Square trig

Geometry Level 4

If a a and b b are positive integers such that

8 + 32 + 768 = a cos ⁡ π b \large \sqrt{8+\sqrt{32+\sqrt{768}}} = a \cos \dfrac{\pi}{b}

Find a + b a+b .


The answer is 28.

Anish Harsha by Anish Harsha , 20, India

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2 solutions

Rishabh Jain
Feb 19, 2016

H = 8 + 32 + 768 \large\mathfrak{H}= \sqrt{8+\sqrt{32+\sqrt{768}}} = 2 2 + 2 + 3 \large =2 \sqrt{2+\sqrt{\color{#D61F06}{2+\sqrt{3}}}} Note : 2 + 3 = 1 2 ( 3 + 1 ) 2 \small{\text{Note}:\color{#D61F06}{2+\sqrt{3}}=\dfrac{1}{2}(\sqrt3+1)^2} ∴ H = 2 2 + 3 + 1 2 \large \therefore \mathfrak{H}=2 \sqrt{2+\dfrac{\sqrt3+1}{\sqrt2}} = 2 2 1 + 3 + 1 2 2 \large =2\sqrt2 \sqrt{1+\dfrac{\sqrt3+1}{2\sqrt2}} Note : 3 + 1 2 2 = cos ⁡ ( π 12 ) 1 + cos ⁡ 2 A = 2 cos ⁡ 2 A \small{\text{Note}:\color{#D61F06}{\dfrac{\sqrt3+1}{2\sqrt2}=\cos (\dfrac{\pi}{12})\\ 1+\cos 2A=2\cos^2 A}} H = 2 2 1 + cos ⁡ π 12 \large \mathfrak{H}=2\sqrt2\sqrt{1+\cos \frac{\pi}{12}} = 4 cos ⁡ π 24 \large =\color{#20A900}{4}\cos \frac{\pi}{\color{#20A900}{24}} 4 + 24 = 28 \huge \color{#20A900}{4+24}=\boxed{\color{#007fff}{28}}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

I did the same way. But I found C o s π 24 = ( 2 + 2 + 3 2 ) Cos\dfrac \pi {24}=\left (\dfrac{\sqrt{2+\sqrt{2+\sqrt3}} }2 \right ) from google.

Niranjan Khanderia - 5 years, 3 months ago
Ahmad Saad
Feb 19, 2016

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Good solution from the fundamentals. Up voted.

Niranjan Khanderia - 5 years, 3 months ago

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