Solve it in Friday 13th - part 3

Geometry Level 5

cos ⁡ ( 2 π 13 ) cos ⁡ ( 6 π 13 ) cos ⁡ ( 8 π 13 ) \large \cos\left(\dfrac{2\pi}{13}\right)\cos\left(\dfrac{6\pi}{13}\right)\cos\left(\dfrac{8\pi}{13}\right)

If the trigonometric expression above can be expressed as a − b c \dfrac{a-\sqrt{b}}{c} for positive integers a , b , c a,b,c with b b is a square-free number and gcd ( a , c ) = 1 \text{gcd}(a,c)=1 , find a + b + c a+b+c .

See Part 1 and part 2 .


The answer is 32.

Alan Enrique Ontiveros Salazar by Alan Enrique Ontiveros S… , 22

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3 solutions

Let w = e 2 π i / 13 w=e^{2 \pi i/13} , a = cos ⁡ ( 2 π 13 ) cos ⁡ ( 6 π 13 ) cos ⁡ ( 8 π 13 ) a=\cos\left(\dfrac{2\pi}{13}\right)\cos\left(\dfrac{6\pi}{13}\right)\cos\left(\dfrac{8\pi}{13}\right) and b = cos ⁡ ( 4 π 13 ) cos ⁡ ( 10 π 13 ) cos ⁡ ( 12 π 13 ) b=\cos\left(\dfrac{4\pi}{13}\right)\cos\left(\dfrac{10\pi}{13}\right)\cos\left(\dfrac{12\pi}{13}\right) .

We know the fact that w + w 2 + w 3 + ⋯ + w 12 = − 1 w+w^2+w^3+\cdots+w^{12}=-1 , which follows inmediatly from factoring w 13 = 1 w^{13}=1 . We are going to use this everytime we see it; and that 1 2 ( w k + 1 / w k ) = cos ⁡ ( 2 π k 13 ) \dfrac{1}{2}(w^k+1/w^k)=\cos\left(\dfrac{2 \pi k}{13}\right) .

Now, express them in terms of w w :

a = ( w + w 12 2 ) ( w 3 + w 10 2 ) ( w 4 + w 9 2 ) b = ( w 2 + w 11 2 ) ( w 5 + w 8 2 ) ( w 6 + w 7 2 ) a=\left(\dfrac{w+w^{12}}{2}\right)\left(\dfrac{w^3+w^{10}}{2}\right)\left(\dfrac{w^4+w^9}{2}\right) \\ b=\left(\dfrac{w^2+w^{11}}{2}\right)\left(\dfrac{w^5+w^8}{2}\right)\left(\dfrac{w^6+w^7}{2}\right) .

Expand them:

a = w 2 + w 5 + w 6 + w 7 + w 8 + w 11 + 2 8 b = w + w 3 + w 4 + w 9 + w 10 + w 12 + 2 8 a=\dfrac{w^2+w^5+w^6+w^7+w^8+w^{11}+2}{8} \\ b=\dfrac{w+w^3+w^4+w^9+w^{10}+w^{12}+2}{8}

With this, find a + b = 3 8 a+b=\dfrac{3}{8} and a b ab , which is a little messy product to expand: a b = − 1 64 ab=-\dfrac{1}{64} .

Finally, use Vieta's formulas with the fact that this time that a < 0 a<0 and b > 0 b>0 : a = 3 − 13 16 a=\dfrac{3-\sqrt{13}}{16} and b = 3 + 13 16 b=\dfrac{3+\sqrt{13}}{16}

Thus, the final answer is 3 + 13 + 16 = 32 3+13+16=\boxed{32} .

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Chew-Seong Cheong
Jun 24, 2015

cos ⁡ 2 π 13 cos ⁡ 6 π 13 cos ⁡ 8 π 13 = cos ⁡ 2 π 13 cos ⁡ 6 π 13 ( cos ⁡ 2 π 13 cos ⁡ 6 π 13 − sin ⁡ 2 π 13 sin ⁡ 6 π 13 ) = cos ⁡ 2 2 π 13 cos ⁡ 2 6 π 13 − sin ⁡ 2 π 13 cos ⁡ 2 π 13 sin ⁡ 6 π 13 cos ⁡ 6 π 13 = 1 4 ( cos ⁡ 4 π 13 + 1 ) ( cos ⁡ 12 π 13 + 1 ) − 1 4 sin ⁡ 4 π 13 sin ⁡ 12 π 13 = 1 4 ( cos ⁡ 4 π 13 cos ⁡ 12 π 13 + cos ⁡ 4 π 13 + cos ⁡ 12 π 13 + 1 − sin ⁡ 4 π 13 sin ⁡ 12 π 13 ) = 1 4 ( cos ⁡ 16 π 13 + cos ⁡ 4 π 13 + cos ⁡ 12 π 13 + 1 ) = 1 4 ( cos ⁡ 10 π 13 + cos ⁡ 4 π 13 + cos ⁡ 12 π 13 + 1 ) = 1 4 ( cos ⁡ 2 π 13 + cos ⁡ 4 π 13 + cos ⁡ 6 π 13 + cos ⁡ 8 π 13 + cos ⁡ 10 π 13 + cos ⁡ 12 π 13 − ( cos ⁡ 2 π 13 + cos ⁡ 6 π 13 + cos ⁡ 8 π 13 ) + 1 ) = 1 4 ( − 1 2 − 13 − 1 4 + 1 ) [ Note 1 ] [ Note 2 ] = 3 − 13 16 \cos{\frac{2\pi}{13}} \cos{\frac{6\pi}{13}} \cos{\frac{8\pi}{13}} \\ = \cos{\frac{2 \pi}{13}} \cos{\frac{6\pi}{13}} \left( \cos{\frac{2\pi}{13}} \cos{\frac{6\pi}{13}} - \sin{ \frac{2\pi}{13}} \sin{\frac{6\pi}{13}} \right) \\ = \cos^2{\frac{2\pi}{13}} \cos^2{\frac {6\pi}{13}} - \sin{\frac{2\pi}{13}} \cos{\frac{2\pi}{13}} \sin{\frac{6\pi}{13}} \cos{\frac{6\pi}{13}} \\ = \frac{1}{4} \left( \cos{\frac{4\pi}{13}} + 1 \right) \left( \cos{\frac{12\pi}{13}} + 1 \right) - \frac{1}{4} \sin{\frac{4\pi}{13}} \sin{\frac{12\pi}{13}} \\ = \frac{1}{4} \left( \cos{\frac{4\pi}{13}} \cos{\frac{12\pi}{13}} + \cos{\frac{4\pi}{13}} + \cos{\frac{12\pi}{13}} + 1 - \sin{\frac{4\pi}{13}} \sin{\frac{12\pi}{13}} \right) \\ = \frac{1}{4} \left( \cos{\frac{\color{#D61F06}{16} \pi}{13}} + \cos{\frac{4\pi}{13}} + \cos{\frac{12\pi}{13}} + 1 \right) \\ = \frac{1}{4} \left( \cos{\frac{\color{#D61F06}{10} \pi}{13}} + \cos{\frac{4\pi}{13}} + \cos{\frac{12\pi}{13}} + 1 \right) \\ = \frac{1}{4} \left( \color{#3D99F6} {\cos{\frac{2\pi}{13}} + \cos{\frac{4\pi}{13}} + \cos{\frac{6\pi}{13}} + \cos{\frac{8\pi}{13}} + \cos{\frac{10\pi}{13}} + \cos{\frac{12\pi}{13}}} \\ \quad \quad - \left(\color{#20A900}{\cos{\frac{2\pi}{13}} + \cos{\frac{6\pi}{13}} + \cos{\frac{8\pi}{13}}} \right) + 1 \right) \\ = \frac{1}{4} \left( \color{#3D99F6}{- \frac{1}{2}} - \color{#20A900}{\frac{\sqrt{13}-1}{4}} + 1 \right) \quad \quad \color{#3D99F6}{[\text{Note 1}]} \color{#20A900}{[\text{Note 2}]} \\ = \boxed{\frac{3-\sqrt{13}}{16}}

Note 1: z 13 = e 2 π 13 i = 1 \color{#3D99F6}{\text{Note 1: }} z^{13} = e^{\frac{2\pi}{13}i} = 1 are the 13 t h ^{th} roots of unity. By Argand's diagram, we can see that: cos ⁡ 2 π 13 + cos ⁡ 4 π 13 + cos ⁡ 6 π 13 + cos ⁡ 8 π 13 + cos ⁡ 10 π 13 + cos ⁡ 12 π 13 = − 1 2 \cos{\frac{2\pi}{13}} + \cos{\frac{4\pi}{13}} + \cos{\frac{6\pi}{13}} + \cos{\frac{8\pi}{13}} + \cos{\frac{10\pi}{13}} + \cos{\frac{12\pi}{13}} = -\frac{1}{2}

Note 2: \color{#20A900}{\text{Note 2: }} See solution of Part 1 .

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@Alan Enrique Ontiveros Salazar Out of curiosity, what was your intended approach for these problems?

Calvin Lin Staff - 5 years, 11 months ago

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For part 1, I let a = cos ⁡ 2 π 13 + cos ⁡ 6 π 13 + cos ⁡ 8 π 13 a=\cos{\frac{2\pi}{13}} + \cos{\frac{6\pi}{13}} + \cos{\frac{8\pi}{13}} and b = cos ⁡ 4 π 13 + cos ⁡ 10 π 13 + cos ⁡ 12 π 13 b=\cos{\frac{4\pi}{13}} + \cos{\frac{10\pi}{13}} + \cos{\frac{12\pi}{13}} . Also, let w = e 2 π i 13 w=e^{\frac{2\pi i}{13}} , so using w k + w 13 − k = 2 cos ⁡ ( 2 π k 13 ) w^k+w^{13-k}=2\cos(\frac{2\pi k}{13}) , we get:

a = w + w 12 + w 3 + w 10 + w 4 + w 9 2 b = w 2 + w 11 + w 5 + w 8 + w 6 + w 7 2 a=\frac{w+w^{12}+w^3+w^{10}+w^4+w^9}{2} \\ b=\frac{w^2+w^{11}+w^5+w^8+w^6+w^7}{2} .

Then, I tried to find a + b a+b and a b ab , the first was easy using w + w 2 + w 3 + ⋯ + w 12 = − 1 w+w^2+w^3+\cdots+w^{12}=-1 :

a + b = − 1 2 a+b=-\frac{1}{2}

But to find a b ab I expanded the whole product and used the fact that w 13 = 1 w^{13}=1 , and I ended with a b = − 3 4 ab=-\frac{3}{4} .

Finally, by the fact that a > 0 a>0 and b < 0 b<0 , use Vieta's formulas to find that a = − 1 + 13 4 a=\frac{-1+\sqrt{13}}{4} and b = − 1 − 13 4 b=\frac{-1-\sqrt{13}}{4} .

For part 2, I only used the product to sum formulas and I ended with: 1 2 [ cos ⁡ ( 8 π 13 ) + cos ⁡ ( 4 π 13 ) + cos ⁡ ( 10 π 13 ) + cos ⁡ ( 6 π 13 ) + cos ⁡ ( 14 π 13 ) + cos ⁡ ( 2 π 13 ) ] \frac{1}{2}[\cos(\frac{8\pi}{13})+\cos(\frac{4\pi}{13})+\cos(\frac{10\pi}{13})+\cos(\frac{6\pi}{13})+\cos(\frac{14\pi}{13})+\cos(\frac{2\pi}{13})] , but cos ⁡ ( 14 π 13 ) = cos ⁡ ( 12 π 13 ) \cos(\frac{14\pi}{13})=\cos(\frac{12\pi}{13}) , and the fact that cos ⁡ ( 2 π 13 ) + ⋯ + cos ⁡ ( 12 π 13 ) = − 1 2 \cos(\frac{2\pi}{13})+\cdots+\cos(\frac{12\pi}{13})=-\frac{1}{2} , we get − 1 4 -\frac{1}{4} .

For part 3, I followed a similar method, let a = cos ⁡ ( 2 π 13 ) cos ⁡ ( 6 π 13 ) cos ⁡ ( 8 π 13 ) a=\cos(\frac{2\pi}{13})\cos(\frac{6\pi}{13})\cos(\frac{8\pi}{13}) and b = cos ⁡ ( 4 π 13 ) cos ⁡ ( 10 π 13 ) cos ⁡ ( 12 π 13 ) b=\cos(\frac{4\pi}{13})\cos(\frac{10\pi}{13})\cos(\frac{12\pi}{13}) . Now, express them in terms of w w :

a = ( w + w 12 2 ) ( w 3 + w 10 2 ) ( w 4 + w 9 2 ) b = ( w 2 + w 11 2 ) ( w 5 + w 8 2 ) ( w 6 + w 7 2 ) a=(\frac{w+w^{12}}{2})(\frac{w^3+w^{10}}{2})(\frac{w^4+w^9}{2}) \\ b=(\frac{w^2+w^{11}}{2})(\frac{w^5+w^8}{2})(\frac{w^6+w^7}{2}) .

Expand them:

a = w 2 + w 5 + w 6 + w 7 + w 8 + w 11 + 2 8 b = w + w 3 + w 4 + w 9 + w 10 + w 12 + 2 8 a=\frac{w^2+w^5+w^6+w^7+w^8+w^{11}+2}{8} \\ b=\frac{w+w^3+w^4+w^9+w^{10}+w^{12}+2}{8}

With this, find a + b = 3 8 a+b=\frac{3}{8} and a b ab , which is a little messy product: a b = − 1 64 ab=-\frac{1}{64} .

Finally, use Vieta's formulas with the fact that this time a < 0 a<0 and b > 0 b>0 :

a = 3 − 13 16 a=\frac{3-\sqrt{13}}{16} and b = 3 + 13 16 b=\frac{3+\sqrt{13}}{16}

These kind of problems are very interesting because when you try to find some expressions involving cos ⁡ ( 2 π k n ) \cos(\frac{2\pi k}{n}) for some prime n n , you, for some reason, find expressions with n \sqrt{n} in them.

Also, with this we can prove that the polynomial P ( x ) = x 6 + 1 2 x 5 − 5 4 x 4 − 1 2 x 3 + 3 8 x 2 + 3 32 x − 1 64 P(x)=x^6+\frac{1}{2}x^5-\frac{5}{4}x^4-\frac{1}{2}x^3+\frac{3}{8}x^2+\frac{3}{32}x-\frac{1}{64} , which has roots cos ⁡ ( 2 π k 13 ) \cos(\frac{2\pi k}{13}) for k k from 1 1 to 6 6 , can be factored like this:

( x 3 + 1 − 13 4 x 2 − 1 4 x − 3 − 13 16 ) ( x 3 + 1 + 13 4 x 2 − 1 4 x − 3 + 13 16 ) (x^3+\frac{1-\sqrt{13}}{4}x^2-\frac{1}{4}x-\frac{3-\sqrt{13}}{16})(x^3+\frac{1+\sqrt{13}}{4}x^2-\frac{1}{4}x-\frac{3+\sqrt{13}}{16})

Alan Enrique Ontiveros Salazar - 5 years, 11 months ago

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Could you add these as solutions to the individual parts? Thanks!

Calvin Lin Staff - 5 years, 11 months ago

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@Calvin Lin – Of course!, I will be posting them soon.

Alan Enrique Ontiveros Salazar - 5 years, 11 months ago

I don't understand how trigonometry can be molded to form such beautiful solutions for such brilliant problems.

Vighnesh Raut - 3 years, 5 months ago

Also, just wandering, will it be possible to find the values of the individual terms?

Vighnesh Raut - 3 years, 5 months ago

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Yes, using the Chebyshev polynomials of the first kind, we can express cos ⁡ ( n θ ) = T n ( cos ⁡ θ ) \cos (n\theta) = T_n (\cos \theta) . For example, T 1 ( cos ⁡ θ ) = cos ⁡ θ T_1 (\cos \theta) = \cos \theta , T 2 ( cos ⁡ θ ) = 2 cos ⁡ 2 θ − 1 T_2(\cos \theta) = 2\cos^2 \theta - 1 , T 3 ( cos ⁡ θ ) = 4 cos ⁡ 3 θ − 3 cos ⁡ θ T_3(\cos \theta) = 4\cos^3 \theta - 3\cos \theta , ⋯ \cdots .

Chew-Seong Cheong - 3 years, 5 months ago
Gari Chua
Jun 24, 2015

From Part 1 and Part 2 ,

cos ⁡ 2 π 13 + cos ⁡ 6 π 13 + cos ⁡ 8 π 13 = 13 − 1 4 \cos \frac{2\pi}{13} + \cos \frac{6\pi}{13} + \cos \frac{8\pi}{13} =\frac{\sqrt{13}-1}{4}

and

cos ⁡ 2 π 13 cos ⁡ 6 π 13 + cos ⁡ 6 π 13 cos ⁡ 8 π 13 + cos ⁡ 8 π 13 cos ⁡ 2 π 13 = − 1 4 \cos \frac{2\pi}{13} \cos \frac{6\pi}{13} + \cos \frac{6\pi}{13} \cos \frac{8\pi}{13} + \cos \frac{8\pi}{13} \cos \frac{2\pi}{13} = - \frac{1}{4} .

We also know that for any x , y , x, y, and z z ,

x 3 + y 3 + z 3 − 3 x y z = ( x + y + z ) ( ( x + y + z ) 2 − 3 ( x y + y z + z x ) ) x^3+y^3+z^3-3xyz = (x+y+z)((x+y+z)^2-3(xy+yz+zx)) .

Thus, letting { x , y , z } = { cos ⁡ 2 π 13 , cos ⁡ 6 π 13 , cos ⁡ 8 π 13 } \{x,y,z\}=\{\cos \frac{2\pi}{13}, \cos \frac{6\pi}{13}, \cos \frac{8\pi}{13} \} , and substituting all above values, we have

cos ⁡ 3 ( 2 π 13 ) + cos ⁡ 3 ( 6 π 13 ) + cos ⁡ 3 ( cos ⁡ 8 π 13 ) − 3 cos ⁡ 2 π 13 cos ⁡ 6 π 13 cos ⁡ 8 π 13 = 7 13 − 13 16 \cos ^3 (\frac{2\pi}{13}) + \cos ^3 (\frac{6\pi}{13}) + \cos ^3 (\cos \frac{8\pi}{13} ) -3\cos \frac{2\pi}{13} \cos \frac{6\pi}{13}\cos \frac{8\pi}{13} = \frac{7\sqrt{13} -13}{16} .

Also, note that from the triple-angle formula, cos ⁡ 3 θ = 4 cos ⁡ 3 θ − 3 cos ⁡ θ \cos 3\theta = 4\cos ^3 \theta -3\cos \theta , we can get an expression for cos ⁡ 3 θ \cos ^3 \theta :

cos ⁡ 3 θ = 1 4 ( cos ⁡ 3 θ + 3 cos ⁡ θ ) \cos ^3 \theta = \frac{1}{4} (\cos 3\theta + 3\cos \theta) .

Thus,

cos ⁡ 3 ( 2 π 13 ) + cos ⁡ 3 ( 6 π 13 ) + cos ⁡ 3 ( cos ⁡ 8 π 13 ) = 1 4 ( cos ⁡ 6 π 13 + cos ⁡ 18 π 13 + cos ⁡ 24 π 13 \cos ^3 (\frac{2\pi}{13}) + \cos ^3 (\frac{6\pi}{13}) + \cos ^3 (\cos \frac{8\pi}{13} ) = \frac{1}{4} (\cos \frac{6\pi}{13} + \cos \frac{18\pi}{13} + \cos \frac{24\pi}{13}

+ 3 ( cos ⁡ 2 π 13 + cos ⁡ 6 π 13 + cos ⁡ 8 π 13 ) ) + 3(\cos \frac{2\pi}{13} + \cos \frac{6\pi}{13} + \cos \frac{8\pi}{13}))

= 1 4 ( cos ⁡ 6 π 13 + cos ⁡ 8 π 13 + cos ⁡ 2 π 13 + 3 ( cos ⁡ 2 π 13 + cos ⁡ 6 π 13 + cos ⁡ 8 π 13 ) ) = \frac {1}{4} ( \cos \frac{6\pi}{13} + \cos \frac{8\pi}{13} + \cos \frac{2\pi}{13} + 3(\cos \frac{2\pi}{13} + \cos \frac{6\pi}{13} + \cos \frac{8\pi}{13}))

= cos ⁡ 2 π 13 + cos ⁡ 6 π 13 + cos ⁡ 8 π 13 = 13 − 1 4 =\cos \frac{2\pi}{13} + \cos \frac{6\pi}{13} + \cos \frac{8\pi}{13} = \frac{\sqrt{13}-1}{4} .

So,

cos ⁡ 2 π 13 cos ⁡ 6 π 13 cos ⁡ 8 π 13 = 1 3 ( 13 − 1 4 − 7 13 − 13 16 ) \cos \frac{2\pi}{13} \cos \frac{6\pi}{13}\cos \frac{8\pi}{13} = \frac{1}{3} (\frac{\sqrt{13}-1}{4} - \frac{7\sqrt{13} -13}{16})

= 3 − 13 16 =\boxed{\frac{3-\sqrt{13}}{16}}

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