If a, b and c are integers not all simultaneously equal and w≠1 is a cube root of unity, then the minimum value of ∣a+bw+cw^2 ∣ is (A) 0 (B) 1 (C) (√3)/2 (D) 1/2


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Question Number 139590 by EnterUsername last updated on 29/Apr/21

$If a, b and c are integers not all simultaneously equal$ $and w \neq 1 is a cube root of unity, then the minimum$ $value of ∣ a + b w + {c w}^{2} ∣ is$ $(A) 0 (B) 1 (C) \sqrt{3} / 2 (D) 1 / 2$
	Answered by MJS_new last updated on 29/Apr/21

	$∣ a + b w + {c w}^{2} ∣=$ $[w = - \frac{1}{2} \pm \frac{\sqrt{3}}{2} i]$ $= \sqrt{a^{2} + b^{2} + c^{2} - (a b + a c + b c)} = A (a, b, c)$ $\frac{d}{d c} [A (a, b, c)] = 0 \Rightarrow c = \frac{a + b}{2}$ $A (a, b, \frac{a + b}{2}) = \frac{\sqrt{3}}{2} ∣ a - b ∣$ $since b = a \Rightarrow c = a is not allowed and a, b \in Z$ $the minimum is at ∣ a - b ∣= 1 \Rightarrow answer is \frac{\sqrt{3}}{2}$
	Commented by EnterUsername last updated on 29/Apr/21

	$Thanks Sir . I^{'} ve understood your method . But {author}^{'} s$ $choice is (B) . Do you think {it}^{'} s a mistake ?$
	Commented by EnterUsername last updated on 29/Apr/21

	$For a = 2, and b = c = 1 we get 1 .$
	Commented by MJS_new last updated on 29/Apr/21

	$oops! I forgot that c = \frac{a + b}{2} \in Z \Rightarrow 2 ∣ (a \pm b)$ $\Rightarrow ∣ a - b ∣\neq 1 so my conclusion is wrong$
	Commented by EnterUsername last updated on 29/Apr/21

	$OK . Thank you Sir$ 😃
	Commented by mr W last updated on 29/Apr/21

	$t h e q u e s t i o n s a i d ‘ ‘ a, b, c a r e n o t a l l$ ${e q u a l}^{″} . t h a t m e a n s t w o o f t h e m m a y$ $b e e q u a l .$ $∣ . . . ∣= \sqrt{a^{2} + b^{2} + c^{2} - (a b + b c + c a)}$ $= \frac{\sqrt{{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}}}{\sqrt{2}}$ $e a c h o f t h e {(. . .)}^{2} c a n b e 0, 1, 4, 9 e t c .$ $s u c h t h a t m i n i m u m o c c u r s, t h e y$ $s h o u l d b e a s s m a l l a s p a s s i b l e .$ $i f t w o o f a, b, c m a y b e e q u a l, t h e n$ $∣ . . . ∣_{m i n} = \frac{\sqrt{0^{2} + 1^{2} + 1^{2}}}{\sqrt{2}} = 1 (\leftarrow a n s w e r)$ $(e . g . a = b = 2, c = 3)$ $i f a l l o f a, b, c m a y n o t b e e q u a l, t h e n$ $∣ . . . ∣_{m i n} = \frac{\sqrt{1^{2} + 1^{2} + 2^{2}}}{\sqrt{2}} = \sqrt{3}$ $(e . g . a = 2, b = 3, c = 4)$
	Commented by EnterUsername last updated on 29/Apr/21

	$Thank you Sir!$
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