A geometry problem by حسن العطية

Geometry Level 4

If sin ⁡ 2 θ = 1 4 \sin2\theta= \dfrac14 , and cos ⁡ 6 θ + sin ⁡ 6 θ = a b 6 \cos^6 \theta + \sin^6 \theta = \dfrac a{b^6} , where a a and b b are coprime positive integers, find b + a 3 \dfrac{b+a} 3 .


The answer is 21.

حسن العطية by حسن العطية , 22, Saudi Arabia

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1 solution

( cos ⁡ θ ) 6 + ( sin ⁡ θ ) 6 = ( ( cos ⁡ θ ) 2 + ( sin ⁡ θ ) 2 ) ( ( cos ⁡ θ ) 4 − ( cos ⁡ θ sin ⁡ θ ) 2 + ( sin ⁡ θ ) 4 ) { (\cos { \theta } ) }^{ 6 }+{ (\sin { \theta } ) }^{ 6 } = \left( { (\cos { \theta } ) }^{ 2 }+{ (\sin { \theta } ) }^{ 2 } \right) \left( { ({ \cos { \theta } }) }^{ 4 }-{(\cos { \theta } \sin { \theta } ) }^{2 }+{ (\sin { \theta } ) }^{ 4 } \right)

we know ( cos ⁡ θ ) 2 + ( sin ⁡ θ ) 2 = 1 { (\cos { \theta } ) }^{ 2 }+{ (\sin { \theta } ) }^{ 2 } = 1 so

( cos ⁡ θ ) 4 − ( cos ⁡ θ sin ⁡ θ ) 2 + ( sin ⁡ θ ) 4 { ({ \cos { \theta } }) }^{ 4 }-{(\cos { \theta } \sin { \theta } ) }^{2 }+{ (\sin { \theta } ) }^{ 4 }

add and subtract 3 ( cos ⁡ θ sin ⁡ θ ) 2 t o ( ( cos ⁡ θ ) 4 − ( cos ⁡ θ sin ⁡ θ ) 2 + ( sin ⁡ θ ) 4 ) 3{\left (\cos { \theta } \sin { \theta } \right) }^{2 } to \left( { ({ \cos { \theta } }) }^{ 4 }-{(\cos { \theta } \sin { \theta } ) }^{2 }+{ (\sin { \theta } ) }^{ 4 } \right)

( cos ⁡ θ ) 4 + 2 ( cos ⁡ θ sin ⁡ θ ) 2 + ( sin ⁡ θ ) 4 − 3 ( cos ⁡ θ sin ⁡ θ ) 2 { ({ \cos { \theta } }) }^{ 4 }+2{(\cos { \theta } \sin { \theta } ) }^{2 }+{ (\sin { \theta } ) }^{ 4 }-3{(\cos { \theta } \sin { \theta } ) }^{2 }

( ( cos ⁡ θ ) 2 + ( sin ⁡ θ ) 2 ) 2 − 3 ( cos ⁡ θ sin ⁡ θ ) 2 ) { \left( { (\cos { \theta } ) }^{ 2 }+{ (\sin { \theta } ) }^{ 2 } \right) }^{ 2 } -3{(\cos { \theta } \sin { \theta } ) }^{2 } )

1 − 3 ( cos ⁡ θ sin ⁡ θ ) 2 1-3{(\cos { \theta } \sin { \theta } ) }^{2 }

we know sin ⁡ 2 θ = 2 sin ⁡ θ cos ⁡ θ \sin { 2\theta } = 2\sin { \theta } \cos { \theta }

2 sin ⁡ θ cos ⁡ θ = 1 4 2\sin { \theta } \cos { \theta } =\frac { 1 }{ 4 }

sin ⁡ θ cos ⁡ θ = 1 8 \sin { \theta } \cos { \theta } =\frac { 1 }{ 8 }

( cos ⁡ θ sin ⁡ θ ) 2 = 1 64 { \left( \cos { \theta } \sin { \theta } \right) }^{ 2 }=\frac { 1 }{ 64 }

So 1 − 3 64 = 61 64 1-\frac { 3 }{ 64 } = \frac { 61 }{ 64 }

So 61 2 6 = a b 6 \frac { 61 }{ { 2 }^{ 6 } } = \frac {a }{ { b}^{ 6 } }

Then 61 + 2 3 \frac { 61+2 }{ 3 }

63 3 \frac { 63 }{ 3 }

21 \boxed{21}

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