Find the Sum

Geometry Level 3

x cos θ = y cos ( θ + 2 π 3 ) = z cos ( θ 2 π 3 ) \large \dfrac x{\cos \theta} = \dfrac y{\cos \left( \theta + \frac{2\pi}3 \right) }= \dfrac z{\cos \left( \theta - \frac{2\pi}3 \right) }

If the equation above is an identity except when cos θ , cos ( θ + 2 π 3 ) , cos ( θ 2 π 3 ) 0 \cos \theta, \cos \left( \theta + \frac{2\pi}3 \right), \cos \left( \theta - \frac{2\pi}3 \right) \ne 0 , find x + y + z x+y+z .


The answer is 0.

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1 solution

Rishabh Jain
Mar 23, 2016

x cos θ = y cos ( θ + 2 π 3 ) = z cos ( θ 2 π 3 ) = k \dfrac x{\cos \theta} = \dfrac y{\cos \left( \theta + \frac{2\pi}3 \right) }= \dfrac z{\cos \left( \theta - \frac{2\pi}3 \right) }=k x + y + z = T x+y+z=\mathfrak T T = k ( cos θ + cos ( θ + 2 π 3 ) + cos ( θ 2 π 3 ) cos C + cos D = 2 cos ( C + D 2 ) cos ( C D 2 ) ) \mathfrak T=k\left( \cos \theta+\underbrace{\cos \left( \theta + \frac{2\pi}3 \right) +\cos \left( \theta - \frac{2\pi}3 \right)}_{\color{#D61F06}{\small{\cos C+\cos D=2\cos\left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)}}} \right) = k ( cos θ + 2 cos θ cos ( 2 π 3 ) ) \large =k\left(\cos \theta +2\cos \theta \cos \left(\dfrac{2\pi}3\right)\right) = 0 ( cos ( 2 π 3 ) = 1 2 ) \large{=0}~~\left(\because \cos\left(\dfrac{2\pi}{3}\right)=-\dfrac 12\right)

Nice solution! I did it in a similar way, in which I used the fact that the equation also represents a line in 3 dimensions, and as ( 0 , 0 , 0 ) \displaystyle (0,0,0) lies on it (along with the fact that the given equation is an identity) the required sum is 0. Is my method even correct, or was I just lucky??

B.S.Bharath Sai Guhan - 5 years, 2 months ago

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