A geometry problem by Refaat M. Sayed

Geometry Level 4

sin ⁡ ( θ 1 ) + sin ⁡ ( θ 2 ) + sin ⁡ ( θ 3 ) = cos ⁡ ( θ 1 ) + cos ⁡ ( θ 2 ) + cos ⁡ ( θ 3 ) = 0 sin ⁡ 2 ( θ 1 ) + sin ⁡ 2 ( θ 2 ) + sin ⁡ 2 ( θ 3 ) = ? \sin \left( \theta _{1}\right) + \sin \left( \theta _{2}\right) + \sin \left( \theta _{3}\right) =\cos \left( \theta _{1}\right) + \cos \left( \theta _{2}\right) + \cos \left( \theta _{3}\right) = 0 \\ \sin ^{2}\left( \theta _{1}\right) + \sin ^{2}\left( \theta _{2}\right) + \sin ^{2}\left( \theta _{3}\right) = \ ?


The answer is 1.5.

Refaat M. Sayed by Refaat M. Sayed , 24, Egypt

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3 solutions

Chew-Seong Cheong
Aug 12, 2015

The general solution of the equations is as follows (see note ):

{ sin ⁡ θ 1 + sin ⁡ θ 2 + sin ⁡ θ 3 = 0 cos ⁡ θ 1 + cos ⁡ θ 2 + cos ⁡ θ 3 = 0 ⇒ { θ 1 = θ θ 2 = θ + 2 π 3 θ 3 = θ + 4 π 3 = θ − 2 π 3 \begin{cases} \sin{\theta_1} + \sin{\theta_2} + \sin{\theta_3} = 0 \\ \cos{ \theta _1} + \cos{\theta_2} + \cos{\theta_3} = 0 \end{cases} \Rightarrow \begin{cases} \theta_1 = \theta \\ \theta_2 = \theta + \frac{2\pi}{3} \\ \theta_3 = \theta + \frac{4\pi}{3} = \theta - \frac{2\pi}{3} \end{cases}

Proof:

sin ⁡ θ 1 + sin ⁡ θ 2 + sin ⁡ θ 3 = sin ⁡ θ + sin ⁡ ( θ + 2 π 3 ) + sin ⁡ ( θ − 2 π 3 ) = sin ⁡ θ − 1 2 sin ⁡ θ + 3 2 cos ⁡ θ − 1 2 sin ⁡ θ − 3 2 cos ⁡ θ = 0 \begin{aligned} \sin{\theta_1} + \sin{\theta_2} + \sin{\theta_3} & = \sin{\theta} + \sin{\left(\theta + \frac{2\pi}{3}\right)} + \sin{\left(\theta - \frac{2\pi}{3}\right)} \\ & = \sin{\theta} - \frac{1}{2}\sin{\theta} + \frac{\sqrt{3}}{2}\cos{\theta} - \frac{1}{2}\sin{\theta} - \frac{\sqrt{3}}{2}\cos{\theta} \\ & = 0 \end{aligned}

cos ⁡ θ 1 + cos ⁡ θ 2 + cos ⁡ θ 3 = cos ⁡ θ + cos ⁡ ( θ + 2 π 3 ) + cos ⁡ ( θ − 2 π 3 ) = cos ⁡ θ − 1 2 cos ⁡ θ − 3 2 sin ⁡ θ − 1 2 cos ⁡ θ + 3 2 sin ⁡ θ = 0 \begin{aligned} \cos{\theta_1} + \cos{\theta_2} + \cos{\theta_3} & = \cos{\theta} + \cos{\left(\theta + \frac{2\pi}{3}\right)} + \cos{\left(\theta - \frac{2\pi}{3}\right)} \\ & = \cos{\theta} - \frac{1}{2}\cos{\theta} - \frac{\sqrt{3}}{2}\sin{\theta} - \frac{1}{2}\cos{\theta} + \frac{\sqrt{3}}{2}\sin{\theta} \\ & = 0 \end{aligned}

Now, we have:

sin ⁡ 2 θ 1 + sin ⁡ 2 θ 2 + sin ⁡ 2 θ 3 = ( sin ⁡ θ 1 + sin ⁡ θ 2 + sin ⁡ θ 3 ) 2 − 2 ( sin ⁡ θ 1 sin ⁡ θ 2 + sin ⁡ θ 2 sin ⁡ θ 3 + sin ⁡ θ 3 sin ⁡ θ 1 ) = 0 − 2 [ sin ⁡ θ sin ⁡ ( θ + 2 π 3 ) + sin ⁡ ( θ + 2 π 3 ) sin ⁡ ( θ − 2 π 3 ) + sin ⁡ ( θ − 2 π 3 ) sin ⁡ θ ] = − 2 [ sin ⁡ θ ( − 1 2 sin ⁡ θ + 3 2 cos ⁡ θ − 1 2 sin ⁡ θ − 3 2 cos ⁡ θ ) + ( − 1 2 sin ⁡ θ + 3 2 cos ⁡ θ ) ( − 1 2 sin ⁡ θ − 3 2 cos ⁡ θ ) ] = − 2 [ − sin ⁡ 2 θ + 1 4 sin ⁡ 2 θ − 3 4 cos ⁡ 2 θ ] = 2 [ 3 4 sin ⁡ 2 θ + 3 4 cos ⁡ 2 θ ] = 3 2 = 1.5 \sin^2{\theta_1} + \sin^2{\theta_2} + \sin^2{\theta_3} \\ = (\sin{\theta_1} + \sin{\theta_2} + \sin{\theta_3}) ^2 - 2 (\sin{\theta_1} \sin{\theta_2} + \sin{\theta_2} \sin{\theta_3} + \sin{\theta_3} \sin{\theta_1}) \\ = 0 - 2 \left[\sin{\theta} \sin{\left(\theta + \frac{2\pi}{3}\right)} + \sin{\left(\theta + \frac{2\pi}{3}\right)} \sin{\left(\theta - \frac{2\pi}{3}\right)} + \sin{\left(\theta - \frac{2\pi}{3}\right)} \sin{\theta}\right] \\ = -2 \left[\sin{\theta} \left( -\frac{1}{2} \sin{\theta} + \frac{\sqrt{3}}{2} \cos{\theta} -\frac{1}{2} \sin{\theta} - \frac{\sqrt {3}} {2} \cos{\theta} \right) \\ \quad \quad \quad \quad + \left( -\frac{1}{2} \sin{\theta} + \frac{\sqrt{3}}{2} \cos{\theta} \right) \left(-\frac{1}{2} \sin{\theta} - \frac{\sqrt {3}} {2} \cos{\theta} \right) \right] \\ = -2 \left[- \sin^2{\theta} + \frac{1}{4} \sin^2{\theta} - \frac{3}{4} \cos^2{\theta} \right] \\ = 2 \left[\frac{3}{4} \sin^2{\theta} + \frac{3}{4} \cos^2{\theta} \right] \\ = \frac{3}{2} = \boxed{1.5}

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Moderator note:

But how do you know that this value must be true for all possible sets of θ i \theta_i that satisfy the conditions?

Good solution sir +1

Refaat M. Sayed - 5 years, 10 months ago

In the note (Argand Diagram and Sum of Cosines of Multiple Angle = 1/2), I showed that for sin ⁡ θ 1 + sin ⁡ θ 2 + sin ⁡ θ 3 = cos ⁡ θ 1 + cos ⁡ θ 2 + cos ⁡ θ 3 = 0 \sin{\theta_1} + \sin{\theta_2} + \sin{\theta_3} = \cos{ \theta _1} + \cos{\theta_2} + \cos{\theta_3} = 0 , which is equivalent to a balanced mechanical system, the three numbers (analogous to "rods") must be symmetrically distributed. In this case 12 0 ∘ 120^\circ apart, otherwise the system is imbalanced or sin ⁡ θ 1 + sin ⁡ θ 2 + sin ⁡ θ 3 ≠ 0 \sin{\theta_1} + \sin{\theta_2} + \sin{\theta_3} \ne 0 or cos ⁡ θ 1 + cos ⁡ θ 2 + cos ⁡ θ 3 ≠ 0 \cos{ \theta _1} + \cos{\theta_2} + \cos{\theta_3} \ne 0 or both ≠ 0 \ne 0 . Therefore, θ + 2 k π 3 \theta + \frac{2k\pi}{3} , where k = 0 , 1 , 2 k = 0,1,2 represent all possible sets of θ i \theta_i .

Chew-Seong Cheong - 5 years, 10 months ago
Sanjeet Raria
Aug 11, 2015

Notation: sin ⁡ θ n = S n , cos ⁡ θ n = C n \sin \theta_n=S_n, \cos \theta_n=C_n

Now, S 1 + S 2 + S 3 = 0 = C 1 + C 2 + C 3 S_1+S_2+S_3=0=C_1+C_2+C_3

Squaring both sides,

⇒ S 1 2 + S 2 2 + S 3 2 + 2 ( S 1 S 2 + S 2 S 3 + S 3 S 1 ) = 0.... ( 1 ) \Rightarrow S^2_1+S^2_2+S^2_3+2(S_1S_2+S_2S_3+S_3S_1)=0 ....(1) , and C 1 2 + C 2 2 + C 3 2 + 2 ( C 1 C 2 + C 2 C 3 + C 3 C 1 ) = 0.... ( 2 ) C^2_1+C^2_2+C^2_3+2(C_1C_2+C_2C_3+C_3C_1)=0 ....(2)

Adding ( 1 ) (1) and ( 2 ) (2) and solving further, ⇒ 3 + 2 ( S 1 S 2 + C 1 C 2 + S 2 S 3 + C 2 C 3 + S 3 S 1 + C 3 C 1 ) = 0 \Rightarrow 3+2(S_1S_2+C_1C_2+S_2S_3+C_2C_3+S_3S_1+C_3C_1)=0 ⇒ cos ⁡ ( θ 1 − θ 2 ) + cos ⁡ ( θ 2 − θ 3 ) + cos ⁡ ( θ 3 − θ 1 ) = − 3 2 \Rightarrow \cos (\theta_1-\theta_2)+\cos (\theta_2-\theta_3)+\cos (\theta_3-\theta_1)=\frac{-3} 2

Now on the right side − 3 2 \frac{-3} 2 reminds us of 3 × cos ⁡ 2 π 3 . 3\times \cos \frac{2\pi}{3}. So we know one such solution set. Based on the observation, we can assume WLOG that

θ 1 = 2 n π \large \boxed{\theta_1=2n\pi}

θ 2 = θ 1 + 2 π 3 \large \boxed{ \theta_2=\theta_1+\frac{2\pi}{3}}

θ 3 = θ 1 − 2 π 3 \large \boxed{\theta_3=\theta_1-\frac{2\pi}{3}}

Therefore one of the solution set is θ 1 = 0 , θ 2 = 2 π 3 , θ 3 = − 2 π 3 \theta_1=0,\theta_2=\dfrac{2\pi}{3}, \theta_3=-\dfrac{2\pi}{3} , ⇒ sin ⁡ 2 ( θ 1 ) + sin ⁡ 2 ( θ 2 ) + sin ⁡ 2 ( θ 3 ) = 3 2 \Rightarrow \sin^2\left(\theta_1\right)+\sin^2\left(\theta_2\right)+\sin^2\left( \theta_3\right) =\boxed{\dfrac{3}{2}}

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A tricky solution:

Choose θ 1 = − 2 π 3 , θ 2 = 0 , θ 3 = 2 π 3 \,\theta_1=-\dfrac{2\pi}{3},\theta_2=0,\theta_3=\dfrac{2\pi}{3} , we get: sin ⁡ 2 ( θ 1 ) + sin ⁡ 2 ( θ 2 ) + sin ⁡ 2 ( θ 3 ) = 3 2 \sin^2\left(\theta_1\right)+\sin^2\left(\theta_2\right)+\sin^2\left( \theta_3\right) =\boxed{\dfrac{3}{2}}

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But how do you know that this value must be true for all possible sets of θ i \theta_i that satisfy the conditions?

Calvin Lin Staff - 5 years, 10 months ago

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