Find-the-minimum-value-of-a-b-c-2-where-a-b-and-c-are-all-not-equal-integers-and-1-is-a-cube-root-of-unity-

Question Number 20551 by Tinkutara last updated on 28/Aug/17

F i n d t h e m i n i m u m v a l u e o f

∣ a + b ω + c ω^{2} ∣, w h e r e a, b a n d c a r e a l l

n o t e q u a l i n t e g e r s a n d ω (\neq 1) i s a c u b e

r o o t o f u n i t y .

Commented by ajfour last updated on 28/Aug/17

Commented by ajfour last updated on 28/Aug/17

l e t d b e t h e m i n i m u m m a g n i t u d e .

\frac{d}{\sin 60 °} + 2 (\frac{b}{2}) = a \dots (i)

a n d \frac{d}{\tan 60 °} + b = c \dots (i i)

\Rightarrow \frac{2 d}{\sqrt{3}} = a - b a n d \frac{d}{\sqrt{3}} = c - b

d_{m i n} = \sqrt{3} (c - b) = \sqrt{3}

a s c, b a r e u n e q u a l i n t e g e r s; t h e i r

m i n i m u m d i f f e r e n c e i s 1 .

Commented by Tinkutara last updated on 28/Aug/17

You all got \sqrt{3}, but in my book answer

is explained and they got 1 . Can you

find the error in my book ?

Commented by Tinkutara last updated on 28/Aug/17

Commented by dioph last updated on 28/Aug/17

I guess we assumed a \neq b, a \neq c and

b \neq c, when in fact the book says

two of them can be equal (just not

all of them at once)

Commented by Tinkutara last updated on 28/Aug/17

So 1 is correct ?

Commented by ajfour last updated on 28/Aug/17

q u e s t i o n s t a t e s^{'} a l l n o t {e q u a l}^{'}

s o 1 i s t r u e . f o r e x a m p l e,

a - 1 = b = c .

Commented by Tinkutara last updated on 28/Aug/17

Thank you very much ajfour Sir!

Answered by dioph last updated on 28/Aug/17

∣ (a - \frac{1}{2} b - \frac{1}{2} c) + i (\frac{\sqrt{3}}{2} b - \frac{\sqrt{3}}{2} c) ∣=

= \sqrt{{(a - \frac{1}{2} b - \frac{1}{2} c)}^{2} + {(\frac{\sqrt{3}}{2} b - \frac{\sqrt{3}}{2} c)}^{2}} =

= \sqrt{a^{2} + \frac{1}{4} b^{2} + \frac{1}{4} c^{2} - a b - a c + \frac{1}{2} b c + \frac{3}{4} b^{2} + \frac{3}{4} c^{2} - \frac{3}{2} b c} =

= \sqrt{a^{2} + b^{2} + c^{2} - a b - a c - b c}

\min (\sqrt{a^{2} + b^{2} + c^{2} - a b - a c - b c}) =

= \sqrt{\min (a^{2} + b^{2} + c^{2} - a b - a c - b c)}

a, b, c are interchangeable, so

assume a < b < c .

Case 1 : (a, b, c) = (b - 1, b, b + 1)

{(b - 1)}^{2} + b^{2} + {(b + 1)}^{2} - (b (b - 1) + (b - 1) (b + 1) + b (b + 1)) =

= b^{2} - 2 b + 1 + b^{2} + b^{2} + 2 b + 1 - (b^{2} - b + b^{2} - 1 + b^{2} + b) =

= 3 b^{2} + 2 - (3 b^{2} - 1) = 3 .

Case 2 : (a, b, c) \neq (k - 1, k, k + 1)

Case 2.1 : a < b - 1

a - b - c < (b - 1) - b - c

a (a - b - c) > (b - 1) ((b - 1) - b - c)

a^{2} - (a b + a c) > {(b - 1)}^{2} - (b - 1) (b + c)

a^{2} + b^{2} + c^{2} - (a b + a c + b c) > {(b - 1)}^{2} + b^{2} + c^{2} - (b (b - 1) + c (b - 1) + b c)

(a, b, c) is not optimal .

Case 2.2 : c > b + 1

c - a - b > (b + 1) - a - b

c (c - a - b) > (b + 1) ((b + 1) - a - b)

c^{2} - (a c + b c) > {(b + 1)}^{2} - (b + 1) (a + b)

a^{2} + b^{2} + c^{2} - (a b + a c + b c) > a^{2} + b^{2} + {(b + 1)}^{2} - (a b + a (b + 1) + b (b + 1))

(a, b, c) is not optimal .

Therefore, the answer is \sqrt{3} .

Commented by Tinkutara last updated on 28/Aug/17

Thank you very much Sir! This

question can also be asked where all

are unequal .