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-

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 .

∣ \dots ∣= \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 {(\dots)}^{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

∣ \dots ∣_{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

∣ \dots ∣_{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!