InterviewSolution
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InterviewSolution
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Let z_(1),z_(2),z_(3) be the three nonzero comple numbers such thatz_(2) ne 1, a= |z_(1)|, b = |z_(2)| and c= |z_(3)|. Let |{:(,a,b,c),(,b,c,a),(,c,a,b):}|= 0 Then |
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Answer» `arg((z_(3))/(z_(2))) = arg((z_(3)-z_(1))/(z_(2)-z_(1)))` `rArr a^(3) + b^(3) + c^(3) - 3abc = 0` `rArr (a+b+c)(a^(2) + b^(2) + c^(2) - ab -ab-ca)=0` `rArr (1)/(2)(a+b+c)[(a-b)^(2) + (c-a^(2)) = 0` `rArr (a-b)^(2) + (b-c)^(2) + (c-a)^(2) = 0` `rArr a=b=c [ therefore a+b + c ne 0, because z_(1) ne 0, therefore|z_(1)| = a ne 0 etc.]` Hence, `OA= OB = OC`, where O is the ORIGIN and A, B, C arethe points represeniting `z_(1),z_(2)` and `z_(3)` respectively . Therefore ,O is circumcenter of `Delta ABC `. Now  `arg((z_(3))/(z_(2)))= /_BOC""(1)` `=2 /_BAC = 2 arg((z_(3)-z_(1))/(z_(2)-z_(1))) ""(2)` `arg ((z_(3)-z_(1))/(z_(2)-z_(1)))""[because /_BOC = 2/_ BAC]` Hence, `arg((z_(3))/(z_(2))) =arg((z_(3)-z_(1))/(z_(2)-z_(1)))^(2)` Also, centroid is `(z_(1) +z_(2) +z_(3))//3`. Since HG: GO -= 2:1 (where H isorthocenter and G is centroid),then orthocentre is `z_(1) + z_(2) +z_(3)` (by sectionformula). When triangle is equilateral centroid conicides with circumcenter, hence `z_(1) + z_(2) +z_(3)= 0` Also, the area for equilateral triangle is `(sqrt(3)//4L)`, where Lis lenghtof side. Since radius is `|z_(1)|,L = sqrt(3)|z_(1)|`, the area is `(3sqrt(3)//4)|z_(1)|` |
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