Pentagonal Diagonal

Geometry Level 3

What is the length of a diagonal of a regular pentagon of unit side length? Give your answer to three decimal places.

Please post your solution!


The answer is 1.618.

Hypergeo H. by Hypergeo H. , 30, Malaysia

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4 solutions

Hypergeo H.
Apr 12, 2020

My solution:

Let ω \omega be the principal fifth root of unity.

ω 5 = 1 1 + ω + ω ² + ω ³ + ω 4 = 0 ( ∗ ) 1 ω ² + 1 ω + 1 + ω + ω ² = 0 ( u ² − 2 ) + u + 1 = 0 ( u = ω + 1 ω ) u ² + u − 1 = 0 ω + 1 ω = u = 5 − 1 2 ( ∗ ∗ ) \begin{aligned} \omega ^5 &= 1 \\ 1+\omega+\omega²+\omega³+\omega^4&=0 &&(*)\\ \frac 1{\omega²}+\frac 1\omega+1+\omega+\omega²&=0\\ (u²-2)+u+1&=0 &&(u=\omega+\frac 1\omega)\\ u²+u-1&=0\\ \omega+\frac 1\omega=u&=\frac {\sqrt {5}-1}2 &&(**)\\ \end{aligned}

Also, from ( ∗ ) (*) ,
1 + ω = − ω ² ( 1 + ω + ω ² ) = − ω ³ ( ω + 1 ω + 1 ) = − ω ³ ( 5 + 1 2 ) (using (**) ) \begin{aligned} 1+\omega&=-\omega²(1+\omega+\omega²)\\ &=-\omega³ (\omega+\frac 1\omega+1)\\ &=-\omega³ \bigg(\frac {\sqrt{5}+1}2\bigg) &&\text{(using (**) )}\\ \end{aligned} .

Length of diagonal is ∣ 1 + ω ∣ = 5 + 1 2 = 1.618 \displaystyle |1+\omega |=\frac {\sqrt{5}+1}2 = \color{#D61F06}{1.618}
as ∣ ω ³ ∣ = 1 |\omega³|=1 .
The solution also happens to be the golden ratio.

2 Helpful 1 Interesting 1 Brilliant 1 Confused

I'm curious: Using roots of unity, for n ⩾ 4 n\geqslant 4 , can we find all the diagonal lengths of any regular n n -gon with side length 1?

Pi Han Goh - 1 year, 2 months ago

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The same method can be applied for regular ( 2 n + 1 ) 2n+1) -gons of unit side lengths but it might not necessarily be easier to compute by hand!

Hypergeo H. - 1 year, 2 months ago
Pi Han Goh
Apr 12, 2020

Here you go

2 Helpful 1 Interesting 1 Brilliant 0 Confused

Haha I'm sure there are many different approaches available online. What's your solution? Anyway, thanks for solving!

Hypergeo H. - 1 year, 2 months ago

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Use cosine rule, and find the exact form of sin ⁡ 5 4 ∘ \sin 54^\circ . Pretty rudimentary stuff.

The interior angle of a regular pentagon is 18 0 ∘ × ( 1 − 2 5 ) = 10 8 ∘ 180^\circ \times \left(1 - \frac25\right) = 108^\circ . Hence by cosine rule, the diagonal in question, D D satisfies D 2 = 2 ⋅ 1 2 ( 1 − cos ⁡ 10 8 ∘ ) D^2 = 2 \cdot1^2 (1 - \cos108^\circ) . Using the double angle identity, 1 − cos ⁡ ( 2 A ) = 2 sin ⁡ 2 A 1 - \cos(2A) = 2\sin^2 A , we have D 2 = 4 sin ⁡ 2 5 4 ∘ ⟹ D = 2 sin ⁡ 5 4 ∘ D^2 = 4 \sin^2 54^\circ \implies D = 2\sin54^\circ .

What's left is to find the exact form of sin ⁡ 5 4 ∘ = cos ⁡ 3 6 ∘ \sin54^\circ = \cos36^\circ . Since 3 6 ∘ × 4 = 18 0 ∘ − 3 6 ∘ 36^\circ \times 4 = 180^\circ - 36^\circ , let B = 3 6 ∘ B= 36^\circ , we have 4 B = 18 0 ∘ − B ⟹ cos ⁡ ( 4 B ) = cos ⁡ ( 18 0 ∘ − B ) = − cos ⁡ ( B ) 4B = 180^\circ - B \implies \cos(4B) = \cos(180^\circ - B) = -\cos(B) .

Apply the double angle formula repeatedly gives 2 ( 2 cos ⁡ 2 B − 1 ) 2 − 1 + cos ⁡ B = 0 2 (2\cos^2 B - 1)^2 - 1 + \cos B = 0 . For simplicities sake, let y = cos ⁡ B y = \cos B . We have 8 y 4 − 8 y 2 + y + 1 = 0 8y^4- 8y^2 + y + 1 = 0 , or ( y + 1 ) ( 2 y − 1 ) ( 4 y 2 − 2 y − 1 ) = 0 (y+1)(2y - 1)(4y^2 - 2y - 1) = 0 .

Since 0 < y = cos ⁡ 3 6 ∘ > cos ⁡ 6 0 ∘ = 1 2 0 < y = \cos36^\circ > \cos60^\circ = \frac12 , then y = 1 + 5 4 y = \frac{1 + \sqrt5}4 only (via quadratic formula).

Thus, the answer is D = 2 sin ⁡ 5 4 ∘ = 2 cos ⁡ 3 6 ∘ = 1 + 5 2 ≈ 1.618 D = 2\sin54^\circ = 2\cos36^\circ = \frac{1+\sqrt5}2 \approx \boxed{1.618} .

Pi Han Goh - 1 year, 2 months ago

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Nice solution!

Hypergeo H. - 1 year, 2 months ago
Kano Boom
Apr 12, 2020

I did it the lazy way hehe... the golden ratio really caught me off guard./

1 Helpful 0 Interesting 0 Brilliant 0 Confused

The length of red diagonal is 1 cos 36 ° + 1 cos 36 ° = 2 cos 36 ° = 5 + 1 2 1\,\text{cos}36\degree + 1\,\text{cos}36\degree = 2\,\text{cos}36\degree = \Large\frac{\sqrt{5}+1}{2} ≈ 1.618 \approx 1.618

0 Helpful 0 Interesting 0 Brilliant 0 Confused

But how would you know that cos ⁡ 30 ° = 5 + 1 2 \cos \ang{30} = \frac {\sqrt{5}+1}2 ?

Hypergeo H. - 1 year, 2 months ago

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cos 36 ° = 1 − 2 sin 2 18 ° \text{cos }36\degree = 1 - 2\text{sin}^2 18\degree

Let θ = 18 ° \theta = 18\degree . Then 3 θ + 2 θ = 5 θ = 90 ° ⇒ 3 θ = 90 ° − 2 θ ⇒ sin 3 θ = cos 2 θ ⇒ 3 sin θ − 4 sin 3 θ = 1 − 2 sin 2 θ ⇒ 4 sin 3 θ − 2 sin 2 θ − 3 sin θ + 1 = 0 ⇒ ( sin θ − 1 ) ( 4 sin 2 θ + 2 sin θ − 1 ) = 0 3\theta + 2\theta = 5\theta = 90\degree\Rightarrow 3\theta = 90\degree - 2\theta\newline \Rightarrow \text{sin }3\theta = \text{cos }2\theta\newline \Rightarrow 3\text{sin }\theta - 4\text{sin}^3 \theta = 1 - 2\text{sin}^2\theta\newline \Rightarrow 4\text{sin}^3\theta - 2\text{sin}^2\theta - 3\text{sin }\theta + 1 = 0 \newline \Rightarrow (\text{sin}\theta - 1)(4\text{sin}^2\theta + 2\text{sin}\theta - 1) = 0

Solving the quadratic equation we get

sin θ = sin 18 ° = 5 − 1 4 \text{sin}\theta = \text{sin}18\degree = \Large\frac{\sqrt{5} - 1}{4}

⇒ sin 2 18 ° = 6 − 2 5 16 = 3 − 5 8 \Rightarrow \text{sin}^2 18\degree = \Large\frac{6 - 2\sqrt{5}}{16} = \frac{3 - \sqrt{5}}{8}

⇒ 2 sin 2 18 ° = 3 − 5 4 \Rightarrow 2\text{ sin}^2 18\degree = \Large\frac{3 - \sqrt{5}}{4}

⇒ cos 36 ° = 1 − 2 sin 2 18 ° = 1 − 3 − 5 4 = 1 + 5 4 \Rightarrow \text{cos }36\degree = 1 - 2\text{ sin}^2 18\degree = 1 - \Large\frac{3 - \sqrt{5}}{4} = \frac{1+ \sqrt{5}}{4}

Shikhar Srivastava - 1 year, 2 months ago

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