Diagonals of pentagon

Geometry Level 3

The diagram below shows a regular pentagon $ABCDE$ .

Without using a calculator, determine which of the following is the ratio of the diagonal length to the side length of $ABCDE$ .

Notation: $\phi=\dfrac{1+\sqrt{5}}{2}$ denotes the golden ratio.

\dfrac{\phi}{2}

\sqrt{5}

\phi

\phi-1

\phi^2

2\phi

1 solution

Steven Liu
Sep 12, 2017

Let the intersection of $\overline{AD}$ and $\overline{CE}$ be $F$ , we have $\bigtriangleup$ $AFC$ $\sim$ $\bigtriangleup$ $EFD$ .

Let the side length of the pentagon be $a$ and length of diagonal be $x$ , note that $\overline{AF} = \overline{CF} = a$ .

By ratio of corresponding sides we have $\frac{a}{x} = \frac{x-a}{a}$ .

Cross multiplying and rearranging gives $x^2-ax-a^2=0$ . Solving using quadratic formula gives $x=\frac{a \pm a \sqrt 5}{2}=a \phi$ . (rejecting the negative value)

The ratio of diagonal length to side length is therefore $a \phi : a = \phi$ .

Diagonals of pentagon

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