This tessellation can be viewed in at least two ways: it can be seen as being composed of overlapping octagons which are equilateral, but not equiangular — or it can be viewed as a periodically-repeating pattern of golden gnomons, as well as golden triangles of two different sizes. Both golden triangles and golden gnomons are isosceles triangles with sides in the golden ratio, but golden triangles are acute, while golden gnomons are obtuse.
The golden ratio, also known as φ, has a value of [1 + sqrt(5)]/2, or ~1.61803. It is associated with a great many figures in geometry, and also appears in numerous other contexts. The most well-known relationship between a geometric figure and the golden ratio is the golden rectangle, which has a length:width ratio equal to the golden ratio. An interesting property of the golden rectangle is that, if a square is removed from it, the remaining portion is simply a smaller golden rectangle — and this process can be continued without limit.
While the golden ratio is related to many polyhedra, this relationship does not always involve golden rectangles, but sometimes it does. For example, it is possible to modify a rhombicosidodecahedron, by replacing that figure’s squares with golden rectangles (with the longest side adjacent to the triangles, not the pentagons), to obtain a “Zomeball” — the node which is at the heart of the Zometool ball-and-stick modeling system for polyhedra, and other phenomena. The entire Zome system is based on the golden ratio. Zome kits are available for purchase at http://www.zometool.com, and this image of a Zomeball was found at http://www.graphics.rwth-aachen.de/media/resource_images/zomeball.png.
In some cases, the relationship between a golden rectangle, and a polyhedron, is more subtle. For example, consider three mutually-perpendicular golden rectangles, each with the same center:
While this is not, itself, a polyhedron, it is possible to create a polyhedron from it, by creating its convex hull. A convex hull is simply the smallest convex polyhedron which can contain a given figure in space. For the three golden rectangles above, the convex hull is the icosahedron, one of the Platonic solids:
In addition to the golden rectangle, there are also other quadrilaterals related to the golden ratio. For example, a figure known as a golden rhombus is formed by simply connecting the midpoints of the sides of a golden rectangle. The resulting rhombus has diagonals which are in the golden ratio.
One of the Archimedean solids, the icosidodecahedron, has a dual called the rhombic triacontahedron. The rhombic triacontahedron has thirty faces, and all of them are golden rhombi.
There are also other polyhedra which have golden rhombi for faces. One of them, called the rhombic hexacontahedron (or “hexecontahedron,” in some sources), is actually the 26th stellation of the rhombic triacontahedron, itself. It has sixty faces, all of which are golden rhombi.
Other quadrilaterals related to the golden ratio can be formed by reflecting the golden triangle and golden gmonon (described in the post right before this one) across each of their bases, to form two other types of rhombus.
In these two rhombi, the golden ratio shows up as the side-to-short-diagonal ratio (in the case of the 36-144-36-144 rhombus), and the long-diagonal-to-side ratio (in the case of the 72-108-72-108 rhombus). These two rhombi have a special property: together, they can tile a plane in a pattern which never repeats itself, but, despite this, can be continued indefinitely. This “aperiodic tiling” was discovered by Roger Penrose, a physicist and mathematician. The image below, showing part of such an aperiodic tiling, was found at https://en.wikipedia.org/wiki/Penrose_tiling.
There are also at least two other quadrilaterals related to the golden ratio, and they are also formed from the golden triangle and the golden gnomon. The procedure for making these figures, which could be called the “golden kite” and the “golden dart,” is similar to the one for making the rhombi for the Penrose tiling above, but has one difference: the two triangles are each reflected over a leg, rather than a base.
In the case of this kite and dart, it is the longer and shorter edges, in each case, which are in the golden ratio — just as is the case with the golden rectangle. Another discovery of Roger Penrose is that these two figures, also, can be used to form aperiodic tilings of the plane, as seen in this image from http://www.math.uni-bielefeld.de/~gaehler/tilings/kitedart.html.
There is yet another quadrilateral which has strong connections to the golden ratio. I call it the golden trapezoid, and this shows how it can be made from a golden rectangle, and how it can be broken down into golden triangles and golden gnomons. However, I have not yet found an interesting polyhedron, not tiling pattern, based on golden trapezoids — but I have not finished my search, either.
[Image credits: see above for the sources of the pictures of the two Penrose tilings, as well as the Zomeball, shown in this post. Other “flat,” nonmoving pictures I created myself, using Geometer’s Sketchpad and MS-Paint. The rotating images, however, were created using a program called Stella 4d, which is available at http://www.software3d.com/Stella.php.]
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.]
The most well-known method for constructing a regular pentagon causes it to be inscribed in a given circle with a known center. It’s easy to find, on-line or in textbooks, and I will not be reproducing that construction here.
Because the diagonals and sides of a regular pentagon are in the golden ratio, it is also not difficult to construct a golden rectangle, and then use it to construct a regular pentagon. Given the ease of finding instructions for constructing a golden rectangle elsewhere, though, I’m not posting that construction here, either. This one begins with two distinct points, A and B, and the segment which connects them. The goal here is to start with only that, and then construct a regular pentagon, with sides of length AB, using only the methods permitted for Euclidean constructions.
After segment AB has been drawn, the next step is to continue it, in both directions, to form line AB. Next, construct two circles, each with radius AB: one centered on A, and the other centered on B. These circles must intersect at two points, labeled C and D, as shown above. The next step is to construct line CD, and label, as point E, the intersection of lines AB and CD. This is the well-known procedure for constructing the perpendicular bisector of a segment, so E is the midpoint of segment AB, and the four angles with vertices at E are each right angles.
Next, construct a circle, centered on E, with radius AE. Label, as point F, the intersection of this circle with segment CE. After than, construct a circle, centered on B, with radius BE. This circle intersects line AB at two points, one of which is already labeled E. Label, as point G, the other such intersection. At this time, because they are all radii of equal-sized circles, the following lengths are equal: AE, BE, BG, and EF. Segment EG, however, is twice as long as any of these shorter and equal-length segments, since it is a straight segment formed by combining non-overlapping segments BE and BG. Since the same can be said for segment AB as well — its length equals AE + BE, or 2AE by substitution and addition — it follows that AB = EG. The next step is to construct line FG, which necessarily includes segment FG. Since segment FG is the hypotenuse of right triangle EFG, which has one leg (segment EG) which is twice as long as the other leg (segment EF), it follows, from the Pythagorean Theorem, that the distance FG is equal to the distance EF times the square root of five. Next, construct a circle which is centered at F, and has segment EF for a radius. Label the intersections of line FG, and this circle, as points H and K, as shown above, with K being the closer of the two to point G. Now there are two new segments, FH and FK, which, because they are radii of the same circle as segment EF, are each equal in length to segment EF, as well as to segments AE, BE, and GB, each of which were already shown to be equal in length to EF, and half as long as either of the equal-length segments EG and AB.
The next step is to construct a circle, centered at G, with radius EG. After that, consider the collinear and adjacent segments FH and FG, which, together, compose segment GH. FH is one of the several segments known to be half as long as AB, and FG’s length is already known to be equal to this shorter length (that of segments EF, FH, AE, etc.) times the square root of five. By the segment addition postulate, then, segment GH’s length is equal to the length of one of the shorter segments AE, EF, etc., times the sum of one and the square root of five. This length, GH, will be used as the diagonal-length of the regular pentagon, because GH’s length, and the edge length originally provided (AB), are in the golden ratio, since segment AB is twice as long as segment EF, and GH is equal to EF times the sum of one and the square root of five. GH is the radius of the next circle to construct, and that circle should be centered at H. This circle, the largest yet constructed, intersects the radius-EG circle centered at G at two points. One of these intersection-points, on the same side of line AB as right triangle EFG, need not be labeled, but the other one, on the other side of line AB, should be labeled as point J, as shown above. Segments GJ and EG, then, have equal lengths, for they are radii of the same circle — and the distance EG has already been shown to be equal to the distance AB, so GJ = AB as well. The segments GJ and AB are highlighted in pink in the diagram above, with their lengths shown, in order to provide additional evidence that their lengths are equal.
In the diagram above, segment GJ is shown in bold, because it is a side of the pentagon being constructed. Segment HJ is then constructed, and is also shown in bold, for it is a diagonal of that same pentagon, having the correct length by virtue of being a radius of the same circle as segment GH, which was already shown to have the correct length for a diagonal of the pentagon. The next step is to locate point L, which will be a vertex of the pentagon. To do that, construct a circle of radius HK, centered on H. Because segment HK is formed by the non-overlapping, adjacent, and collinear segments FH and FK, each of which has the same length as AE or AB, it follows that HK, the radius of the last circle constructed, is equal in length to both AB and GJ, with this distance being the side-length of the pentagon being constructed.
Next, construct a circle which has radius GH (the diagonal-length of the pentagon), and is centered at G. This circle, and the already-constructed circle with radius HK which is centered at H, intersect at two points. Label one of these points of intersection (the one on the same side of line AB as point J) as point L, as shown above, and then construct pentagon-sides HL and JL, also shown in bold in the figure above. (Also shown: measured lengths of all segments rendered in bold thus far, as well as AB.) Of these last two circle-intersection points, one has now been labeled as point L. The next step is to label the other such intersection as point M, as shown in the diagram below.
Once point M is located, the remaining sides and diagonals of regular pentagon GJLHM can be constructed, as shown above, in bold, simply by connecting each remaining unconnected pair of pentagon-vertices with a segment. The lengths of all bold segments, plus the original segment AB, are also shown above. As you can see, all five sides have length equal to the length of segment AB, while all five diagonals have a length greater than that of AB, but one which is constant throughout the set of diagonals.
In the diagram above, no new objects have been constructed using the Euclidean tools, but the pentagon has been given color, and a check of the diagonal-length to side-length ratio has been performed. As you can see, in the uppermost calculation, the decimal approximation for this distance-ratio is given as 1.61803. The same result is obtained, in the second calculation shown, when the sum of one, and the square root of five, is divided by two — and, since this is the definition of the golden ratio, obtaining the same result, for both of these calculations, provides supporting evidence for the validity of this construction.
To perform one more test of this construction — one only possible with a virtual compass and straightedge, such as Geometer’s Sketchpad, the one I used here — points A and/or B can be moved around. I chose to move them both, and I also increased the distance between them. As you can see, this increased all the side and diagonal lengths, as well as the length of the original segment, AB. However, segment AB still has the same length as all five pentagon sides. Also, all five diagonals, while they do have a longer length than any side, still have lengths equal to each other. Also, the diagonal-length to side-length ratio remains constant, at the value of the golden ratio, even though the actual distances all changed when A and B were moved. The construction, and its explanation, are now complete.
As you can see, this can be continued indefinitely from the center.
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).
Alternating sides of these hexagons have lengths in the Golden Ratio. I created this as a faceting of the rhombicosidodecahedron. It also has faces which are star pentagons and triangles, but those faces are hidden in this view.
(Created using software from www.software3d.com/stella.php)
The Platonic Icosahedron has twenty faces which are equilateral triangles. In the Golden Icosahedron, twelve of those triangles (the yellow ones) have been replaced by acute, isosceles triangles with a leg:base ratio which is the Golden Ratio.
To try the software I used to make this, just visit http://www.software3d.com/stella.php.
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