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Algebra Level 4

How many values of K K are there, such that there exists distinct complex numbers a , b , a, b, and c c which satisfy a 1 − b = b 1 − c = c 1 − a = K ? \frac{a}{1-b} = \frac{b}{1-c} = \frac{c}{1-a} = K?


The answer is 2.

Vishal Yadav by Vishal Yadav , 20, India

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1 solution

Vishal Yadav
Mar 26, 2017

a 1 − b = k ⟹ a = k − b k ⟹ a = k − k ( k − c k ) [ ∵ b = k − c k ] ⟹ a = k − k ( k − k ( k − a k ) ) [ ∵ c = k − a k ] ⟹ a = k − k 2 − k 3 − a k 3 \frac{a}{1-b} = k \implies a = k - b k \\ \implies a = k - k(k - ck) \quad [\because b = k - ck] \\ \implies a = k - k (k - k(k - a k) ) \quad [\because c = k - ak] \\ \implies a = k - k^2 - k^3 - a k^3 \\

Similarly for b , c ∵ RHS independent of k ⟹ k 3 + 1 = 0 ⟹ k 3 = − 1 which has three solutions: − 1 , ω , ω 2 \\ \text{Similarly for } b, c \\ \\ \\ \because \text{ RHS independent of } k \\ \implies k^3 + 1 = 0 \implies k^3 = -1 \\ \text{which has three solutions: } -1, \omega, \omega^2 \\

Check ‾ For k = − 1 , a + b + c = 3 k − k ( a + b + c ) ⟹ a + b + c = a + b + c − 3 Hence, k ≠ − 1 Thus, two solutions are ω , ω 2 \\ \\ \underline{\text{Check}} \\ \text{For } k = -1, \\ a + b + c = 3 k - k (a + b + c) \\ \implies a + b + c = a + b + c - 3 \\ \\ \text{Hence, } k \neq -1 \\ \\ \text{Thus, two solutions are } \omega, \omega^2

16 Helpful 0 Interesting 0 Brilliant 0 Confused

Note that from "RHS independent of K", we not only have 1 + k 3 = 0 1 + k^3 = 0 , we also must have k − k 2 − k 3 = 0 k - k^2 - k^3 = 0 . This is why we have to reject k = − 1 k = -1 .

Calvin Lin Staff - 4 years, 2 months ago

The solution is correct but wrongly derived. It should be as below: a ( K 3 + 1 ) = K − K 2 + K 3 a(K^3 + 1) = K - K^2 + K^3 {not K − K 2 − K 3 K - K^2 - K^3 as written in the solution} ⇒ a ( K + 1 ) ( K 2 − K + 1 ) − K ( K 2 − K + 1 ) = 0 \Rightarrow a(K+1)(K^2 - K + 1) - K(K^2 - K + 1) = 0 ⇒ ( K 2 − K + 1 ) [ a ( K + 1 ) − K ] = 0 \Rightarrow (K^2 - K + 1) [a(K + 1) - K] = 0 Hence either K 2 − K + 1 = 0 K^2 - K + 1 = 0 or a ( K + 1 ) − K = 0 a(K + 1) - K = 0

If a ( K + 1 ) − K = 0 a(K + 1) - K = 0 then a = K K + 1 a = \frac{K}{K + 1} . So the equations are satisfied for all values of K K except [-1]. But then we solve for b b and c c using above value of a a and we get b = K K + 1 b = \frac{K}{K + 1} and c = K K + 1 c = \frac{K}{K + 1} . Hence a , b , a, b, and c c are all equal and not distinct and this is not valid solution.

If K 2 − K + 1 = 0 K^2 - K + 1 = 0 then K = − ω , − ω 2 K = -\omega, -\omega^2 So there are two values of K K for which the mentioned conditions are satisfied.

Manish Tiwari - 4 years, 2 months ago

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I solved it in the same way.Moreover we have a^2 +b^2 +c^2=ab+bc + ca and a,b,c are distinct complex numbers, also seen to be non collinear, hence a,b,c are vertices of a non degenerate equilateral triangle in the complex plane...

vinod trivedi - 4 years, 2 months ago

Use the Add image button:

Agnishom Chattopadhyay - 4 years, 2 months ago

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I'm on Android.. No such option.

Vishal Yadav - 4 years, 2 months ago

how rhs is independ of K??

Nivedit Jain - 4 years, 2 months ago

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My mistake.. It actually says R.H.S is independent of a,b,c.

Vishal Yadav - 4 years, 2 months ago

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ok. Nice question Bro

Nivedit Jain - 4 years, 2 months ago

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