# Show that a diagonal in a pentagon is the golden ratio

I just learned that the diagonal of a pentagon (size 1) is the golden ratio (https://twitter.com/fermatslibrary/status/1210561047154872320)

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?

• Dec 27, 2019 at 15:54
• I guess you mean a regular pentagon? Dec 27, 2019 at 15:55
• @Bernard All sides equal Dec 27, 2019 at 15:57

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

• How can we know these two triangles are isoscele? Dec 27, 2019 at 16:43
• For $Ebc$ it is obvious, as it is symmetrically aligned. For $AEd$ you just have to observe that $ED$ and $AC$ are parallel, just as $EB$ and $DC$. Accordingly $Ed=DC=1$. Dec 27, 2019 at 16:49
• And why is 1=x+y? Dec 27, 2019 at 16:52
• $x+y=Ec+cd=Ed=DC=1$. Dec 27, 2019 at 16:53
• I don't get that last step. Why is Ed=DC? Dec 27, 2019 at 16:55 $$\triangle AA'C\sim\triangle BB'C\quad\to\quad\frac{|A'C|}{|A'A|}=\frac{|B'C|}{|B'B|}\quad\to\quad \underbrace{\frac{a+b}{a}=\frac{a}{b}}_{=\,\phi\;\text{(by def'n)}} \quad\to\quad \frac{\text{diagonal}}{\text{side}}= \phi$$