Question

Let $z_1,z_2$ and $z_3$ be non-zero complex numbers representing vertices of $△ABC$ and $|z_1-z_2|=a,$
$|z_2-z_3|=b,$ $|z_3-z_1|=c.$ If $P=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$ is singular and $Q=\left[\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right],$ then

• A. area of circumcircle of $△ABC$ is two times the area of incircle of $△ABC$
• B. area of circumcircle of $△ABC$ is four times the area of incircle of $△ABC$
• C. $det(adj\left(adjQ\right))=(abc{)}^4$
• D. $det(adj\left(adjQ\right))=0$

Answer 1

Answer: (B), (D)

$det(adj\left(adjQ\right))=0$

$P=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$

$|P|=-(a^3+b^3+c^3-3abc)=0\Rightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$
$\Rightarrow \frac{1}{2}(a+b+c)\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]=0$
$a,b,c>0$
So, $a+b+c\ne 0$
$\therefore a=b=c$
Hence, $△ABC$ is equilateral.
$\therefore$ Circumradius $=2\times \text{inradius}$
$\therefore$ Area of circumcircle $=4\times \text{area of incircle}$

$|Q|=(a-b)(b-c)(c-a)=0$