# New very simple golden ratio construction incorporating a triangle, square, and pentagon all with sides of equal length. Is there any prior art?

Consider three regular polygons with 3, 4, and 5 sides wherein all the polygons have sides of equal length X throughout, as illustrated below. The ratio of the red line segment a to the blue line segment t is the golden ratio PHI or 1.6181....

The construction is made as follows.

1. Begin with a triangle, square, and pentagon where all the sides are X.

2. Arrange the polygons so that the square rests upon the triangle and the pentagon rests upon the square as drawn.

3. Draw a line segment from the very top point H to the far-away bottom point A.

4. The side of the square will then cut the line segment in the golden section at point I, so that the ratio of the red segment a to the blue segment t is the golden ratio PHI.

As always, geometric and algebraic proofs are welcome! And too, if anyone has seen any prior art relating to this particular golden ratio construct, please do share! I have searched long and hard and could find none.

Is there any pertinent prior art?

• I do not believe there is any prior art for this. – PhysicsTheory May 25 '16 at 1:15
• How do you know your result is correct? – Joe May 25 '16 at 1:23
• @Joe I drew it in Geogebra and Geogebra gave me PHI to fifteen decimal places. :) – Astrophysics Math May 25 '16 at 1:35
• This has to stop. – Pedro Tamaroff May 25 '16 at 4:21
• I have no doubt that someone has published something at some time that used the fact that $\sin(\pi/6):\sin(\pi/10)$ is the golden ratio. But the argument over how profoundly original these constructions are misses the point of how surprising, beautiful, and instructive they are. Those latter qualities are valuable too and (I think) are why someone suggested submitting the earlier ones to The American Mathematical Monthly (published by the MAA). I think this one also has those qualities, and Mathematics Magazine (also from MAA) would be another likely outlet. – David K May 25 '16 at 14:16

With $A$ the origin, rotate everything so that $AG$ lies on the $x$-axis. Drop a perpendicular from $H$ to meet the $x$-axis at $J$. Extend segment $CE$ to meet segment $HJ$ at point $K$. The key to the proof is the fact that that angle $\angle HEK$ has measure $18^\circ$, since each interior angle in a regular pentagon is $108^\circ$. Picture (not to scale): For simplicity of calculation, assume all edges in the regular $n$-gons have length 2. Look at triangle $AJH$ and note that segments $IK$ and $AJ$ are parallel. Since segment $KJ$ has length 1, the side-splitter theorem gives $$\frac at=\frac 1h,\tag1$$ i.e. $\frac1h$ is the ratio of segment $a$ to segment $t$. But $h=2\sin18^\circ$, and $\sin 18^\circ=(\sqrt 5-1)/4$ so $$\frac1h=\frac1{2\sin 18^\circ}=\frac2{\sqrt 5-1},$$ the golden ratio.

• thanks so much @grand_chat for the fine proof! illustrations are always welcome too! :) – Astrophysics Math May 25 '16 at 1:50

Migrating my thoughts from comments below my answer to another question ...

The square and triangle have little to do with the appearance of the golden ratio here. The essence of the construction is this:

The figure has two salient characteristics:

1. $O$ lies on the perpendicular bisector of edge $\overline{RS}$ of the regular pentagon; and,
2. $B$ lies on the perpendicular to $\overline{RS}$ through $S$.

This information is all we need to find $\phi$.

Since $\triangle OAM \sim \triangle OBN$, we have $$\frac{a}{b} = \frac{\frac{1}{2}|\overline{AQ}|}{\frac{1}{2}|\overline{RS}|} = \frac{\text{diagonal of regular pentagon}}{\text{edge of regular pentagon}} = \phi = 1.618\dots$$

(leveraging a well-known property of regular pentagons), and then the golden-ness of the $a/b$ ratio passes to the target $b/(a-b)$ ratio, because that's exactly how the golden ratio works. :) $\qquad\square$

As I write in my referenced comments:

$$\text{[T]he construction is } \textit{sneaky} \text{, in that it makes one think} \\ \text{ that the square and triangle matter, when they don't.}$$

Sure, the even-ness and odd-ness of the square and triangle's respective edge counts naturally guarantee that the target segment has an endpoint on the perpendicular bisector, as per "salient characteristic" $(1)$; and, having the square appropriately sized and positioned makes the perpendicular in $(2)$ a natural part of the construction, as well. (Plus, as OP mentions, the $3$-$4$-$5$ progression has some appeal. I'll also say that I like the sneaky aspect. :) Apart from that, nothing about the particular geometry of those elements comes into play: the triangle could be merely isosceles; the square could be merely rectangular; and/or, any number of additional figures could join them (or replace them) in the chain. It just doesn't matter; so long as $(1)$ and $(2)$ hold, the construction yields the golden ratio.

For example, here's a comparably-sneaky construction that might give the (false) impression that the $5$-$6$-$7$ progression in the component edge counts has some special connection with $\phi$:

• Thanks @Blue! But in your 5,6,7 construction, you must draw a line in the hexagon that is not naturally present in the polygons themselves. Compared with my simple 3,4,5 construction, you need infinitely times as many extra lines. And thus I would argue that the square and rectangle matter very, very much. Sure, one could perhaps use a rectangle and isosceles triangle, but then the simple condition that all sides of all the polygons must be of the same length would be violated. :) I maintain that the 3,4,5 progression is the only one which naturally presents PHI without extra lines. :) – Astrophysics Math May 26 '16 at 15:10

Inspired by a comment posted by OP and by the answer given by @Blue, here are three other ways to combine the regular pentagon, square, and equilateral triangle of equal sides in such a way as to divide a segment between vertices in the ratio $\phi:1$.

The basic idea is simply to place the square so that it provides part of the perpendicular to one edge of the pentagon at one end of that edge, then place the triangle so that one vertex is on the perpendicular bisector of the same edge of the pentagon and so that a segment from that vertex of the triangle to a suitable vertex of the pentagon intersects an edge of the square; the segment so constructed will be cut in the ratio $\phi:1$ by the edge of the square. I rather like the last version, which constructs all three polygons on the same edge, "facing" the same direction.

Just for good measure, here is a construction without the triangle, just the square and pentagon. The square has been placed so one edge is collinear with an edge of the pentagon and overlapping that edge, but another edge of the square passes through a different vertext of the pentagon. This is just another "sneaky" way of using the fact that the ratio of the diagonal of a pentagon to its edge is $\phi:1$. The only purpose of the square here is to mark off a segment equal to the pentagon's edge along the pentagon's diagonal, something that could be done more simply by a circle using one edge of the pentagon as one of its radii.

• Nice @DavidK! Also, might there be any other way to distribute the shapes so that they are all exterior to one another and the golden ratio naturally emerges, as I did in the first construction? – Astrophysics Math May 28 '16 at 18:54
• Since the first three figures basically just shift the polygons left or right following the ideas in @Blue's answer, there isn't a lot to play with. I suspect these may be the only four arrangements of the three figures following that pattern. In order to avoid overlapping figures, I think one would have to come up with a pattern that used a different way to extract the ratio from the pentagon. – David K May 28 '16 at 19:03
• As the Pythagoreans well knew (even before PHI had been defined by Euclid) the pentagon and all its associated diagonals and internal shapes are so overflowing with golden ratios and golden harmonies, that we should probably not be too surprised to find golden ratios inside a pentagon based on polygons with the same size lengths as the sides of said pentagon. "Low hanging fruit," I believe is what they say. :) – Astrophysics Math May 28 '16 at 19:12
• @David K Just stretching it into 3D...Referring to third figure, if the equilateral triangle and pentagon (EP) are fixed on a common fulcrum, square side rotated until it cuts an EPLine (without skew).. may be $\phi$ is in there too? – Narasimham May 28 '16 at 19:35

This is not a construction of the golden ratio.

It is division of a segment into parts whose ratio is golden, but the segment is incommensurable with the rest of the figure, so it is not a construction in the classical sense. It also assumes as given a regular pentagon, which is practically the same thing as the golden ratio. This is why in the figure can find the golden number without using a compass. What the diagram describes is rather a "relative construction" based on some starting figure whose construction is not specified.

Given a square and regular pentagon with some segment in common (side, inradius, circumradius) the golden ratio can be always squeezed out of the diagram eventually, because the ratio of sides is $a + b \sqrt{5}$.

The usual Euclidean meaning of "construction" of a number is to exhibit it as a length when another segment has been specified as the unit of length. I would call this diagram an "appearance" of the ratio in a diagram based on a pentagon.

• I would argue that this is a construction of the golden ratio. It is a unique construction with interesting properties, but it is, nonetheless, a construction of the golden ratio. You write, "This is not a construction of the golden ratio, but of two segments with that ratio of lengths." If you can provide us with a golden ratio construction that does not include, " two segments with that ratio of lengths," we would very much like to see it! Even Euclid used two segments to describe the golden ratio. :) – Astrophysics Math May 28 '16 at 20:25
• you writes, "The usual Euclidean meaning of "construction" of a number is to exhibit it as a length when another segment has been specified as the unit of length. I would call this diagram an "appearance" of the ratio in a diagram based on a pentagon." Yes! I use a number X which represents the sides of all the polygons. When the polygons are arranged as shown, the golden ratio emerges as shown. There are other ways to construct the golden ratio, but that does not mean that this is not "a golden ratio construction." It is a unique construction with the three simplest polygons. – Astrophysics Math May 28 '16 at 22:46
• Yes. But the golden ratio does not emerge as a length $\phi X$ that is golden-ratio times the length of X. You construct a segment incommensurable with X that is divided into two parts whose ratio is the golden number. Your observation is a very nice one but it is something different from a construction (in the absolute sense). It is a relative construction as I mentioned. The classical construction game has particular rules, which are not sacred, but are different from the ones that you followed. – zyx May 29 '16 at 0:20
• thanks for your insights @zyx !! :) i am actually amazed by the sometimes mysterious definitions of axioms, postulates, scoliums, theorems, theories, laws, principles, and more! :) for instance we have kepler's laws and newton's laws. but then einstein's general relativity, which supersedes both, is only called "a theory." thanks again and any more information on the particular use and definition of the word "construction" would be great! :) – Astrophysics Math May 29 '16 at 0:58