Five Pointed Star Reference Shapes
Figure 1: The Five Pointed Star
The Five Pointed Star has become a common symbol in Western Culture, most notably introduced on the flag of the United States. But it’s also a very interesting geometric shape which is worthy of further study.
In Sacred Geometry, one of the most fascinating concepts that intrigues us the most is the Golden Ratio, usually denoted by the Greek symbol “Phi” or $\varphi$. This ratio appears over and over again throughout nature, and the five pointed star contains numerous instances of it.
In Figure 1, notice that we’ve connected the lines of the star and color coded some of the interesting shapes that we’ll be discussing below. The top blue triangle is known as the “Golden Triangle”, and the green triangle is often known as the “Phi Triangle,” the “Divine Triangle,” or the “Golden Gnomon.”
The Golden Ratio
Before we discuss shapes involving Phi, we should first define what Phi means. Phi is the Golden Ratio, which is perhaps easiest to see in another shape known as the Golden Rectangle. The golden rectangle is a rectangle whose sides are in the golden ratio as shown in Figure 2, that is,
\begin{align}
\frac{a + b}{a} = \frac{a}{b} = \varphi
\end{align}
where $a$ is the width of the rectangle, $a + b$ is the length, and $\varphi$ or “Phi” is the golden ratio:
\begin{align}
\varphi = \frac{1 + \sqrt{5}}{2}
\end{align}
To see a more formal definition of the Golden Rectangle and a proof of the Phi derivation shown above, see the following article,
Department of Mathematics Golden Rectangle » University of Washington
Golden Rectangle Construction
Another nice way to see a derivation of Phi from the golden rectangle is from the construction method used by Euclid as shown in Figure 3.
We start with the square $ABDC$, and set $E$ as a midpoint between $A$ and $C$. Now we have the following:
\begin{align}
AE = EC = x
\end{align}
If we use $E$ as the center point to form an arch from point $B$, where the arch intersects a line extended from $CA$ completes the golden rectangle. Now we can draw the radius of this arch $BE$ and calculate its length as follows:
\begin{align}
BE^2 = AB^2 + AE^2
\end{align}
Substituting our values for $x$, this can be shown as,
\begin{align}
BE^2 = \left(2x\right)^2 + x^2 = 5x^2 \\\\
BE = EF = x\sqrt{5}
\end{align}
We can now see that rectangle $CFGD$ is golden since,
\begin{align}
\varphi = \frac{FC}{CD} = \frac{EF + CE}{CD} = \frac{x\left(\sqrt{5} + 1\right)}{2x} \\\\
\varphi = \frac{1}{2}\left(\sqrt{5} + 1\right)
\end{align}
Five Pointed Star Angles
By connecting the lines of a five pointed star, we can see that the center shape is actually a pentagon. So if we can find the interior and exterior angels of a pentagon, that will tell us everything we need to know about the angles of a five pointed star!
Interior Angles of a Pentagon
For any polygon, the number of interior angles is equal to the number of sides. So in a pentagon, there are 5 sides and 5 interior angles. And for any polygon, each interior angle can be calculated by the following formula:
\begin{align}
\frac{\left(\left(n – 2\right) \times 180^{\circ} \right)}{n}
\end{align}
where $n$ is the number of sides. For our pentagon $n = 5$, so this formula becomes:
\begin{align}
\frac{\left(\left(5 – 2\right) \times 180^{\circ} \right)}{5} = \frac{\left(3 \times 180^{\circ} \right)}{5} = 108^{\circ}
\end{align}
Exterior Angles of a Pentagon
In figure 2, we can see that all of the exterior angles of the pentagon will be the same.
\begin{align}
180^{\circ} \; – \; 108^{\circ} = 72^{\circ}
\end{align}
Another way to calculate the exterior angle of a pentagon is to observe that the sum of all exterior angles of a regular polygon are equal to $360^{\circ}$. So the exterior angle of a regular polygon is $360^{\circ} \div n$, where $n$ is the total number of sides in a polygon. So the formula for the exterior angles of our 5 sided pentagon becomes the following,
\begin{align}
\frac{360^{\circ} }{n} = \frac{360^{\circ} }{5} = 72^{\circ}
\end{align}
The Golden Triangle
From Figure 1 above, we’ve pulled out the top blue triangle ABC from the five pointed star to derive a formula showing its relationship to Phi.
Angles of the Golden Triangle
From Figure 4 above, we know that all of the exterior angles of a pentagon are $72^{\circ}$. So the base angles of the golden triangle in Figure 5 are $72^{\circ}$. For any triangle, all of the interior angles add up to $180^{\circ}$. So the apex angle of the golden triangle is given by,
\begin{align}
180^{\circ} \; – \; 2\left(72^{\circ} \right) = 36^{\circ}
\end{align}
Golden Triangle Relationship to Phi
Our claim is that a golden triangle is isosceles with an apex angle of 36 degrees, and the ratio of the long side over its base is the golden ratio.
In Figure 5, we can see that $AB = y$, and $AC = x + 1$.
But since $\bigtriangleup ABC$ is isosceles, $AB = AC$, and $\; y = x + 1$.
Also notice that $\bigtriangleup BCD$ is similar to $\bigtriangleup ABC$. So we can observe the following,
\begin{align}
\frac{y}{1} = \frac{1}{x} \\\\
x + 1 = \frac{1}{x} \\\\
x^2 + x = 1
\end{align}
Golden Triangle Relationship to Multiplicative Inverse
One very important side note here is that we can see a direct relationship between between the golden triangle and the Multiplicative Inverse in the $\; y = 1/x$ equation above. This also relates to our article about the Chestahedron Seven Sided Geometry, since that shape is formed by folding in three points of a five pointed star — so it contains three golden triangles. Then we have the “Painter’s Paradox” geometry known as Gabriel’s Horn which is formed by spinning the $y = 1/x$ graph around the $x$ axis. This has turned out to be a fascinating line of research which ties in with the creation of perfectly formed eggs and seed shapes, the Reciprocal Mandelbrot Set, and the Vesica Piscis. So we’ve written a new article featuring an Interactive Gabriel’s Horn App at the link below,
Completing the Square From Above
Using another old trick from geometry for an equation like $\; x^2 + x = 1$, we can complete the square as follows,
\begin{align}
x^2 + x + \frac{1}{4} = 1 + \frac{1}{4} \\\\
\left(x + \frac{1}{2}\right)^2 = \frac{5}{4} \\\\
\end{align}
Taking the square root of both sides will give us,
\begin{align}
x + \frac{1}{2} = \frac{\sqrt{5}}{2} \\\\
x = \frac{\sqrt{5}}{2} \; – \; \frac{1}{2} = \frac{\sqrt{5} \; – \; 1}{2} \\\\
\end{align}
Now if we go back and solve for $y = x + 1$, we have,
\begin{align}
y = \frac{\sqrt{5} \; – \;1}{2} + 1 \\\\
y = \frac{\sqrt{5} \; – \; 1 + 2}{2} = \frac{\sqrt{5} + 1}{2} = \varphi \\\\
\end{align}
Notice that we also have our final requirement for a Golden Triangle. The ratio of the long side over its base is the golden ratio,
\begin{align}
\frac{AB}{BC} = \frac{\sqrt{5} + 1}{2} = \varphi \\\\
\end{align}
Calculating the Height of a Golden Triangle
The golden triangle is a very key shape in the five pointed star configuration. So it would be nice to know the height of this triangle, indicated by the vertical line $h$ in Figure 6.
This would be a good time to mention that another interesting property of Phi is that it’s the only number for which adding one will produce its square, while subtracting one will produce its inverse. In other words, it can easily be shown that both of the following statements are true,
\begin{align}
\frac{1}{\varphi } = \; \varphi \; – \; 1 \\\\
\varphi^2 = \; \varphi + 1
\end{align}
Referring again to Figure 6, by the Pythagorean Theorem we can see that,
\begin{align}
\varphi^2 = \left(\frac{1}{2}\right)^2 + \; h^2 \\\\
h^2 \; = \; \varphi^2 \; – \; \frac{1}{4} \; = \; \left(\varphi + 1\right) \; – \; \frac{1}{4} \\\\
= \; \frac{4\varphi \; + \; 4 \; – \; 1}{4} = \frac{4\varphi \; + \; 3}{4} \\\\
\end{align}
At this point we can substitute for $\varphi$ and get,
\begin{align}
h^2 = \frac{4\left(\frac{1 \; + \; \sqrt{5}}{2}\right) + 3}{4} \\\\
= \left(\frac{1}{4}\right)\left(\frac{4 + 4\sqrt{5} + 6}{2}\right) = \frac{10 + 4\sqrt{5}}{8} \\\\
h^2 = \frac{5 + 2\sqrt{5}}{4}
\end{align}
Now we can take the square root of both sides and solve for $h$,
\begin{align}
h = \frac{\sqrt{5 + 2\sqrt{5}}}{2}
\end{align}
We can also get exact values for $\sin \left(72^{\circ} \right)$ and $\cos \left(18^{\circ} \right)$ from this by observing that,
\begin{align}
\sin \left(72^{\circ} \right) = \cos \left(18^{\circ} \right) = \frac{{h}}{\varphi} = \frac{\sqrt{5 + 2\sqrt{5}}}{1 + \sqrt{5}}
\end{align}
Now that we have a value for $h$, it’s easy to solve for the area of a golden triangle $A_{gt}$,
\begin{align}
A_{gt} = \frac{1}{2}bh = \frac{\sqrt{5 + 2\sqrt{5}}}{4} \\
\end{align}
The Divine Triangle
The Divine Triangle is an obtuse isosceles triangle which can be constructed from a regular pentagon, and this is illustrated as the two green triangles shown in Figure 7. This triangle is also known as the Golden Gnomon with angles $108^{\circ} \; 36^{\circ} \; 36^{\circ}$, and its ratio of side to base lengths is given by the following,
\begin{align}
\frac{1}{\varphi } = \; \varphi \; – \; 1 \\\\
\end{align}
In Figure 8, we’re also showing a second larger green triangle which is formed by joining any two tips of the five pointed star. We already know the side lengths of this triangle share common sides with two golden triangles. So these lengths are equal to Phi.
Since this larger green triangle also has the same angles as the divine triangle in the pentagon, this means it’s similar. To find the base, we can multiply all of the sides by Phi.
So the larger divine triangle in Figure 8 has side lengths equal to $\varphi$ and a base length of $\varphi^2$.
We can also see from Figure 8 that triangle $\bigtriangleup ABC$ forms another golden triangle with side lengths = $2\varphi + 1$ and a base length of $\varphi^2$.
Another way to evaluate this is by noticing the blue golden triangle seen in Figure 7 is similar. So we can multiply the base and side lengths of that triangle by $\varphi^2$ to get triangle $\bigtriangleup ABC$.
In that case, the side lengths will be $\varphi^3$ while the base is $\varphi^2$. Since we already know the side lengths of $\bigtriangleup ABC$ are $2\varphi + 1$, we can see that,
\begin{align}
\varphi^3 = 2\varphi + 1
\end{align}
This is easy to check by multiplying $\varphi$ through both sides of our equation for $\varphi^2 = \varphi + 1$,
\begin{align}
\varphi \varphi^2 = \left(\varphi + 1\right)\varphi = \varphi^2 + \varphi \\\\
\varphi^3 = (\varphi + 1) + \varphi = 2\varphi + 1
\end{align}
It’s also interesting that a pentagon with side lengths equal to one has lengths of $\varphi$ from one corner to either of its opposite corners. So the five pointed star is absolutely crawling with instances of the golden ratio, and so far we’ve only uncovered a small handful of them!
Calculating the Height of a Divine Triangle
There has been some dispute online about the height of a divine triangle, and it’s an interesting exercise to calculate an exact value for that number.
We can see from Figure 9 that the vertical line $h$ for the height divides the triangle in half, so the base of one side is $\varphi \div 2$. One way to calculate this is with Trigonometry since,
\begin{align}
\sin \left(36^{\circ} \right) = \frac{h}{1}
\end{align}
While it is possible to find an exact value with trig, it’s probably easier to use algebra. With the Pythagorean Theorem, we can find the value of $h$.
\begin{align}
1^2 = \left(\frac{\varphi }{2}\right)^2 + \; h^2 \\\\
1 = \frac{\varphi^2}{4} + \; h^2 \\\\
\end{align}
Now we can substitute $\varphi^2 = \; \varphi + 1$ and solve for $h^2$,
\begin{align}
h^2 = 1 \; – \left(\frac{\varphi + 1}{4}\right) \\\\
= \frac{4 \; – \; \varphi \; – \; 1}{4} = \frac{3 \; – \; \varphi }{4}
\end{align}
Next, we can substitute $\varphi = \left(\sqrt{5} + 1\right) \div 2$,
\begin{align}
h^2 = \frac{3 \; – \left(\frac{\sqrt{5} \; + \; 1}{2}\right)}{4} \\\\
= \frac{\frac{6 \; – \; \sqrt{5} \; – \; 1}{2}}{4} = \frac{5 \; – \; \sqrt{5}}{8}
\end{align}
Taking the square root of both sides and rationalizing the denominator will give us,
\begin{align}
h = \frac{\sqrt{5 \; – \; \sqrt{5}}}{\sqrt{8}}\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{4}\sqrt{5 \; – \; \sqrt{5}}
\end{align}
We also now have an exact value for $\sin \left(36^{\circ} \right)$ due to the fact that,
\begin{align}
h = \sin \left(36^{\circ} \right) = \frac{\sqrt{2}}{4}\sqrt{5 \; – \; \sqrt{5}}
\end{align}
While we’re at it, we can spot a very simple way to get an exact value for $\cos \left(36^{\circ} \right)$ as well,
\begin{align}
\cos \left(36^{\circ} \right) = \frac{\frac{\varphi }{2}}{1} = \frac{\varphi }{2} = \frac{1 + \sqrt{5}}{4}
\end{align}
The “Divine Triangle” should not be confused with the Kepler Triangle, which is another fascinating shape in its own right and we’ve written about it in a separate article.
Pentagonal Triangles
Some more interesting shapes to look at from the five pointed star are the triangles formed by joining the center point of the pentagon to its outer vertices. This results in dividing the pentagon into 5 isosceles and congruent pentagonal triangles as shown in Figure 10.
Since the central angle of a regular pentagon is $360^{\circ}$, the angle at each vertex per triangle would be,
\begin{align}
360^{\circ} \div 5 = 72^{\circ}
\end{align}
We know the interior angles of a pentagon are $108^{\circ}$. And since the pentagonal triangles are isosceles and congruent to each other, these interior angles of the pentagon will be bisected. So the base angles of the pentagonal triangles will be,
\begin{align}
108^{\circ} \div 2 = 54^{\circ}
\end{align}
We can see some interesting relationships between the angles of the pentagonal triangles, the golden triangles, the divine triangles, and obviously with the pentagon. But what’s really interesting is what happens when we stack two pentagonal triangles as shown in Figure 11.
By duplicating and inverting one pentagonal triangle and stacking it at the base of the first one, this gives the same result as following the same procedure with two divine triangles, i.e., duplicating, inverting, and stacking one at the base of the other!
In Figure 11 we can see the two divine triangles with bases oriented vertically, while at the same time we see the stacked pentagonal triangles with bases oriented horizontally.
Calculating the Height and Side Length of a Pentagonal Triangle
So far we only have the angles and base lengths of pentagonal triangles, but we don’t know the side lengths and heights. In Figure 12, we still need to calculate the height $h$ and side length $s$.
Another important feature worth recognizing is that a line extending from the side of a pentagonal triangle will be the perpendicular bisector of the opposite side of the pentagon. Where any two of these lines intersect will define the center point of the pentagon, and we should also know the length of this line if needed, given by $h + s$.
We can see from Figure 12 that the vertical line $h$ for the height divides the triangle in half. So since our pentagon has sides equal to one, half the base of one side forming a right triangle is equal to one half. In this case we could use a little help from Trigonometry since we don’t have side lengths yet. We also have an exact value for $\sin \left(36^{\circ} \right)$ from the divine triangle equations we derived earlier, so we can use that as well.
\begin{align}
\sin \left(36^{\circ}\right) = \frac{\frac{1}{2}}{s} = \frac{1}{2s}
\end{align}
We can solve for side length $s$ as follows,
\begin{align}
s = \frac{1}{2\sin \left(36^{\circ}\right)} = \frac{1}{\frac{\sqrt{2}}{4}\sqrt{5 \; – \; \sqrt{5}}}\left(\frac{1}{2}\right) \\\\
s = \frac{2}{\sqrt{2}\sqrt{5 \; – \; \sqrt{5}}} \\\\
\end{align}
Since we also have an exact value for $\cos \left(36^{\circ} \right)$ from the divine triangle equations, we can use that to solve for $h$,
\begin{align}
\cos \left(36^{\circ}\right) = \frac{h}{s} \\\\
h = s\cos \left(36^{\circ}\right) = \frac{2}{\sqrt{2}\sqrt{5 \; – \; \sqrt{5}}}\left(\frac{1 + \sqrt{5}}{4}\right) \\\\
h = \frac{1 + \sqrt{5}}{2\sqrt{2}\sqrt{5 \; – \; \sqrt{5}}} = \frac{1 + \sqrt{5}}{2\sqrt{10 \; – \; 2\sqrt{5}}}
\end{align}
Now that we have the height $h$, we can easily calculate the area of a pentagonal triangle $A_{pt}$,
\begin{align}
A_{pt} = \frac{1}{2}bh = \frac{1 + \sqrt{5}}{4\sqrt{10 \; – \; 2\sqrt{5}}} \\
\end{align}
While it’s important to get exact numbers for $h$ and $s$ from our original model, we can get more elegant numbers for pentagonal triangles by stacking the two divine triangles shown in Figure 9. If we use that model, we get side lengths $s = 1$ along with these values for $h$ and the $base$,
\begin{align}
h = \frac{\varphi }{2} = \frac{1 + \sqrt{5}}{4} \; = \; \cos \left(36^{\circ}\right) \\\\
base = \frac{\sqrt{2}}{2}\sqrt{5 \; – \; \sqrt{5}} \; = \; 2 \; \sin \left(36^{\circ}\right)
\end{align}
Conclusion
Our research into the Five Pointed Star is still ongoing, and it’s lead us to some fascinating discoveries lately. The first of these was noticing the Chestahedron’s significant relationship to the five pointed star, and this inspired us to develop a new math model that we’re calling the “Chestahedron Phi Model.” Virtually all of the math we’ve outlined for the five pointed star was referenced for calculating the Chestahedron’s spectacular golden ratio measurements and angles.
This lead to an even more recent discovery of the Star Fort Crystal Geometry which is constructed using a secondary five pointed star shape that you’ll find in the ancient star fort geometry. The axiomatic definition of this new form is the same as that of the Chestahedron, with only the star points being different. So we recommend reading the Chestahedron article first, since that’s where we’ve outlined all of the math for proving the legitmacy of these incredible new and never-before-seen structures.
We’ve also developed a new research page for the Heptagon Seven Pointed Star Geometry, and the math involved in this shape is every bit as interesting as the five pointed star!
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The Math Zone. “Five Pointed Star Reference Shapes.” From MathZone.io — A Modern Exploration of Ancient Mathematics. https://mathzone.io/five-pointed-star-reference-shapes/
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