How Long Can The Product Go!!!

Geometry Level pending

Suppose a 3 D 3\text{D} vector v ⃗ = ( a , b , c ) \vec{v}=(a,b,c) with a , b , c ≥ 0 a,b,c\geq 0 makes angles α , β , γ \alpha,\beta,\gamma with the x , y , z x,y,z axes respectively. The direction cosines of v ⃗ \vec{v} is then defined by the followings: cos ⁡ α = a a 2 + b 2 + c 2 \cos \alpha=\dfrac{a}{\sqrt{a^2+b^2+c^2}} cos ⁡ β = b a 2 + b 2 + c 2 \cos\beta=\dfrac{b}{\sqrt{a^2+b^2+c^2}} cos ⁡ γ = c a 2 + b 2 + c 2 \cos \gamma=\dfrac{c}{\sqrt{a^2+b^2+c^2}} The maximum value of cos ⁡ α cos ⁡ β cos ⁡ γ \cos\alpha \cos\beta\cos \gamma can be expressed as m n \dfrac{\sqrt{m}}{n} where ( m , n ) ∈ N 2 (m,n)\in\mathbb{N}^2 . Find the value of m + n m+n .


The answer is 12.

Jubayer Nirjhor by Jubayer Nirjhor , 22, Bangladesh

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1 solution

Jubayer Nirjhor
Jun 18, 2014

Note that cos ⁡ 2 α + cos ⁡ 2 β + cos ⁡ 2 γ = 1 \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1 . Hence by AM-GM inequality, we have: cos ⁡ 2 α cos ⁡ 2 β cos ⁡ 2 γ ≤ ( cos ⁡ 2 α + cos ⁡ 2 β + cos ⁡ 2 γ 3 ) 3 = ( 1 3 ) 3 = 3 3 4 ∴ cos ⁡ α cos ⁡ β cos ⁡ γ ≤ 3 3 4 = 3 9 \begin{aligned} &\cos^2 \alpha~\cos^2\beta~\cos^2\gamma ~~ &\le ~~\left(\dfrac{\cos^2\alpha+\cos^2\beta+\cos^2\gamma}{3}\right)^3 \\ &~ &=~~\left(\dfrac{1}{3}\right)^3 \\ &~&=~~\dfrac{3}{3^4} \\ &\therefore ~\cos\alpha~\cos\beta~\cos\gamma ~~&\le ~~\sqrt{\dfrac{3}{3^4}} \\ &~ &=~~ \dfrac{\sqrt 3}{9}\\ \end{aligned} Hence, m = 3 , n = 9 m=3, n=9 and the answer is m + n = 3 + 9 = 12 m+n=3+9=\fbox{12} .

Note: Equality occurs when cos ⁡ 2 α = cos ⁡ 2 β = cos ⁡ 2 γ \cos^2\alpha=\cos^2\beta=\cos^2\gamma , that is, when a = b = c a=b=c and α = β = γ = ± arccos ⁡ ( 1 3 ) \alpha=\beta=\gamma=\pm\arccos{\left(\dfrac{1}{\sqrt 3}\right)} .

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