Consider a regular pentagon circumscribed in a circle. Connect each vertex of this pentagon to every other not adjacent to it with a straight line segment to obtain a pentagram which contains a smaller pentagon. What is the ratio of the area of the original (large) pentagon to the small one in terms of the golden ratio? (Synthetic area proofs preferred)
2026-04-18 05:01:18.1776488478
The Regular Pentagram
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An isosceles triangle with angles $36^\circ,72^\circ,72^\circ$ appears in many places in the configuration. By similarity and the sine theorem, we have that the ratio between the side lengths of the bigger and smaller pentagons is just:
$$\left(\frac{\sin 72^\circ}{\sin 36^\circ}\right)^2 = \left(2\cos\frac{\pi}{5}\right)^2 = \frac{3+\sqrt{5}}{2}=\phi^2.$$ Obviously, the ratio of the areas is just the square of $\phi^2=\phi+1$, i.e.: $$\phi^4 = (\phi+1)^2 = \phi^2+2\phi+1 =3\phi+2.$$