# The Golden Ratio and Construction of the Pentagon

## Introduction

The geometric construction of the regular pentagon is an interesting problem. To be able to construct the pentagon, we need to know the angle measures of the vertices and be able to construct the vertex angle or the diagonlas of the pentagon.

Coincidentally, the pentagon is related to the Golden Ratio and the 36-72-72 triangle. We will first find the trigonometric values of the 36-72-72 triangle before proceeding to construct the pentagon.

## The Pentagon

In a regular pentagon - one in which all the sides and angles are equal - each vertex of the pentagon has an angle measure of 108°. Since there are 5 vertices, the sum of all the angles is 540°. The formula to determine the sum of the angles, *S*, of any regular polygon is \( S = 180n - 360 \), where *n* is the number of sides. Therefore, if we divide by *n*, we get the angle measure of 1 vertex, *V*: \( V = 180 - \frac{360}{n} \). For a pentagon, *n* = 5. Therefore, the angle measure at 1 vertex is \( V = 180 - \frac{360}{5} = 108 \).

The 36-72-72 isosceles triangle is created when we draw the diagonals of the pentagon. This is the special isosceles triangle because that is associated with the Golden Ratio.

## The Golden Ratio

Figure 1 shows the 36-72-72 isosceles triangle. If we bisect one of the obtuse angles, which are equal to 72°, we get a similar smaller triangle, which is isosceles by similarity. If we let *x* be the length of the base, then we can determine the lengths of the legs by similar triangles.

By similar triangles, we have the following relationship:

(i) \( \frac{x}{y-x} = \frac{y}{x} \)

(ii) \( x^2 = y^2 - xy \)

(iii) \( y^2 - xy - x^2 =0 \)

(iv) \( y = \frac{x \pm \sqrt{x^2 + 4x^2}}{2} \) (Solving the quadratic equation for *y*)

(v) \( y = \frac{x \pm x\sqrt{5}}{2} \)

(vi) \( y = \frac{1 + \sqrt{5}}{2}x \) (Rejecting the negative length and factoring *x*)

The value \( \frac{1 + \sqrt{5}}{2} \) is the Golden Ratio. In the 36-72-72 triangle, the ratio of the side to the base (*y* to *x*) is the Golden Ratio.

If we solve for *x* instead, we get \( x = \frac{-1 + \sqrt{5}}{2}y \). The value \( \frac{-1 + \sqrt{5}}{2} \) is the inverse of the Golden Ratio.

### The Trigonometric Values

Although not necessary for the construction, we can determine the sine and cosine of 36°. We will use the Law of Cosines to find the cosine first. The Law of Cosines gives us the relationship: \(c^2 = a^2 + b^2 - 2ab\cos C\). We will let *a* = *b* be the legs and *c* be the base of the triangle.

(i) \( x^2 = \left( \frac{1+\sqrt{5}}{2}x \right)^2 + \left( \frac{1+\sqrt{5}}{2}x \right)^2 - 2\left( \frac{1+\sqrt{5}}{2}x \right)\left( \frac{1+\sqrt{5}}{2}x \right)\cos 36^{\circ}\)

(ii) \( 1 = \frac{1}{4}(6+2\sqrt{5}) + \frac{1}{4}(6+2\sqrt{5}) - 2\cdot\frac{1}{4}(6+2\sqrt{5})\cos 36^{\circ} \)

(iii) \( 2 = 5+2\sqrt{5} - (6+2\sqrt{5})\cos 36^{\circ} \)

(iv) \( \cos 36^{\circ} = \frac{4+2\sqrt{5}}{6+2\sqrt{5}} \)

(v) \( \cos 36^{\circ} = \frac{4+4\sqrt{5}}{16} = \frac{1+\sqrt{5}}{4} \) (Rationalize the denominator.)

The cosine of 36° is half the Golden Ratio. Using the identity \( \cos^{2}x + \sin^{2}x = 1\), we can find sine to be: \( \sin 36^{\circ} = \frac{\sqrt{10-2\sqrt{5}}}{4} \).

We can also find the values for 18°, 54°, and 72° which are all multiples or divisors of 36°. The exact values are listed in the table here: Angle Meausres and Area.

### Construction of the Golden Ratio

We must be able to construct the 36-72-72 triangle to construct the pentagon. In order to do that, we need to construct the Golden Ratio.

Here are the steps to construct the Golden Ratio in brief before we contruct the pentagon:

- Construct a rectangle with sides of lengths 1 and 2.
- Construct a diagonal of this rectangle. The diagonal will have a length of \(\sqrt{5}\).
- Extend the diagonal by 1 unit, so that total length is \( 1 + \sqrt{5} \).
- Find the midpoint so the segment is bisected into lengths \(\frac{1+\sqrt{5}}{2} \).

We will now go through the construction steps. Note that instead of using actual compass and straight edge, I will be using Geogebra for the images. Therefore, to construct equivalent lengths, I will draw full circles. If doing this on paper, we would actually draw partial circles or arcs of the circle instead of the full circle.

Also, some basic construction steps are not shown in the images. It is implied that the reader is familiar with the basics of geometric constructions.

## Construction Steps

**Step Set 1.** In the first set of steps, construct a rectangle with sides of length 1 and 2. Then, construct one diagonal of the rectangle, OA, which will have a length of \(\sqrt{5}\). Then, at one corner (A in the image), extend the diagonal by 1 unit using the side of the rectangle with length 1 unit. We will call that point B. Therefore, OB will have length \(1+\sqrt{5}\). Lastly, bisect segment OB by finding its midpoint. We will call the midpoint C. Therefore, OC will have a length of \(\frac{1+\sqrt{5}}{2}\), the Golden Ratio.

**Step Set 2:** In the next set of steps, draw a circle with radius OC and center at O. Then draw another circle with radius of 1 unit and center at C. You can use the side AV_{1} for the length of 1. The two circles will cross at point D. OC and OD will be the legs of the 36-72-72 triangle and CD will be the base.

**Step Set 3:** Next, draw the segments DC and OD. Then, draw circle with radius of DC and center at D. This circle will intersect side OC at E, making DC and DE equal. Alternatively, you can simply bisect the angle ODC. Either way, you get two similar 36-72-72 triangles as in Figures 1 and 2.

**Step Set 4:** Next, draw a circle with radius of DO and center at D. The circle will intersect the line containing the segment ED at F. Then connect F and H, the intersection of circle D with the side OD. It will intersect D at G. FG will be parallel to CD.

After this step, the vertices of the pentagon should be apparent. And we can actually draw the 5-pointed star.

**Step Set 5:** Lastly, to finish the construction, we simply connect segments OF, FC, DG, and GO. I have shown the angle measures of all the vertices using Geogebra, confirming we have a regular pentagon. Step Set 5 shows the pentagon with the original rectangle we started with.