I just learned that the diagonal of a pentagon (size 1) is the golden ratio ()
I tried to verify that, but ended up having to show that: $$\cos{\frac{2\pi}{5}}=\frac{1}{\phi}$$
Question: is there a way to calculate the diagonal of a pentagon without having to do any relatively complex trigonometric calculations? For example only by drawing some smart supportive lines and using the Pythagorean Theorem?
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$\begingroup$Consider a regular pentagon $ABCDE$ and a vertex inscribed pentagram $ACEBD$ with the additional intersection points $a,...,e$ somewhat closer to the center but on the oposite ray. Then the pentagram will have sides $AdeC, CabE, EcdB, BeaD, DbcA$.
By assumption you have $AB=BC=CD=DE=EA=1$. Let further be $Ad=Bd=Be=Ce=Ca=Da=Db=Eb=Ec=Ac=:x$ and $ab=bc=cd=de=ea=:y$.
Now consider the isoscele triangle $AEd$. Thus you get $1=x+y$. Its base is $x$. Then consider the scaled down isoscele triangle $Ebc$ with sides $x, x, y$ (its tip angle clearly is the same). From those you get the scaling ratio$$\varphi:=\frac1x=\frac xy$$Together with the above ($y=1-x$) the equation for the golden ratio follows:$$1-x=x^2$$or, when dividing by $x^2$ and re-inserting $\varphi$:$$\varphi^2=\varphi+1$$--- rk
$\endgroup$ 10 $\begingroup$$$\triangle AA'C\sim\triangle BB'C\quad\to\quad\frac{|A'C|}{|A'A|}=\frac{|B'C|}{|B'B|}\quad\to\quad \underbrace{\frac{a+b}{a}=\frac{a}{b}}_{=\,\phi\;\text{(by def'n)}} \quad\to\quad \frac{\text{diagonal}}{\text{side}}= \phi$$
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