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

$$Findtheminimumvalueof$$$$\mid a+b\omega +c{\omega }^2\mid ,wherea,bandcareall$$$$notequalintegersand\omega (\ne 1)isacube$$$$rootofunity.$$

Commented by ajfour last updated on 28/Aug/17

Commented by ajfour last updated on 28/Aug/17

$$let\mathit{d}betheminimummagnitude.$$$$\frac{d}{\sin 60°}+2\left(\frac{b}{2}\right)=a...\left(i\right)$$$$and\frac{d}{\tan 60°}+b=c...\left(ii\right)$$$$\Rightarrow \frac{2d}{\sqrt{3}}=a-band\frac{d}{\sqrt{3}}=c-b$$$${\mathit{d}}_{\mathit{m}\mathit{i}\mathit{n}}=\sqrt{3}(\mathit{c}-\mathit{b})=\sqrt{3}$$$$asc,bareunequalintegers;their$$$$minimumdifferenceis1.$$

Commented by Tinkutara last updated on 28/Aug/17

$$\mathrm{You}\mathrm{all}\mathrm{got}\sqrt{3},\mathrm{but}\mathrm{in}\mathrm{my}\mathrm{book}\mathrm{answer}$$$$\mathrm{is}\mathrm{explained}\mathrm{and}\mathrm{they}\mathrm{got}1.\mathrm{Can}\mathrm{you}$$$$\mathrm{find}\mathrm{the}\mathrm{error}\mathrm{in}\mathrm{my}\mathrm{book}?$$

Commented by Tinkutara last updated on 28/Aug/17

Commented by dioph last updated on 28/Aug/17

$$\mathrm{I}\mathrm{guess}\mathrm{we}\mathrm{assumed}a\ne b,a\ne c\mathrm{and}$$$$b\ne c,\mathrm{when}\mathrm{in}\mathrm{fact}\mathrm{the}\mathrm{book}\mathrm{says}$$$$\mathrm{two}\mathrm{of}\mathrm{them}\mathrm{can}\mathrm{be}\mathrm{equal}(\mathrm{just}\mathrm{not}$$$$\mathrm{all}\mathrm{of}\mathrm{them}\mathrm{at}\mathrm{once})$$

Commented by Tinkutara last updated on 28/Aug/17

$$\mathrm{So}1\mathrm{is}\mathrm{correct}?$$

Commented by ajfour last updated on 28/Aug/17

$$questionstates{}^{\prime }allnot{equal}^{\prime }$$$$so1istrue.forexample,$$$$a-1=b=c.$$

Commented by Tinkutara last updated on 28/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{ajfour}\mathrm{Sir}!$$

Answered by dioph last updated on 28/Aug/17

$$\mid (a-\frac{1}{2}b-\frac{1}{2}c)+i(\frac{\sqrt{3}}{2}b-\frac{\sqrt{3}}{2}c)\mid =$$$$=\sqrt{{(a-\frac{1}{2}b-\frac{1}{2}c)}^2+{(\frac{\sqrt{3}}{2}b-\frac{\sqrt{3}}{2}c)}^2}=$$$$=\sqrt{a^2+\frac{1}{4}{b}^2+\frac{1}{4}{c}^2-ab-ac+\frac{1}{2}bc+\frac{3}{4}{b}^2+\frac{3}{4}{c}^2-\frac{3}{2}bc}=$$$$=\sqrt{a^2+b^2+c^2-ab-ac-bc}$$$$\min \left(\sqrt{a^2+b^2+c^2-ab-ac-bc}\right)=$$$$=\sqrt{\min (a^2+b^2+c^2-ab-ac-bc)}$$$$a,b,c\mathrm{are}\mathrm{interchangeable},\mathrm{so}$$$$\mathrm{assume}a<b<c.$$$$\mathbf{Case}1:(a,b,c)=(b-1,b,b+1)$$$${(b-1)}^2+b^2+{(b+1)}^2-(b(b-1)+(b-1)(b+1)+b(b+1))=$$$$=b^2-2b+1+b^2+b^2+2b+1-(b^2-b+b^2-1+b^2+b)=$$$$=3b^2+2-(3b^2-1)=3.$$$$\mathbf{Case}2:(a,b,c)\ne (k-1,k,k+1)$$$$\mathbf{Case}2.1:a<b-1$$$$a-b-c<(b-1)-b-c$$$$a(a-b-c)>(b-1)((b-1)-b-c)$$$$a^2-(ab+ac)>{(b-1)}^2-(b-1)(b+c)$$$$a^2+b^2+c^2-(ab+ac+bc)>{(b-1)}^2+b^2+c^2-(b(b-1)+c(b-1)+bc)$$$$(a,b,c)\mathrm{is}\mathrm{not}\mathrm{optimal}.$$$$\mathbf{Case}2.2:c>b+1$$$$c-a-b>(b+1)-a-b$$$$c(c-a-b)>(b+1)((b+1)-a-b)$$$$c^2-(ac+bc)>{(b+1)}^2-(b+1)(a+b)$$$$a^2+b^2+c^2-(ab+ac+bc)>a^2+b^2+{(b+1)}^2-(ab+a(b+1)+b(b+1))$$$$(a,b,c)\mathrm{is}\mathrm{not}\mathrm{optimal}.$$$$\mathrm{Therefore},\mathrm{the}\mathrm{answer}\mathrm{is}\sqrt{3}.$$

Commented by Tinkutara last updated on 28/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!\mathrm{This}$$$$\mathrm{question}\mathrm{can}\mathrm{also}\mathrm{be}\mathrm{asked}\mathrm{where}\mathrm{all}$$$$\mathrm{are}\mathrm{unequal}.$$

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