible parts of partially-hidden faces) are not uncommon, as previously mentioned. The three most well-known examples are three of the four Kepler-Poinsot solids. In the first two shown below, the small stellated dodecahedron and the great stellated dodecahedron, the actual faces are regular star pentagons which interpenetrate, but the facelets are golden triangles.

The next example is also a Kepler-Poinsot solid: the great dodecahedron. Its actual faces are simply regular pentagons, not star pentagons, but, again, they interpenetrate, hiding much of each face from view. The visible parts, or “facelets,” are golden gnomons.

For another example of a polyhedron made of golden gnomons, I made one myself — meaning that if anyone else has ever seen this polyhedron before, this fact is unknown to me, although I cannot rule it out. I have not given it a name. It has thirty-six faces, all of which are golden gnomons. There are twelve of the larger ones, shown in yellow, and twenty-four of the smaller ones, shown in red. This polyhedron has pyritohedral symmetry (the same type of symmetry seen in the seam-pattern of a typical volleyball), and its convex hull is the icosahedron.

[Picture credits: to create the images in this post, I used both *Geometer’s Sketchpad* and *MS-Paint* for the two still, flat pictures found at the top. To make images of the four rotating polyhedra, I used a different program, *Stella 4d: Polyhedron Navigator. Stella* is available for purchase, with a free trial download available, at http://www.software3d.com/Stella.php.]

# On Triangle Congruence, and Why SSA Does Not Work

Those who have taught geometry, when teaching triangle congruence, go through a familiar pattern. SSS (side-side-side) triangle congruence is usually taught first, as a postulate, or axiom — a statement so obvious that it requires no proof (although demonstrations certainly do help students understand such statements, even if rigorous proof is not possible). Next, SAS (side-angle-side) and ASA (angle-side-angle) congruence are taught, and most textbooks also present them as postulates. AAS (angle-angle-side) congruence is different, however, for it need not be presented without proof, for it follows logically from ASA congruence, paired with the Triangle Sum Theorem. With such a proof, of course, AAS can be called a theorem — and one of the goals of geometricians is to keep the number of postulates as low as possible, for we dislike asking people to simply accept something, without proof.

At about this point in a geometry course, because the subject usually is taught to teenagers, some student, to an audience of giggling and/or snickering, will usually ask something like, “When are we going to learn about angle-side-side?”

The simple answer, of course, is that there’s no such thing, but there’s a much better reason for this than simple avoidance of an acronym which many teenagers, being teenagers, find amusing. When I’ve been asked this question (and, yes, it has come up, *every* time I have taught geometry), I accept it as a valid question — since, after all, it is — and then proceed to answer it. The first step is to announce that, for the sake of decorum, we’ll call it SSA (side-side-angle), rather than using a synonym for a donkey (in all caps, no less), by spelling the acronym in the other direction. Having set aside the silliness, we can then tackle the actual, valid question: why does SSA not work?

This actually is a question worth spending class time on, for it goes to the heart of what conjectures, theorems, proof, and disproof by counterexample actually mean. When I deal with SSA in class, I refer to it, first, as a conjecture: that two triangles can be shown to be congruent if they each contain two pairs of corresponding, congruent sides, and a pair of corresponding and congruent angles which are not included between the congruent sides, of either triangle. To turn a conjecture into a theorem requires rigorous proof, but, if a conjecture is false, only one counterexample is needed to disprove its validity. Having explained that, I provide this counterexample, to show why SSA does not work:

In this figure, A is at the center of the green circle. Since segments AB and AC are radii of the same circle, those two segments must be congruent to each other. Also, since congruence of segments is reflexive, segment AD must be congruent to itself — and, finally, because angle congruence is also reflexive, angle D must also be congruent to itself.

That’s two pairs of corresponding and congruent segments, plus a non-included pair of congruent and corresponding angles, in triangle ABD, as well as triangle ACD. If SSA congruence worked, therefore, we could use it to prove that triangle ABD and triangle ACD are congruent, when, clearly, they are not. Triangle ACD contains all the points inside triangle ABD, plus others found in isosceles triangle ABC, so triangles ABD and ACD are thereby shown to have different sizes — and, by this point, it has already been explained that two triangles are congruent if, and only if, they have the same size *and* shape. This single counterexample proves that SSA does not work.

Now, can this figure be modified, to produce an argument for a different type of triangle congruence? Yes, it can. All that is needed is to add the altitude to the base of isosceles triangle ABC, and name the foot of that altitude point E, thereby creating right triangle AED.

It turns out that, for right triangles only, SSA actually *does* work! The relevant parts of the right triangle, shown in red, are segment DA (congruent to itself, in any figure set up this way), segment AE (also congruent to itself), and the right angle AED (since all right angles are congruent to each other). However, as I’ve explained to students many times, we don’t *call* this SSA congruence, since SSA only works for right triangles. To call this form of triangle congruence SSA (forwards or backwards), when it only works for some triangles, would be confusing. We use, instead, terms that are specific to right triangles — and that’s how I introduce HL (hypotenuse-leg) congruence, which is what SSA congruence for right triangles is called, in order to avoid confusion. Only right triangles, of course, contain a hypotenuse.

This is simply one example of how to use a potentially-disruptive student question — also known as a teenager being silly — and turn it around, using it as an opportunity to teach something. Many other examples exist, of course, in multiple fields of learning.

# A Polyhedral Demonstration of the Fact That Twenty Times Four Is Eighty

The Platonic solid known as the icosahedron has twenty triangular faces. This polyhedron resembles the icosahedron, but with each of the icosahedron’s triangles replaced by a panel of four faces: three isosceles trapezoids surrounding a central triangle. Since (20)(4) = 80, it is possible to know that this polyhedron has eighty faces — without actually counting them.

To let you see the interior structure of this figure, I next rendered its triangular faces invisible, to form “windows,” and then, just for fun, put the remaining figure in “rainbow color mode.”

I perform these manipulations of polyhedra using software called *Stella 4d*. If you’d like to try this program for yourself, the website to visit for a free trial download is http://www.software3d.com/Stella.php.

# The 9-81-90 Triangle

In a previous post (right here), I explained the 18-72-90 triangle, derived from the regular pentagon. It looks like this:

I’m now going to attempt derivation of another “extra-special right triangle” by applying half-angle trigonometric identities to the 18º angle. After looking over the options, I’m choosing cot(θ/2) = csc(θ) + cot(θ). By this identity, cot(9°) = csc(18°) + cot(18°) = 1 + sqrt(5) + sqrt[2sqrt(5) + 5].

Since cotangent equals adjacent over opposite, this means that, in a 9-81-90 triangle, the side adjacent to the 9° angle has a length of 1 + sqrt(5) + sqrt[2sqrt(5) + 5], while the side opposite the 9° angle has a length of 1. All that remains, now, is to use the Pythagorean Theorem to find the length of the hypotenuse.

By the Pythagorean Theorem, and calling the hypotenuse h, we know that h² = (1)² + {[1 + sqrt(5)] + sqrt[2sqrt(5) + 5]}² = 1 + {2[(1 + √5)/2] + sqrt[(2√5) + 5]}² = 1 + {2φ + sqrt[(2√5) + 5]}², where φ = the Golden Ratio, or (1 + √5)/2, since I want to use the property of this number, later, that φ² = φ + 1.

Solving for h, h = sqrt(1 + {2φ + sqrt[(2√5) + 5]}²) = sqrt{1 + 4φ² + (2)2φsqrt[(2√5) + 5] + (2√5) + 5} = sqrt{6 + 4(φ + 1) + 4φsqrt[(2√5) + 5] + (2√5)} = sqrt{6 + 4φ + 4 + 4φsqrt[(2√5) + 5] + (2√5)} = sqrt{10 + 4[(1 + √5)/2] + 4φsqrt[(2√5) + 5] + (2√5)} = sqrt{10 + 2 + (2√5) + 4φsqrt[(2√5) + 5] + (2√5)} = sqrt{12 + (4√5) + 4[1 + √5)/2]sqrt[(2√5) + 5]} = sqrt{12 + (4√5) + (2 + 2√5)sqrt[(2√5) + 5]}, the length of the hypotenuse. Here, then, is the 9-81-90 triangle: