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)
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1Could you include a picture/sketch? – coffeemath Aug 05 '15 at 16:35
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$\phi^4 = 3\phi + 2$ But I'm to lazy to write a proof without a labeled diagram. :) – PM 2Ring Aug 05 '15 at 17:26
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1@PM2Ring: If you know of a book that has the solution, please let me know its title. Nice Fibonacci numbers, btw. – Aug 05 '15 at 18:22
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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.: $$\phi^4 = (\phi+1)^2 = \phi^2+2\phi+1 =3\phi+2.$$
Jack D'Aurizio
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