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In both triangles, the ratio of a big side to a small one is of course $\phi,$ what else? The resulting pentagon is very nearly regular, but not quite.) In the above mentioned article, K.But this is not the only occurence of the golden ratio in regular pentagon. Hofstetter, offered another elegant way of obtaining the Golden Ratio.

Bùi Quang Tuån found a simple construction: if $\displaystyle\frac=3,$ then $\displaystyle\frac=\phi .$ I placed a proof in a separate file.54]: The golden rectangle is easily constructed from a square as shown in the diagram below: Changing the order of operations, the problem of inscribing a square into a semicircle has a golden rectangle as a biproduct.(The problem is easily solved using homothety, the same way as inscribing a square into a triangle.) Tran Quang Hung has prepared a graphical illustration below where the ratio of the blue to a red segment is Golden.In the third paper in the series, Hofstetter mentions having been alerted to the fact that the above proof had been discovered previously by E. Draw $A(B)$ and $B(A)$ and find $C$ and $D$ at their intersection.Let $M$ be the midpoint of $AB$ found at the intersection of $AB$ and $CD.$ Construct $C(M, AB),$ a circle with center $M$ and radius $AB.$ Let it intersect $B(A)$ in $F$ and another point, $F$ being the farthest from $D.$ Define $G$ as the intersection of $AB$ and $DG.$ $G$ then is the sought point.