A geometry problem by Mateus Gomes

Geometry Level 4

sin ⁡ ( 2 π 7 ) + sin ⁡ ( 4 π 7 ) + sin ⁡ ( 8 π 7 ) \sin \left(\dfrac{2\pi}{7}\right)+\sin \left(\dfrac{4\pi}{7}\right)+\sin \left(\dfrac{8\pi}{7}\right)

If the value of the expression above is equal to A B \dfrac {\sqrt A}B , where A A and B B are positive integers with A A square-free, find A + B A+B .


The answer is 9.

Mateus Gomes by Mateus Gomes , 22, Brazil

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1 solution

Aditya Kumar
Apr 11, 2016

Let x = 2 π 7 x=\frac{2\pi}{7}

We get: 7 x = 2 π 7x=2\pi

sin ⁡ ( 4 x ) = sin ⁡ ( 2 π − 3 x ) \sin(4x)=\sin(2\pi-3x) ⟹ \Longrightarrow sin ⁡ ( 4 x ) = − sin ⁡ ( 3 x ) \sin(4x)=-\sin(3x) .

On simplyfying, we get 4 cos ⁡ ( x ( 1 − 2 sin ⁡ 2 ( x ) ) ) = 4 sin ⁡ 2 ( x ) − 3 4\cos { \left( x\left( 1-2\sin ^{ 2 }{ \left( x \right) } \right) \right) } =4\sin ^{ 2 }{ \left( x \right) } -3

On squaring both the sides and simplifying, we get: 64 sin ⁡ 6 ( x ) − 112 sin ⁡ 4 ( x ) + 56 sin ⁡ 2 ( x ) − 7 = 0 64\sin ^{ 6 }{ \left( x \right) } -112\sin ^{ 4 }{ \left( x \right) } +56\sin ^{ 2 }{ \left( x \right) } -7=0

The roots of this equation are: sin ⁡ 2 ( 2 π 7 ) , sin ⁡ 2 ( 4 π 7 ) , sin ⁡ 2 ( 6 π 7 ) \sin ^{ 2 }{ \left( \frac { 2\pi }{ 7 } \right) } ,\sin ^{ 2 }{ \left( \frac { 4\pi }{ 7 } \right) } ,\sin ^{ 2 }{ \left( \frac { 6\pi }{ 7 } \right) }

On further simplification (using Vieta's formula and doing some basic algebra) we get: sin ⁡ ( 2 π 7 ) + sin ⁡ ( 4 π 7 ) + sin ⁡ ( 8 π 7 ) = 7 2 \boxed{\sin \left(\dfrac{2\pi}{7}\right)+\sin \left(\dfrac{4\pi}{7}\right)+\sin \left(\dfrac{8\pi}{7}\right) =\frac{\sqrt{7}}{2}}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

I'm really confused trying to read this. Several issues that I noticed:

  1. You are only showing that x = 2 π 7 x = \frac{2 \pi}{7} is a root of the polynomial that you created. You have not yet shown that the other values are also roots.
  2. I don't understand how sin ⁡ ( 7 x ) = sin ⁡ ( 2 π − 3 x ) \sin(7x)=\sin(2\pi-3x) for x = 2 π 7 x = \frac{ 2 \pi}{7} . Are you saying that sin ⁡ ( 2 π ) = sin ⁡ 8 π 7 \sin (2 \pi ) = \sin \frac{8\pi}{7} ?
  3. Because of that, I don't quite understand what else you are doing from the third line onwards.
  4. If the roots are sin ⁡ 2 θ \sin ^2 \theta , how did you find ∑ sin ⁡ θ \sum \sin \theta ? The "basic algebra" part should be elaborated on.

@Calvin Lin :

  1. x = 2 π 7 x=\frac{2\pi }{7} . It is just a cubic equation in sin ⁡ 2 x \sin ^2x which can be solved.

  2. That was a typo.

  3. ....

  4. That can be found by completing the square. ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( a b + b c + a c ) (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac) . The values can easily be found by using Vieta's formula.

Aditya Kumar - 5 years, 1 month ago

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  1. You still need to show that the roots of the cubic equation 64 y 3 − 122 y 2 + 56 y − 7 = 0 64y^3 -122 y^2 +56 y -7 = 0 are sin ⁡ 2 4 π 7 \sin^2 \frac{4 \pi}{7} and sin ⁡ 2 6 π 7 \sin^2 \frac{6\pi}{7} . Your work has only established that sin ⁡ 2 2 π 7 \sin^2 \frac{2\pi}{7} is a root, and has made no mention of these 2 other values.
  2. This makes more sense now.
  3. Now I understand what you are doing.
  4. You should show this step in full. Note that we do not yet have the cross product a b + b c + c a ab+bc+ca .

Calvin Lin Staff - 5 years, 1 month ago

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