# The Pentagon Files

## Introduction

The pentagon is a special polygon, partly because a regular pentagon is related to the Golden Ratio. The pentagon has 5 diagonals and the diagonals make a 5-pointed star, the shape that all kids learn to draw. Triangles have no diagonals. Squares have 2 diagonals and one cannot make any stars with a square. One cannot draw a continuous 6-pointed star with a hexagon without lifting the pencil. A 6-pointed star is two equilateral triangles drawn independently. Therefore, the pentagon is special in this way.

## Diagonals of a Polygon

Let’s make a ‘tangent’ from this topic and talk about diagonals. How many diagonals does a polygon with *n* sides have? Let’s think of the diagonals as lines drawn between 2 points or vertices of a polygon. There are *n* vertices to choose from and we have to choose 2 vertices. Therefore, this is a combinations problem. The number of lines that can be drawn with *n* vertices taking 2 vertices at a time is: \(C_2^n = \frac{n!}{2!(n-2)!}\), where *C* is the combination function. This will give us the sides of the polygon also, so we need to subtract *n* from this result. So our actual formula is \(C_2^n - n = \frac{n!}{2!(n-2)!} - n\).

Expanding this, we get \(\frac{n!}{2!(n-2)!} - n \) = \( \frac{n(n-1)(n-2)!}{2!(n-2)!} - n \) = \( \frac{n^2-n-2n}{2} = \frac{n(n-3)}{2} \).

A polygon with *n* number of sides has \(\frac{n(n-3)}{2}\) number of diagonals.

A pentagon, with *n* = 5, has \(\frac{5\cdot 2}{2} = 5 \) diagonals. A square with *n* = 4, has \(\frac{4\cdot 1}{2} = 2 \) diagonals.

## Back to Pentagons

The ratio of the diagonal of a pentagon to its side is the Golden Ratio, *φ*: \(\phi = \frac{1+\sqrt{5}}{2} \).

We will find some of the measurements related to the pentagon.

## The Area

The area of a regular pentagon with side length of *s* is \(\dfrac{\sqrt{25+10\sqrt{5}}}{4}s^2 \). The approximate area is 1.72*s*².

To prove the area formula, we need to find the measure of segment OF = *a*. The segment OF is referred to as the apothem. The area of triangle AOB is 1/5 of the pentagon. The area of AOB is \(\frac{1}{2}sa\).

Using the tangent relationship, we can calculate the apothem to be \(\tan 54^{\circ} = \dfrac{a}{s/2} = \dfrac{2a}{s}\). The tangent of 54° is \(\frac{\sqrt{25+10\sqrt{5}}}{5}\), which is given in the reference table here: Areas and Angle Measures. Therefore, \(a = \frac{\sqrt{25+10\sqrt{5}}}{10}s \). Therefore, the area of the pentagon, *A*, is 5 times area of AOB:

(i) \(A = 5\cdot\frac{1}{2}\cdot s \cdot \frac{\sqrt{25+10\sqrt{5}}}{10}s \)

(ii) \(A = \frac{\sqrt{25+10\sqrt{5}}}{4}s^2 \)

## The Inradius

In Figure 1, you may have noticed that the apothem is also the “inradius” of the pentagon, which is the circle that is tangent to all 5 sides of the pentagon at the midpoint of the sides. We already derived the formula for the inradius in terms of the side of the pentagon: \(a = \frac{\sqrt{25+10\sqrt{5}}}{10}s \) ≈ \(0.688s\).

### The Area in Terms of the Apothem

If the apothem or the inradius is known, we can derive the area of the pentagon based on *a* instead. Solving for *s* in the apothem-to-side relationship: \(s = \frac{10}{\sqrt{25+10\sqrt{5}}}a\).

(i) \(A = \frac{\sqrt{25+10\sqrt{5}}}{4}s^2 \)

(ii) \(A = \frac{\sqrt{25+10\sqrt{5}}}{4}\left( \frac{10}{\sqrt{25+10\sqrt{5}}}a \right)^2 \)

(iii) \(A = \frac{\sqrt{25+10\sqrt{5}}}{4}\cdot \frac{100}{25+10\sqrt{5}}a^2 \)

(iv) \(A = \frac{25}{\sqrt{25+10\sqrt{5}}}a^2 \)

Rationalizing the denominator, the area is \(A = 5\sqrt{5-2\sqrt{5}}\cdot a^2 \) ≈ \(3.63a^2\).

## The Circumradius

The circumradius is the radius of the circle that passes through the 5 vertices of the pentagon.

From triangle AFO, we have the following relationship: \(\cos 54^{\circ} = \frac{s/2}{r}\) and \(r = \frac{s}{2\cos 54^{\circ}} = \frac{1}{2}(\sec 54^{\circ})s\). From the references table, \(\cos 54^{\circ} = \frac{\sqrt{10-2\sqrt{5}}}{4}\). The inverse of cosine is the secant, which is \(\sec 54^{\circ} = \frac{\sqrt{50+10\sqrt{5}}}{5}\).

Therefore, the circumradius is \(r = \frac{\sqrt{50+10\sqrt{5}}}{10}s\) ≈ \(0.85s\). The circumradius is indeed smaller than the side of the pentagon.

### The Area in Terms of the Circumradius

The area of the pentagon is 5 times the area of AOB. We can use the sine formula for the area of a triangle: \(A = \frac{1}{2}ab\sin C\). Then, multiply by 5 for the area of the pentagon.

(i) \(A = 5\cdot\frac{1}{2}ab\sin C\)

(ii) \(A = \frac{5}{2}r\cdot r\cdot \sin 72^{\circ}\)

(iii) \(A = \frac{5}{2}r^2\cdot \frac{\sqrt{10+2\sqrt{5}}}{4}\)

(iv) \(A = \frac{5\sqrt{10+2\sqrt{5}}}{8}r^2 \approx 2.378r^2\)

Therefore, the area of the pentagon is \(A = \frac{5\sqrt{10+2\sqrt{5}}}{8}r^2\) in terms of the circumradius. We can verify this formula but substituting the circumradius in terms of *s*: \(r = \frac{\sqrt{50+10\sqrt{5}}}{10}s\). We should get the original formula for the area in terms of the length of the side.

(i) \(A = \frac{5\sqrt{10+2\sqrt{5}}}{8}r^2\)

(ii) \(A = \frac{5\sqrt{10+2\sqrt{5}}}{8}\left( \frac{\sqrt{50+10\sqrt{5}}}{10}s \right)^2\)

(iii) \(A = \frac{5\sqrt{10+2\sqrt{5}}}{8}\cdot \frac{50+10\sqrt{5}}{100}s^2\)

(iv) \(A = \frac{5\sqrt{10+2\sqrt{5}}}{8}\cdot \frac{5+\sqrt{5}}{10}s^2\)

(v) \(A = \frac{\sqrt{10+2\sqrt{5}}\cdot(5+\sqrt{5})}{16}s^2\)

(vi) \(A = \frac{\sqrt{(10+2\sqrt{5})(5+\sqrt{5})^2}}{16}s^2\)

(vii) \(A = \frac{\sqrt{400+160\sqrt{5}}}{16}s^2\)

(viii) \(A = \frac{4\sqrt{25+10\sqrt{5}}}{16}s^2\)

(ix) \(A = \frac{\sqrt{25+10\sqrt{5}}}{4}s^2\)

And we have the original formula for the area of a pentagon in terms of the side. So the circumradius formula verifies.

## Lengths

We will now find the lengths of some segments of the pentagon. Let’s bring back the pentagon image.

Letting the side of the pentagon equal *s*, we will find the length of EG = CH = DH, which is the length of the side of the smaller pentagon.

Recall that the ratio of DB to AB is equal to the Golden Ratio. Therefore, \(\frac{DB}{AB} = \frac{1+\sqrt{5}}{2}\) and \(DB = \frac{1+\sqrt{5}}{2}s\). Also, BH is equal to AB. Therefore, \(DH = \frac{1+\sqrt{5}}{2}s - s\) = \(\frac{\sqrt{5}-1}{2}s\).

Now, to find GH, we note that DHG is similar to DBA. Therefore, we divide DH by the Golden Ratio to get GH: \(GH = \frac{\frac{\sqrt{5}-1}{2}s}{\frac{\sqrt{5}+1}{2}} \) = \(\frac{3-\sqrt{5}}{2}s\) ≈ \(0.382s\).

### The Geometric Ratio of the Lengths

The lengths of the segments of the pentagon form a geometric ratio with the ratio being the Golden Ratio.

The length of DB = DA = \(\frac{1+\sqrt{5}}{2}s\). Let’s use the variable *φ* for the Golden Ratio. Therefore \(DB = \phi s\).

We divide this by *φ* and get the length AB: \(\frac{DB}{\phi} = \frac{\phi s}{\phi} = s\).

We divide the side by *φ* and get DH: \(\frac{s}{\phi} = \frac{1}{\phi}s = \frac{\sqrt{5}-1}{2}s \).

We divide DH by *φ* and get GH: \(\frac{DH}{\phi} = \frac{1}{\phi^2}s = \frac{3-\sqrt{5}}{2}s \).

If we draw the diagonals of the smaller pentagon, then we can continue this geometric sequence forever.

The sum of all these segments converges. The first term is \(\phi s\) and the ratio is \(\frac{1}{\phi}\). Therefore, the sum, *S*, is:

(i) \(S = \frac{\phi s}{1-\frac{1}{\phi}} \)

(ii) \(S = \frac{\phi^2}{\phi - 1}s \)

(iii) \(S = (\sqrt{5} + 2)s = (2\phi + 1)s \)

Do you see the significance of the length \((2\phi + 1)s \)?

If we write \((2\phi + 1)s \) as \(2\phi s + s \), then it becomes more apparent. The length \(\phi s\) is one side of the 36-72-72 triangle DAB. Twice that would be the sum of both legs: DB + DA. Finally, we add *s* which is the side AB of triangle DAB. The sum of the geometric series is the perimeter of triangle DAB.

## The Circumradius of the Smaller Pentagon

The circumradii of the two pentagon would have the same ratio as the sides. We can find the circumradius of the smaller pentagon using this fact. Let’s just call the circumradius of the smaller pentagon *x*.

(i) \( \frac{s}{\frac{\sqrt{50+10\sqrt{5}}}{10}s} = \frac{\frac{3-\sqrt{5}}{2}s}{x} \)

(ii) \( x = \frac{3-\sqrt{5}}{2}\cdot \frac{\sqrt{50+10\sqrt{5}}}{10}s \)

(iii) \( x = \frac{\sqrt{(14-6\sqrt{5})(50+10\sqrt{5})}}{20}s \)

(iii) \( x = \frac{\sqrt{700+140\sqrt{5}-300\sqrt{5}-300}}{20}s \)

(iii) \( x = \frac{4\sqrt{25-10\sqrt{5}}}{20}s \)

(iii) \( x = \frac{\sqrt{25-10\sqrt{5}}}{5}s \) ≈ \(0.325s\)

The circumradius of the smaller pentagon is \(\frac{\sqrt{25-10\sqrt{5}}}{5}s \) ≈ \(0.325s\).

## The Ratios

The ratio of the sides of the two pentagons is \(1 + \phi\). All of the other lengths will also be in the same ratio.