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The Golden Ratio: The Most Elegant Construction

Here is the previous post concerning the Golden Ratio. For a collection of all the posts concerning the Golden Ratio, click #GoldenRatio below.

Sorry for the long wait! This is the final post about the geometry of the golden ratio, and to celebrate that, I will present to you, with proof, the most elegant construction of the golden ratio. This construction is so awesome, it can even be performed without a compass, paper, or pencil.


Take three segments of equal length (whether they be toothpicks, playing cards, or a drawing on a piece of paper). Stand one upright perpendicular to a flat surface. Now place the next one so that it touches the flat surface and the midpoint of the other segment. Place the final one so that it touches the flat surface and the midpoint of the middle segment. The line formed on the flat surface is divided into the golden ratio. See the above picture.


Let the segments have length 2. Then \(A_1B_2 = 1\), making \(A_1A_2 = \sqrt{2^2-1^2} = \sqrt{3}\). Draw the perpendicular from \(B_3\) to the surface. It has length \(\frac{1}{2}\). Then \(A_2A_3 = \sqrt{2^2 - \left(\frac{1}{2}\right)^2} - \sqrt{1^2 - \left(\frac{1}{2}\right)^2} = \sqrt{\frac{15}{4}} - \sqrt{\frac{3}{4}} = \sqrt{3} \frac{\sqrt{5}-1}{2}\). Then we can take the ratio:

\[\frac{A_1A_2}{A_2A_3} = \frac{\sqrt{3}}{\sqrt{3} \frac{\sqrt{5}-1}{2}} = \frac{1+\sqrt{5}}{2} = \phi\]

Be prepared for an even more exciting quest into the algebraic mysterious of the golden ratio!


Note by Bob Krueger
2 years, 9 months ago

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That is AWESOME!!!!! Finn Hulse · 2 years, 9 months ago

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