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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-

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Question Number 20551 by Tinkutara last updated on 28/Aug/17
Find the minimum value of ∣a + bω + cω^2 ∣, where a, b and c are all not equal integers and ω(≠1) is a cube root of unity.
$${Find}\:{the}\:{minimum}\:{value}\:{of} \\ $$$$\mid{a}\:+\:{b}\omega\:+\:{c}\omega^{\mathrm{2}} \mid,\:{where}\:{a},\:{b}\:{and}\:{c}\:{are}\:{all} \\ $$$${not}\:{equal}\:{integers}\:{and}\:\omega\left(\neq\mathrm{1}\right)\:{is}\:{a}\:{cube} \\ $$$${root}\:{of}\:{unity}. \\ $$
Commented by ajfour last updated on 28/Aug/17
Commented by ajfour last updated on 28/Aug/17
let d be the minimum magnitude. (d/(sin 60°))+2((b/2))=a ...(i) and (d/(tan 60°))+b=c ...(ii) ⇒ ((2d)/( (√3)))=a−b and (d/( (√3)))=c−b d_(min) =(√3)(c−b) = (√3) as c, b are unequal integers; their minimum difference is 1 .
$${let}\:\boldsymbol{{d}}\:{be}\:{the}\:{minimum}\:{magnitude}. \\ $$$$\frac{{d}}{\mathrm{sin}\:\mathrm{60}°}+\mathrm{2}\left(\frac{{b}}{\mathrm{2}}\right)={a}\:\:\:\:…\left({i}\right) \\ $$$${and}\:\:\frac{{d}}{\mathrm{tan}\:\mathrm{60}°}+{b}={c}\:\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow\:\:\:\:\frac{\mathrm{2}{d}}{\:\sqrt{\mathrm{3}}}={a}−{b}\:\:\:{and}\:\:\:\frac{{d}}{\:\sqrt{\mathrm{3}}}={c}−{b} \\ $$$$\:\:\boldsymbol{{d}}_{\boldsymbol{{min}}} =\sqrt{\mathrm{3}}\left(\boldsymbol{{c}}−\boldsymbol{{b}}\right)\:=\:\sqrt{\mathrm{3}} \\ $$$${as}\:{c},\:{b}\:{are}\:{unequal}\:{integers};\:{their} \\ $$$${minimum}\:{difference}\:{is}\:\mathrm{1}\:. \\ $$
Commented by Tinkutara last updated on 28/Aug/17
You all got (√3), but in my book answer is explained and they got 1. Can you find the error in my book?
$$\mathrm{You}\:\mathrm{all}\:\mathrm{got}\:\sqrt{\mathrm{3}},\:\mathrm{but}\:\mathrm{in}\:\mathrm{my}\:\mathrm{book}\:\mathrm{answer} \\ $$$$\mathrm{is}\:\mathrm{explained}\:\mathrm{and}\:\mathrm{they}\:\mathrm{got}\:\mathrm{1}.\:\mathrm{Can}\:\mathrm{you} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{error}\:\mathrm{in}\:\mathrm{my}\:\mathrm{book}? \\ $$
Commented by Tinkutara last updated on 28/Aug/17
Commented by dioph last updated on 28/Aug/17
I guess we assumed a≠b, a≠c and b≠c, when in fact the book says two of them can be equal (just not all of them at once)
$$\mathrm{I}\:\mathrm{guess}\:\mathrm{we}\:\mathrm{assumed}\:{a}\neq{b},\:{a}\neq{c}\:\mathrm{and} \\ $$$${b}\neq{c},\:\mathrm{when}\:\mathrm{in}\:\mathrm{fact}\:\mathrm{the}\:\mathrm{book}\:\mathrm{says} \\ $$$$\mathrm{two}\:\mathrm{of}\:\mathrm{them}\:\mathrm{can}\:\mathrm{be}\:\mathrm{equal}\:\left(\mathrm{just}\:\mathrm{not}\right. \\ $$$$\left.\mathrm{all}\:\mathrm{of}\:\mathrm{them}\:\mathrm{at}\:\mathrm{once}\right) \\ $$
Commented by Tinkutara last updated on 28/Aug/17
So 1 is correct?
$$\mathrm{So}\:\mathrm{1}\:\mathrm{is}\:\mathrm{correct}? \\ $$
Commented by ajfour last updated on 28/Aug/17
question states ′all not equal′ so 1 is true.for example, a−1=b=c.
$${question}\:{states}\:'{all}\:{not}\:{equal}' \\ $$$${so}\:\mathrm{1}\:{is}\:{true}.{for}\:{example}, \\ $$$${a}−\mathrm{1}={b}={c}. \\ $$
Commented by Tinkutara last updated on 28/Aug/17
Thank you very much ajfour Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{ajfour}\:\mathrm{Sir}! \\ $$
Answered by dioph last updated on 28/Aug/17
∣(a−(1/2)b−(1/2)c)+i(((√3)/2)b−((√3)/2)c)∣= =(√((a−(1/2)b−(1/2)c)^2 +(((√3)/2)b−((√3)/2)c)^2 ))= =(√(a^2 +(1/4)b^2 +(1/4)c^2 −ab−ac+(1/2)bc+(3/4)b^2 +(3/4)c^2 −(3/2)bc))= =(√(a^2 +b^2 +c^2 −ab−ac−bc)) min((√(a^2 +b^2 +c^2 −ab−ac−bc)))= = (√(min(a^2 +b^2 +c^2 −ab−ac−bc))) 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 −2b+1+b^2 +b^2 +2b+1−(b^2 −b+b^2 −1+b^2 +b)= = 3b^2 +2−(3b^2 −1) = 3. Case 2: (a,b,c)≠(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 −(ab+ac) > (b−1)^2 −(b−1)(b+c) a^2 +b^2 +c^2 −(ab+ac+bc) > (b−1)^2 +b^2 +c^2 −(b(b−1)+c(b−1)+bc) (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 −(ac+bc) > (b+1)^2 −(b+1)(a+b) a^2 +b^2 +c^2 −(ab+ac+bc) > a^2 +b^2 +(b+1)^2 −(ab+a(b+1)+b(b+1)) (a,b,c) is not optimal. Therefore, the answer is (√3).
$$\mid\left({a}−\frac{\mathrm{1}}{\mathrm{2}}{b}−\frac{\mathrm{1}}{\mathrm{2}}{c}\right)+{i}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{b}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{c}\right)\mid= \\ $$$$=\sqrt{\left({a}−\frac{\mathrm{1}}{\mathrm{2}}{b}−\frac{\mathrm{1}}{\mathrm{2}}{c}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{b}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{c}\right)^{\mathrm{2}} }= \\ $$$$=\sqrt{{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}{b}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}{c}^{\mathrm{2}} −{ab}−{ac}+\frac{\mathrm{1}}{\mathrm{2}}{bc}+\frac{\mathrm{3}}{\mathrm{4}}{b}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}{c}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}{bc}}= \\ $$$$=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{ac}−{bc}} \\ $$$$\mathrm{min}\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{ac}−{bc}}\right)= \\ $$$$=\:\sqrt{\mathrm{min}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{ac}−{bc}\right)} \\ $$$${a},{b},{c}\:\mathrm{are}\:\mathrm{interchangeable},\:\mathrm{so} \\ $$$$\mathrm{assume}\:{a}<{b}<{c}. \\ $$$$\boldsymbol{\mathrm{Case}}\:\mathrm{1}:\:\left({a},{b},{c}\right)=\left({b}−\mathrm{1},{b},{b}+\mathrm{1}\right) \\ $$$$\left({b}−\mathrm{1}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} +\left({b}+\mathrm{1}\right)^{\mathrm{2}} −\left({b}\left({b}−\mathrm{1}\right)+\left({b}−\mathrm{1}\right)\left({b}+\mathrm{1}\right)+{b}\left({b}+\mathrm{1}\right)\right)= \\ $$$$=\:{b}^{\mathrm{2}} −\mathrm{2}{b}+\mathrm{1}+{b}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{b}+\mathrm{1}−\left({b}^{\mathrm{2}} −{b}+{b}^{\mathrm{2}} −\mathrm{1}+{b}^{\mathrm{2}} +{b}\right)= \\ $$$$=\:\mathrm{3}{b}^{\mathrm{2}} +\mathrm{2}−\left(\mathrm{3}{b}^{\mathrm{2}} −\mathrm{1}\right)\:=\:\mathrm{3}. \\ $$$$\boldsymbol{\mathrm{Case}}\:\mathrm{2}:\:\left({a},{b},{c}\right)\neq\left({k}−\mathrm{1},{k},{k}+\mathrm{1}\right) \\ $$$$\boldsymbol{\mathrm{Case}}\:\mathrm{2}.\mathrm{1}:\:{a}\:<\:{b}−\mathrm{1} \\ $$$${a}−{b}−{c}\:<\:\left({b}−\mathrm{1}\right)−{b}−{c} \\ $$$${a}\left({a}−{b}−{c}\right)\:>\:\left({b}−\mathrm{1}\right)\left(\left({b}−\mathrm{1}\right)−{b}−{c}\right) \\ $$$${a}^{\mathrm{2}} −\left({ab}+{ac}\right)\:>\:\left({b}−\mathrm{1}\right)^{\mathrm{2}} −\left({b}−\mathrm{1}\right)\left({b}+{c}\right) \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\left({ab}+{ac}+{bc}\right)\:>\:\left({b}−\mathrm{1}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\left({b}\left({b}−\mathrm{1}\right)+{c}\left({b}−\mathrm{1}\right)+{bc}\right) \\ $$$$\left({a},{b},{c}\right)\:\mathrm{is}\:\mathrm{not}\:\mathrm{optimal}. \\ $$$$\boldsymbol{\mathrm{Case}}\:\mathrm{2}.\mathrm{2}:\:{c}\:>\:{b}+\mathrm{1} \\ $$$${c}−{a}−{b}\:>\:\left({b}+\mathrm{1}\right)−{a}−{b} \\ $$$${c}\left({c}−{a}−{b}\right)\:>\:\left({b}+\mathrm{1}\right)\left(\left({b}+\mathrm{1}\right)−{a}−{b}\right) \\ $$$${c}^{\mathrm{2}} −\left({ac}+{bc}\right)\:>\:\left({b}+\mathrm{1}\right)^{\mathrm{2}} −\left({b}+\mathrm{1}\right)\left({a}+{b}\right) \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\left({ab}+{ac}+{bc}\right)\:>\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\left({b}+\mathrm{1}\right)^{\mathrm{2}} −\left({ab}+{a}\left({b}+\mathrm{1}\right)+{b}\left({b}+\mathrm{1}\right)\right) \\ $$$$\left({a},{b},{c}\right)\:\mathrm{is}\:\mathrm{not}\:\mathrm{optimal}. \\ $$$$\mathrm{Therefore},\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\sqrt{\mathrm{3}}. \\ $$
Commented by Tinkutara last updated on 28/Aug/17
Thank you very much Sir! This question can also be asked where all are unequal.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}!\:\mathrm{This} \\ $$$$\mathrm{question}\:\mathrm{can}\:\mathrm{also}\:\mathrm{be}\:\mathrm{asked}\:\mathrm{where}\:\mathrm{all} \\ $$$$\mathrm{are}\:\mathrm{unequal}. \\ $$

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