This Euclidean construction was made using Geometer’s Sketchpad, and colored using that program, as well as MS-Paint.
Start with points A and B. Construct a circle, centered on A, with radius AB. Draw line AB, which intersects this circle at B and C. Construct a second line which is perpendicular to the first line, intersecting it at A. Let the two intersections of the circle and this second be named points D and E.
Bisect segment AB, and call its midpoint F. Construct a line containing D and F. If the circle’s radius is two, then AF = FB = 1, while AD = 2. By the Pythagorean Theorem applied to right triangle DAF, then, DF = sqrt(5). Construct a second circle centered on F, with radius DF. Construct point G where this circle intersects segment AC. It follows that FG, being another radius of this second circle, has a length of sqrt(5). This makes BG = FB + FG = 1 + sqrt(5).
Construct the line which is perpendicular to line AE and passes through point E. Next, construct the two lines perpendicular to line AB and passing through points G and B. These two lines intersect the first line described in this paragraph at two points: H (below G) and I (below B). ABIE is a square with edge length two, and both GBIH and AEHG are golden rectangles.
This particular tessellation is full of angles measuring 20 degrees, 40 degrees, and other angles which are not constructable using the traditional rules of Euclidean constructions. This is because this tessellation is based on a matrix which includes regular enneagons.
Steps of this construction:
- Use the green circles and blue lines to construct the yellow pentagon, along with its green inscribed pentagram.
- Construct the equilateral triangle shown in gray. This is needed to obtain a twelve degree angle. The triangle is needed for its sixty degree angle, because 72 – 60 = 12. (The 72 degree angle is found inside the pentagon.)
- Identify the twelve degree angle shown in bold. A twelve degree angle is needed because 360 / 30 = 12.
- Use the red circles to complete the thirty sides of the regular triacontagon, which is shown with bold black segments, inscribed inside a large, bold, red circle.
Because I did not start this blog until mid-2012, I sometimes encounter things I made before then, but have not yet posted here. I made this image in 2011, after reading that the ancient Greeks discovered how to combine the Euclidean constructions of the regular pentagon and the equilateral triangle, in order to create a construction for the regular pentadecagon. Having read this, I felt compelled to try this for myself, without researching further how the Greeks did it — and, as evidenced by the image above, I successfully figured it out, using the Euclidean tools embedded in a computer program I often use, Geometer’s Sketchpad.
What I did not do at that time was show the pentagon’s sides (so it is rather hard to find in the image above, but its vertices are there), nor record step-by-step instructions for the construction. For those who wish to try this themselves, I do have some advice: construct the pentagon before you construct the triangle, and not the other way around, and you are likely to find this puzzle easier to solve than it would be, if this polygon-order I recommend were reversed.
I also have two more hints to offer: 108º – 60º = 48º, and half of 48º is 24º. Noticing this was, as I recall, the key to cracking the puzzle.
There is more than one way to construct a golden rectangle using the Euclidean rules, but all the ones I have seen before use circles with irrational radii. This construction, which I believe to be new, does not use that shortcut, which helps explain its length. The cost of avoiding circles of irrational radius is decreased efficiency, as measured by the number of steps required for the entire construction.
In the diagram below, the distance between points A and B is set at one. All of the green circles have this radius, while the magenta circles have a radius exactly twice as long.
To make following the construction from the diagram above easier, I named the points in alphabetical order, as they appear, as the construction proceeds. The yellow rectangle is the resulting golden rectangle. The blue right triangle is what I used to get a segment with a length equal to the square root of five, which is a necessary step, given that this irrational number is part of the numerical definition of the exact value of the golden ratio (one-half of the sum of one and the square root of five). In order to make the hypotenuse have a length equal to the square root of five, by the Pythagorean Theorem, the two legs of this triangle have lengths of one and two.
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.
I have just completed my first construction of the regular heptadecagon — a construction that even the ancient Greeks were never able to figure out. They did figure out how to construct a regular pentadecagon (by combining the constructions for the regular pentagon and triangle), and I once replicated that discovery, meaning that I figured it out independently.
The regular heptadecagon construction, however, I did not figure out independently. I used instructions found here (http://www.mathpages.com/home/kmath487.htm), which built on the work of Carl Friedrich Gauss, who, in 1796, at the age of 19, became the first person in history to determine that such a construction is possible with the traditional Euclidean tools.
A word of warning, if you attempt to replicate this construction yourself: points M and G are merely close together, but are not in the same place. Point M is the center of the circle which passes through points D and V17, while point G is one of the two points of intersection of (1) the line passing through points O and V17, and (2) the circle centered at C, and passing through E.
Gauss (and other mathematicians, building on his work) also showed, later, that constructions are possible for regular polygons with 257 sides, as well as 65,537 sides. I might, someday, replicate the construction of the regular polygon with 257 sides.
A man named Johann Gustav Hermes once spent ten years completing a 200-page manuscript showing how to construct the regular polygon with 65,537 sides, and I believe he actually performed the construction, as well. I will not be constructing this polygon — ever. I will, however, figure out a proper name for it. Let’s see . . . it’s the heptakaitriacontakaipentacosioikaipentachilikaihexamyriagon. Try saying that five times in a row, quickly!
The Ancient Greeks figured out how to combine the Euclidean constructions of the regular pentagon and triangle to obtain constructions for the regular pentadecagon, which has central angles (between adjacent radii) of 360/15 = 24 degrees. Here’s an example, showing how this can be performed:
Also, it’s easy to construct an equilateral triangle, and then bisect an angle of it, to obtain a 30 degree angle.
The existence of angle difference identities in trigonometry is tied to the fact that you can subtract angles, on paper, with Euclidean constructions. Therefore, an angle of 24 degrees may be subtracted from a 30 degree angle to obtain a 6 degree angle. This can be bisected to get a 3 degree angle, and then bisected again to obtain a 1.5 degree angle, then a 0.75 degree angle, and so on.
However, a one degree angle is impossible to construct. Were this not the case, a 24 degree angle’s constructibility would imply that of the 23 degree angle, by subtraction of a one degree angle. After that, subtract three degrees more, and you have a 20 degree angle . . . and with that, you can construct a regular enneagon, also known an a nonagon. But we know — it has been proven — that regular enneagons have no valid Euclidean constructions. Therefore, one degree angles are also non-constructible, by reductio ad absurdam.
Carl Friedrich Gauss’s much more recent proof (1796; he was 19 years old) that a regular polygon of 17 sides can also be constructed — the first significant advance in this field since the time of the ancient Greeks — adds more constructible angles. Building on his work, other mathematicians have also shown that regular polygons with 257 and 65,537 sides can also be constructed, adding yet more constructible angles, but they are all for angles measuring fractional numbers of degrees, since none of these numbers are factors of 360, which equals (2³)(3²)(5). It’s also possible to combine these possible constructions to construct more regular polygons, as was shown above for the pentadecagon. For example, one can construct a regular pentagon with 51 sides, since 51 = (17)(3) — but, again, combinations of this type only lead to possible constructions of angles with measures which are fractional numbers of degrees. For angles with degree measures which are integers, it’s multiples of three — and that’s it.
[Note regarding images: the photograph of a compass at the top of this page was not taken by me, but simply found with a Google image-search. The pentadecagon-construction image, though, I did make, using both Geometer’s Sketchpad and MS-Paint.]