Math Series Part IV: The Number Phi

Updated on November 30, 2016 Michael has been an online freelancer and writer for many years and loves discovering and sharing about new experiences and opportunities.

The Pentagonal Golden Ratio: Finding Phi in a Pentagon Inscribed in Circle of Radius 1.

We begin exploring how the Golden Ratio comes about in a pentagon inscribed in a circle with radius one by first constructing the image. If we take our circle and draw two diameter lines on the y- and x-axes, we are then able to take the midpoint of OB, M, and join it to the point C above. A line is then drawn from the middle of the angle of OMC to meet the line OC at Q. Q is then connected by a straight line to point Z on the circle, and the resulting line CZ is one side of our pentagon. If we repeat this process for the remaining four corners of the pentagon and connect the bottom two vertices together instead of having them meet at point D, we end up with the full regular pentagon displayed in green above, where all of its sides are equal in length at 1 unit.

If we now join all of the corners of the pentagon together, we get a pentagram: a shape that includes the number phi, also known as the Golden Ratio. What’s amazing about the pentagram is that the lengths of each of the lines connecting two diagonal corners of the pentagon are phi. Indeed, with the length of each of the edges of the pentagon having length 1, we can compute these diagonal lengths through the sine formula. For instance, if we take the pentagon below and let AE = AB = 1, we can calculate EB, hence the length of the diagonal, by using: EB / sin(EAB) = AB / sin(AEB). As this is a regular pentagon, angle EAB = 108 degrees, meaning angle AEB = ABE = 36 degrees. Therefore, we can use EB = [sin(108)*1] / sin(36) = 1.6180339… The number phi is: 1.6180339887... Hence, we have proved the existence of the Golden Ratio inside the regular pentagon.

In a similar way, we can prove the existence of 1/phi in the pentagon as well. The triangles that make up these smaller sections of the pentagram within the pentagon, within our circle of radius 1, also have a remarkable connection to phi. The fatter isosceles triangle at the bottom has base 1 and angles ABE = EAB = 36 degrees. By the same logic as above, AF = [sin(36)*1] / sin(108) = 0.618033… This is equal to 1/phi. This proves that the other isosceles at the top of our pentagram also has sides of 1/phi. We know, by symmetry, that the lengths EF = GC = 1/phi and that EC = phi. Thus, we can now also determine the length of line FG: phi = 1/phi + 1/phi + FG. FG = phi - 2/phi = (phi2 -2)/phi = 1/phi2. Therefore, FG = 1/phi2. And, if we look closely at the seemingly hidden, inverted pentagon in the middle, we will find that this regular pentagon has sides of 1/phi2. The Golden Ration is thus a very interesting Phenomenon that occurs for a pentagon and pentagram inscribed within a circle of radius 1. And what we also find in this way is that another pentagram can be drawn in the smaller pentagon to initiate the process again, and then to continue forever and ever.

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