Phi in the Pentagon
 | This article needs to be tidied. In particular: Much of this is probably already in existence, in the coverage of Euclid. Also see Pentalpha. Please fix formatting and $\LaTeX$ errors and inconsistencies. It may also need to be brought up to our standard house style. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Tidy}} from the code. |
Theorem
Let $ABCDE$ be a regular pentagon.
The chords and sides of $ABCDE$ form three kinds of similar triangles.
The sides of the first type are in the ratio $\phi : 1$, where $\phi$ is the golden mean.
The sides of the second type are in the ratio $1 : \phi$.
In the words of Euclid:
- If in an equilateral and equiangular pentagon straight lines subtend two internal angles taken in order, they cut one another in extreme and mean ratio, and their greater segments are equal to the side of the pentagon.
(The Elements: Book $\text{XIII}$: Proposition $8$)
Proof
by hypothesis five sides of $ABCDE$ are equal.
Draw all five chords.
From symmetry, the chords are all equal.
$\leadsto$:
- $\triangle ACD$ is an isosceles triangle.
$\leadsto$:
- $\triangle BCD$ is an isosceles triangle
| \(\ds \angle DBC\) | \(=\) | \(\ds \angle CDB = \alpha\) | Isosceles Triangle has Two Equal Angles | |||||||||||
| \(\ds \angle ACD\) | \(=\) | \(\ds \angle ADC\) | Isosceles Triangle has Two Equal Angles | |||||||||||
| \(\ds \triangle ACD\) | \(=\) | \(\ds 3 \beta + 2 \alpha = 180 \degrees\) | Sum of Internal Angles of Triangle | |||||||||||
| \(\ds \triangle BCD\) | \(=\) | \(\ds 1 \beta + 4 \alpha = 180 \degrees\) | Sum of Internal Angles of Triangle | |||||||||||
| \(\ds 2 \beta - 2 \alpha\) | \(=\) | \(\ds 0\) | subtracting | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \alpha\) | \(=\) | \(\ds \beta\) | |||||||||||
| \(\ds \angle CA'D\) | \(=\) | \(\ds \angle BCD\) | Sum of Internal Angles of Triangle | |||||||||||
| \(\ds \angle B'A'E'\) | \(=\) | \(\ds \angle CA'D\) | Vertical Angle Theorem | |||||||||||
| \(\ds \angle B'A'E'\) | \(=\) | \(\ds 3 \beta = \angle BCD\) | Common Notion $1$ |
Hence $A'B'C'D'E'$ has all internal angles equal and is Similar to $ABCDE$.
| \(\ds \alpha\) | \(=\) | \(\ds \beta = \dfrac \pi 5 = 72^{\circ}\) | Sum of Internal Angles of Polygon |
By Quadrilateral is Parallelogram iff Both Pairs of Opposite Angles are Equal:
- $BCDC'$ is a parallelogram.
Also by symmetry:
- $\triangle AD'C'$ is isosceles
| \(\ds \triangle AD'C'\) | \(\sim\) | \(\ds \triangle ACD\) | Equiangular Triangles are Similar |
Let the length of $D'C' = B'C' = 1$.
Let the length of $EB' = AD' = AC' = \phi$, where $\phi$ is to be determined.
| \(\ds \leadsto \ \ \) | \(\ds \dfrac \phi 1\) | \(=\) | \(\ds \dfrac {2 \phi + 1} {\phi + 1}\) | |||||||||||
| \(\ds \phi^2\) | \(=\) | \(\ds \phi + 1\) | rearranging | |||||||||||
| \(\ds \phi^2 - \phi - 1\) | \(=\) | \(\ds 0\) |
By the quadratic equation:
- $\phi = \dfrac{ 1 \pm \sqrt{5} } 2$
We take $\phi > 0$ so:
- $\phi = \dfrac{ 1 + \sqrt{5} } 2$
By definition of golden mean:
- $\phi$ is the golden mean
The length of the sides of the pentagon is $\phi + 1$.
By Equiangular Triangles are Similar:
- $\triangle ADE \sim \triangle B'ED$
$\leadsto$:
- $\dfrac {\phi + 1} { 2 \phi + 1 } = \dfrac {\phi} {\phi + 1}$
Working with the right hand side and substituting from above:
- $\dfrac {\phi} {\phi + 1} = \dfrac {\phi} {\phi^2}$
- $= \dfrac 1 \phi$
The result follows.
$\blacksquare$
Also known as
- Definition:Golden Mean: a classical name for $\phi$