# Pentagon A **pentagon** is a [[polygon]] with five sides. ## Regular Pentagon * The [[exterior angle]] in a regular pentagon is $72^\circ$. * The [[interior angle]] in a regular pentagon is $108^\circ$. * The length of the line across a regular pentagon is the golden ratio times the side length. ## Lengths and Angles in a Regular Pentagon +-- {.image} [[RegularPentagon.png:pic]] =-- The [[exterior angle]] of a regular pentagon is $\frac{360^\circ}{5} = 72^\circ$, so the [[interior angle]] is $180^\circ - 72^\circ = 108^\circ$. With the points labelled as in the above diagram, this means that triangle $A B E$ is an [[isosceles]] triangle with apex angle $108^\circ$ so its base angles are $\frac{180^\circ - 108^\circ}{2} = 36^\circ$. Therefore, angles $B \hat{A} F$ and $J \hat{A} E$ are both $36^\circ$, meaning that angle $F \hat{A} J = 108^\circ - 2 \times 36^\circ = 36^\circ$. So triangle $A F J$ is also isosceles with base angles $\frac{180^\circ - 36^\circ}{2} = 72^\circ$. Triangle $A C D$ is also isosceles with angles $36^\circ$, $72^\circ$, and $72^\circ$. Then triangle $A B J$ has angles $J \hat{B} A = 36^\circ$, $B \hat{A} J = 108^\circ - 36^\circ = 72^\circ$, and angle $A \hat{J} B = 72^\circ$ so is also isosceles. These three triangles are therefore all [[similar]], and the scale factors are such that the short side of $A C D$ has the same length as the long sides of $A B J$, and the short side of $A B J$ has the same length as the long sides of $A F J$. Let $a$ be the length of $F J$, and let $\phi$ be the [[length scale factor]] from $A F J$ to $B J A$, so then $A F$ has length $\phi a$, and $A B$ has length $\phi^2 a$. Since $A B J$ is [[isosceles]], $A B$ and $B J$ have the same length, and this is equal to the lengths of $B F$ and $F J$ which is $\phi a + a$. Hence $\phi^2 a = \ph a + a$, or equivalently $\phi^2 = \phi + 1$. Together with the condition that $\phi \gt 1$, this establishes $\phi$ as the [[golden ratio]]. Now consider the two radii of the pentagon: the radius of the [[incircle]] and the [[circumcircle]], equivalently the lengths of $O L$ and $O D$. Write these as $r$ and $R$, respectively. Let $b$ be the side length of the pentagon. Viewing $C D$ as the base of triangle $C D O$, its area is $\frac{1}{2} b r$. Viewing $O D$ as its base, its height is half the length of $C E$ which is $\phi b$ so its area is $\frac{1}{4} \phi b R$. Equating these establishes that $2 r = \phi R$. Lastly, the total height of the pentagon, which is the length of $A L$, relates to the side length via [[Pythagoras' theorem]] applied to the triangle $A L C$. The length of $C L$ is half a side-length, $\frac{1}{2} b$, and the length of $A C$ is $\phi b$. So $A L$ has length: $$ \sqrt{ (\phi b)^2 - \left(\frac{1}{2} b\right)^2} = \frac{b}{2} \sqrt{4 \phi^2 - 1} = b \frac{\sqrt{ 4\phi + 3}}{2} $$ [[!redirects regular pentagon]] [[!redirects angles in a regular pentagon]] [[!redirects lengths in a regular pentagon]]