# New Golden Ratio Construction with Two Adjacent Squares and Circle. Have you seen anything similar?

The below Golden Ratio Construction results in a ratio of PHI (1.6180...) between the blue line and red line, as found in Geogebra. This seems like a simple construction of the golden ratio, yet so far I have not found anything similar. If you know of a similar golden ratio construction, please do share! Thanks!

This Golden Ratio construction is made in the following manner:

1. Draw two adjacent squares Square1 and Square2.
2. Draw a circle with radii equal to the side of the Square, placed at the corner of Square2 as drawn below.
3. A line passing through the two opposing corners of the two adjacent squares will then define the golden cut in conjunction with the circle, as shown below. The ratio of segment t to segment s, (or the blue segment to the red segment), is the golden ratio PHI. Has anyone come across anything the same or similar?

And of course trigonometric and geometric proofs of the golden ratio construction are always welcome!

• I haven't seen such a construction before, but just a quick analysis - normalizing the side length of the squares to 1, we have the diagonal of the rectangle with length $\sqrt{1^2+2^2}=\sqrt{5}$, giving the red line length $\sqrt{5}-1$. The diameter of the circle (i.e. the blue line) has length $2$, so the ratio of the blue line to red line is $\frac{2}{\sqrt{5}-1}=\frac{2\left(\sqrt{5}+1\right)}{4}=\frac{\sqrt{5}+1}{2}$ as desired. – Peter Woolfitt May 23 '16 at 21:31
• @Peter Woolfitt That could be an answer to the follow on question. – user301988 May 24 '16 at 0:46
• It is very similar to one of the steps for the usual construction of the regular pentagon (en.wikipedia.org/wiki/Pentagon). – Jack D'Aurizio May 24 '16 at 2:40
• which step @JackD'Aurizio? could you please be more specific and perhaps even provide a figure or link to a figure or drawing? thanks! :) – Astrophysics Math May 24 '16 at 2:41
• en.wikipedia.org/wiki/File:Richmond_pentagon_1.PNG – Jack D'Aurizio May 24 '16 at 2:42

Your construction in this question occurs as a subobject in the construction here. Your particular result does not appear explicitly there, but is produced by projecting Molokach's points $A$, $B$, and $H$ through an unconstructed point interior to the triangle $ACH$.