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

golden 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.

golden triangles 2

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.

Small Stellated Dodeca

Great Stellated Dodeca

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.

Great Dodeca

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.

36 golden gnomons

[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:

no SSA

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

20 times 4 is 80

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.”

20 times 4 is 80 version twoI 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:

18-72-90 triangle

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:

9-81-90 triangle


A Comparison of the Areas of Some of the Triangles Formed By Connecting Three Noncollinear Triangle Centers

The five most well-known triangle centers are the centroid (where a triangle’s medians meet), the orthocenter (where the lines containing the altitudes meet), the incenter (where a triangle’s three interior angle bisectors meet), the circumcenter (where the perpendicular bisectors of a triangle’s three sides meet), as well as the center of a triangle’s 9-point circle (see https://en.wikipedia.org/wiki/Nine-point_circle for more information on this circle, and how it is defined). In the diagram below, the constructions for all five of these triangle centers have been performed, for obtuse, scalene triangle ABC.


The thick pink line is called the Euler line, and four of the five triangle centers mentioned above — all of them except the incenter — are always located on this line, no matter a triangle’s shape or size. The incenter, however, is only found on the Euler line for isosceles or equilateral triangles, so, for such triangles, all five of these triangle centers are collinear — and, as a consequence, no triangles can be made by connecting any set of three of them. If the triangle is scalene, however, the incenter will leave the Euler line, and these triangles may then be defined (with construction-clutter removed, but for the same triangle ABC as shown in the first diagram):


If A, B, and/or C are moved around, the area of triangle ABC changes, as do, of course, the areas of the colored triangles above, of which there are six:  yellow, red, blue, yellow and red together, blue and red together, and all three taken as one triangle. For the original configuration of triangle ABC, you can see those triangle areas on the right side of the image above. On the left side, various ratios are given:

  • The triangle which joins the incenter, 9-point circle center, and circumcenter has the same area as the triangle joining the incenter, 9-point circle center, and the orthocenter.
  • The triangle joining the incenter, centroid, and orthocenter has twice the area of the triangle joining the incenter, centroid, and circumcenter — and this latter triangle, itself, has twice the area of the triangle joining the incenter, centroid, and 9-point circle center.
  • The area of the triangle connecting the incenter, orthocenter, and circumcenter has an area three times as large as the triangle connecting the incenter, centroid, and circumcenter.
  • As a consequence of the last two bulleted statements, the area of the triangle connecting the incenter, orthocenter, and circumcenter is six times the area of the triangle connecting the incenter, centroid, and 9-point circle center.

In both diagrams above, the original triangle ABC is scalene and obtuse. If A, B, and/or C are moved around, but the triangle remains scalene (so that the five triangle centers in question remain noncollinear), all six of the colored triangles described above still exist — and the area ratios given in the bulleted statements above remain constant, also. I do not yet have proofs for the constancy of these area ratios, but am confident that it is possible to write them.

If A, B, and C are positioned in such a way that triangle ABC is almost equilateral, the five triangle centers discussed here get very close together — because for a triangle which actually is regular, all five are located in exactly the same spot. Here’s what the almost-regular case looks like:


As you can see, the area ratios described above (left side of diagram) remain the same, even as the actual colored-triangle areas (right side) all approach zero. If I complete a proof for the constancy of any or all of these area ratios, I’ll post such proofs in subsequent posts on this blog — or readers are welcome to write their own proofs, and are invited to leave them as comments on this post.

A Triangle’s Tridpoint-Hexagon


A Triangle's Tridpoint-Hexagon

Any triangle may be named triangle ABC. Each of its sides will contain exactly two points — called “tridpoints” — which divide that side into three segments of equal length. In triangle ABC above, the tridpoints are named in such a way that two of them, E and F, are encountered, in that order, if one moves from A to B. On the way from B to C, two more tridpoints are encountered: first F, and then G. Finally, going from C back to A, the last two tridpoints are found: first H, and then J. If a polygon is formed using those six tridpoints in alphabetical order (matching the order of their placement), that polygon is a convex hexagon, DEFGHJ. Another name for it is hexagon DJHGFE, which I mention only because Geometer’s Sketchpad called it that, in the picture above, when I asked it for the area of this hexagon, shown in green. The original triangle, ABC, includes both the yellow and green regions, and I asked Sketchpad for the area of this triangle, also, as well as the hexagon-to-triangle area ratio, which is shown above as the familiar “decimalized” version of the fraction 2/3.

A nice feature of Sketchpad is that you can do things like this — and then move points around, to see what effect that has on measured and calculated values. When I move points A, B, and/or C, the triangle and hexagon areas, of course, change. Their area ratio, however, remains at a decimal which is a rounded-off version of 2/3. It doesn’t change at all, no matter where A, B, and C are placed. Any triangle’s tridpoint-hexagon has an area exactly 2/3 that of the original triangle.

This is not yet a theorem — because what is written above is an explanation, not a proof. I’ve started working on a proof for this conjecture in my head, and will post it on this blog when/if I successfully complete it.

[Later edit — one of my readers provided a proof, so now it’s a theorem. For his proof, see the first comment on this post.]

Triangle Investigations

I just spent a couple of hours playing with triangles, their centers, and related things, using Geometer’s Sketchpad. Much of what I found was already known, but some of the things I found may not be. It will take more research to sort out which is which.


This is where I started. My original triangle, ABC, has heavy black sides and a yellow interior. The sides of ΔABC are extended as black dashed lines. I constructed ΔABC’s three angle bisectors to find its incenter, then constructed perpendicular lines to each side of the triangle from the incenter to find the three “touchpoints,” called A’, B’, and C’, allowing me to construct the incircle of my starting triangle. I also wanted the three excircles, and they are centered at A”, B”, and C” — the three excenters of triangle ABC. These points are found by bisecting all exterior angles of the original triangle, and then looking for the three point of concurrency among themselves, and the three interior angle bisectors previously used to locate the incenter. The heavy green, large triangle is the triangle which connects the three excenters, and probably has an established name, but I couldn’t find it — so I’ll just be calling it the “green triangle.” It is not to be confused with the small triangle in the diagram’s center, which has red sides and a green interior. This small triangle is simply formed by connecting the touchpoints A’, B’, and C’, and has several names already:  the Gergonne triangle, the contact triangle, and the intouch triangle of ΔABC.

I set this up in such a way that I could move points A, B, and C around, and watch what happened to angle measures, segment lengths, areas, and area ratios. The picture above shows what happens near regularity, with all three interior angles very close to 60°. This is how I learned that certain area ratios are at a minimum when the triangle is equilateral — as you’ll see, they are all larger in subsequent pictures. At regularity, the green triangle has an area exactly four times that of the original triangle, which, in turn, has an area four times larger than the Gergonne triangle — and, away from regularity, these two area ratios remain equal to each other.

I also found that the green triangle’s incircle’s area is, at minimum, four times that of the original triangle’s incircle — but if you deviate from regularity even slightly, it can be seen that this “just above four” is not equal to the “just above four” area ratios described at the end of the last paragraph.

I also tried adding the areas of all three excircles, then dividing this sum by the area of the incircle. At regularity, this ratio is 27, so each excircle would have exactly nine times the area of an incircle for an equilateral triangle. From that, it follows that an equilateral triangle’s excircle’s radius is three times that of the radius of the incircle. These numbers are minimums; the get bigger if the triangle deviates from regularity in any way.


This second picture shows what happens when regularity is abandoned. I noticed that the green triangle’s incenter, the original triangle’s incenter, and a third point of concurrency I am calling the “x-point” (because I suspect it already has a name, but I don’t know what that name is) appeared as if they might be collinear — but might not be. I checked, and they are non-collinear. I then decided to investigate this x-point further. The x-point is the one point of concurrency of the three lines, one for each excircle, which contain that excircle’s center, as well as the point of tangency between that excircle and the original triangle.


To investigate this x-point further, I started locating other triangle centers, and also continued moving A, B, and C around. The picture above shows the Gergonne point added to the diagram. The Gergonne point is a point of concurrency of three lines — each being a line containing a triangle’s vertex, and the touchpoint opposite that vertex. As you can see, these points still refused to line up. I therefore decided to go all-out, and add many more triangle centers.


In this last picture, I’ve added the circumcenter (point of concurrence of the triangle’s sides’ three perpendicular bisectors), centroid (point of concurrence of the three medians), orthocenter (point of concurrence of the lines containing the triangle’s three altitudes), and several other things for which you can easily find definitions on Wikipedia — such as the 9-point circle, its center, the Nagel point, the Nagel line, and the Euler line. As you can see, the Euler Line (shown as a heavy orange line) passes through the orthocenter, the center of the nine-point circle, the centroid, and the circumcenter. It does not, however, contain the incenter. What does? The Nagel line (shown in heavy purple), for one thing, which also holds the Nagel point, as well as the centroid, where it intersects the Euler line.

Neither the Euler line nor the Nagel line contains the x-point I was investigating, though — but I found a line which does. It is shown in heavy grey, and passes through (in addition to the x-point) the circumcenter (where it intersects the Euler line) and the Gergonne point, which was completely unexpected. I suspect this line has already been discovered and named, but, until I find out what that name is, I’m calling it the “x-line.”

If anyone who reads this knows more about the x-point or x-line, such as their already-existing names (assuming they’ve been discovered before), please leave this information in a comment here.