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The Golden Ratio: Constructing the Golden Rectangle

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

Today we will learn how to construct the famous Golden Rectangle. Remember that constructions only allow the use of a compass and a straightedge (not a ruler). Simple things that we can do with these devices are make perpendicular lines, find midpoints, draw circles, and bisect angles. It's a fun exercise to follow along with my instructions below.

Here is what I think is the most straightforward way to construct the Golden Rectangle. Take a unit segment. Make a segment perpendicular at the end of the unit segment two units long. Complete the right triangle. We now have a segment measuring \(\sqrt{5}\) units. Take this segment and extend it one unit. Then draw the segment that is the average of these two (to do this, find the midpoint of that segment). Then we have a segment measuring \(\phi\) units. We can then easily construct a rectangle with dimensions \(\phi\) and \(1\).

There is, however, a more elegant way. Reference the picture above. Begin by constructing square \(ABCD\). Then, construct midpoint \(M\) of \(CD\). Draw the circle with radius \(MB\). Extend \(DC\) until it intersects the circle. Call this point of intersection \(X\). Finish the rectangle with vertices \(ADX\). This rectangle is a Golden Rectangle. Note also that you have created another Golden Rectangle in the process.

Can you prove that this second construction actually produces a golden rectangle? How is this similar to my construction? Post your answers or inquiries below.

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Note by Bob Krueger
2 years, 9 months ago

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since triangleBMC is congruent AND EQUAL to triangle BCX,the new rectangle thus formed is also a goldenrectangle,which is of unit lengthg Sunitha Bhadragiri · 2 years, 9 months ago

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