Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I would like to find the apothem of a regular pentagon. It follows from

$$\cos \dfrac{2\pi }{5}=\dfrac{-1+\sqrt{5}}{4}.$$

But how can this be proved (geometrically or trigonometrically)?

share|improve this question

7 Answers 7

up vote 29 down vote accepted

Since $x := \cos \frac{2 \pi}{5} = \frac{z + z^{-1}}{2}$ where $z:=e^{\frac{2 i \pi}{5}}$, and $1+z+z^2+z^3+z^4=0$ (for $z^5=1$ and $z \neq 1$), $x^2+\frac{x}{2}-\frac{1}{4}=0$, and voilà.

share|improve this answer
    
Very clever: the real part of $z$ is indeed $2\cos \frac{2\pi }{5}+4\left( \cos \frac{2\pi }{5}\right) ^{2}-1$ –  Américo Tavares Oct 24 '10 at 21:16
    
@MFV: By Vieta, yes. (The coefficient of $z^{n-1}$ is zero). –  J. M. Oct 24 '10 at 23:48
1  
A tiny correction @MFV: $z^5-1=(z-1)(1+z+z^2+z^3+z^4)$. Here, the Vieta relations become even more evident. –  J. M. Oct 25 '10 at 3:15
    
Where exactly did the polynomial in $x$ come from? –  Geano Oct 3 at 7:34
    
Since $$z \neq 0$$, we have $$z^{-2}+z^{-1}+1+z+z^2=0$$. Expand $$z^{-2}+z^{-1}+1+z+z^2 - 4x^2$$. –  Plop Oct 3 at 20:46

diagram

Consider a $\triangle ABC$ with $AB=1$, $\mathrm{m}\angle A=\frac{\pi}{5}$ and $\mathrm{m}\angle B=\mathrm{m}\angle C=\frac{2\pi}{5}$, and point $D$ on $\overline{AC}$ such that $\overline{BD}$ bisects $\angle ABC$. Now, $\mathrm{m}\angle CBD=\frac{\pi}{5}$ and $\mathrm{m}\angle BDC=\frac{2\pi}{5}$, so $\triangle ABC\sim\triangle BCD$. Also note that $\triangle ABD$ is isosceles so that $BC=BD=AD$.

Let $x=BC=BD=AD$. From the similar triangles, $\frac{AB}{BC}=\frac{BC}{CD}$ or $\frac{1}{x}=\frac{x}{1-x}$, so $1-x=x^2$ and $x=\frac{\sqrt{5}-1}{2}$ (the other solution is negative and lengths cannot be negative).

Now, apply the Law of Cosines to $\triangle ABC$: $$\begin{align} \cos\frac{2\pi}{5}=\cos C&=\frac{a^2+b^2-c^2}{2ab} \\ &=\frac{(\frac{\sqrt{5}-1}{2})^2+1^2-1^2}{2\cdot\frac{\sqrt{5}-1}{2}\cdot 1} \\ &=\frac{\frac{3-\sqrt{5}}{2}}{\sqrt{5}-1} \\ &=\frac{-1+\sqrt{5}}{4}. \end{align}$$

share|improve this answer
1  
@MFV: Given the context of your question, it seemed that you'd want a proof that was more geometrically oriented. In my head, the triangle I used is the triangle-equivalent of the golden rectangle. Also, if you draw a regular pentagon and all its diagonals, you'll see the diagram I drew as a small part of the picture (AB is a side, C and D are inside the pentagon at intersections of diagonals). –  Isaac Oct 24 '10 at 21:51
    
This was the proof I always used to quickly remember the value of $\cos \frac{2 \pi}{5}$. –  Beni Bogosel Apr 17 '12 at 17:38

How about combinatorially? This follows from the following two facts.

  • The eigenvalues of the adjacency matrix of the path graph on $n$ vertices are $2 \cos \frac{k \pi}{n+1}, k = 1, 2, ... n$.

  • The number of closed walks from one end of the path graph on $4$ vertices to itself of length $2n$ is the Fibonacci number $F_{2n}$.

The first can be proven by direct computation (although it also somehow falls out of the theory of quantum groups) and the second is a nice combinatorial argument which I will leave as an exercise. I discuss some of the surrounding issues in this blog post.

share|improve this answer
3  
Also look up Kastelyn's method on counting the number of tilings of a 2mx2n chessboard with 2x1 rectangles. Choose m=1 and n=2 for this particular problem. –  Aryabhata Oct 24 '10 at 20:58

Look up the "construction of a regular pentagon" using the straightedge and compass. If you keep track of each step in this construction, you will find that the angle $72^\circ$ comes up in a few places, and this expression follows from it.

It's a fun exercise-- you should do it.

share|improve this answer
1  
@MFV: cut-the-knot.org/pythagoras/cos36.shtml might give you some help if you get stuck. But I agree with the suggestion to work it out yourself; that's much more interesting than trying to decipher someone else's "AB=AE=PQ=etc" description. –  Hans Lundmark Oct 24 '10 at 17:37
    
One of the first things I did upon learning coordinate geometry was to prove to myself that the compass-straightedge construction of the regular pentagon works as advertised. Do it when you can, it's fun. :) –  J. M. Oct 24 '10 at 23:51

Note that $$2\cdot \dfrac{2\pi}{5} + 3\cdot \dfrac{2\pi}{5} = 2\pi,$$ therefore $$\cos\left(2\cdot \dfrac{2\pi}{5}\right) = \cos\left(3\cdot \dfrac{2\pi}{5}\right).$$ Put $\cos \dfrac{2\pi}{5} = x$. Using the formulas \begin{equation*} \cos 2x = 2\cos^2 x - 1, \quad \cos 3x = 4\cos^3 x - 3\cos x, \end{equation*} we have \begin{equation*} 4x^3 - 2x^2 -3x + 1 = 0 \Leftrightarrow (x - 1)(4x^2 + 2x - 1) = 0. \end{equation*} Because $\cos \dfrac{2\pi}{5} \neq 1$, we get \begin{equation*} 4x^2 + 2x - 1 = 0. \end{equation*} Another way, $\cos \dfrac{2\pi}{5} > 0$, then $\cos \dfrac{2\pi}{5} = \dfrac{-1 + \sqrt{5}}{4}$.

share|improve this answer

Or you can go to Mathworld for $\pi/5$ and use the multiple angle formula

share|improve this answer

for any $\theta \in \mathbb{R}$ the transformation $\psi: x \rightarrow 2x^2 -1$ sends $cos \theta$ to $cos 2\theta$ . hence $\psi^2 $ sends $cos \theta$ to $cos4\theta$

if $\alpha = \frac{2\pi}5$ then $cos 4\alpha = cos \alpha$ so that $cos \alpha$ is a fixed point for $\psi$ and if $c=cos \alpha$ we have $$ \psi^2(c) = c $$ i.e. $$ 2(2c^2-1)^2-1= c $$ or $$ (c-1)(8c^3-1) = 0 $$ i.e. $$ (c-1)(2c-1)(4c^2+2c-1) = 0 $$ if we disregard the roots corresponding to angles with rational cosines we have: $$ 4c^2+2c-1 =0, $$ giving $$ c = \frac14\left(-1 \pm \sqrt{5}\right) $$ the negative root corresponds to the angle $\frac{4\pi}5$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.