Question

If $p=a+bw+c{w}^2,q=b+cw+a{w}^2$ and $r=c+aw+b{w}^2$ where a, b, c $\ne$ and w is the complex cube root of unity, then

• A. If p, q, r lie on the circle |z| = 2, the triangle formed by these points is equilateral
• B. $p^2+q^2+r^2=a^2+b^2+c^2$
• C. $p^2+q^2+r^2=2(pq+qr+rp)$
• D. none of these

Answer 1

Answer: (A), (C)

$p^2+q^2+r^2=2(pq+qr+rp)$
If p, q, r lie on the circle |z| = 2, the triangle formed by these points is equilateral

$p+q+r=a+b\omega +c{\omega }^2$

$+b+c\omega +a{\omega }^2$

$+c+a\omega +b{\omega }^2$

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$\therefore p+q+r=(a+b+c)(1+\omega +{\omega }^2)=0$ (1)

p, q, r lie on the circle $\mid z\mid$ = 2, whose circumcentre is origin. Also, (p + q + r)/3 = 0. Hence the centroid coincides with circumcentre. So, the triangle is equilateral. Now,

$(p+q+r{)}^2=0$

$\Rightarrow p^2+q^2+r^2=-2pqr[\frac{1}{p}+\frac{1}{q}+\frac{1}{r}]$

$=-2pqr[\frac{1}{a+b\omega +c{\omega }^2}+\frac{1}{b+c\omega +a{\omega }^2}+\frac{1}{c+a\omega +b{\omega }^2}]$

$=-2pqr[\frac{1}{{\omega }^2(a\omega +b{\omega }^2+c)}+\frac{1}{\omega (b{\omega }^2+c+a\omega )}+\frac{1}{c+a\omega +b{\omega }^2}]$

$=\frac{-2pqr}{a\omega +b{\omega }^2+c}[\frac{1}{{\omega }^2}+\frac{1}{\omega }+\frac{1}{1}]=0$

Hence,

$p^2+q^2+r^2=2(pq+qr+rp)$