Question

The maximum area of the triangle formed by the complex coordinates $z,z_1,z_2$ which satisfy the relations $|z-z_1|=|z-z_2|$ and $|z-(z_1+z_2\left)/2\right|\le r$ where $r>|z_1-z_2|$, is

• A. $\frac{1}{2}{|z_1-z_2|}^2$
• B. $\frac{1}{2}|z_1-z_2|r$
• C. $\frac{1}{2}{|z_1-z_2|}^2{r}^2$
• D. $\frac{1}{2}|z_1-z_2|r^2$

Answer 1

Answer: (B)

$\frac{1}{2}|z_1-z_2|r$

Given that $|z-z_1|=|z-z_2|$

and $|z-\frac{(z_1+z_2)}{2}|=r$ volume $r>|z_1-z_2|$

$\Rightarrow$ Let $z=x+iy$ $z_1=x_1+{iy}_1$ $z_2=x_2+{iy}_2$

we have $|z_1-z_1|=|z-z_2|$

${\Rightarrow |(x-x_1)+i(y-y_1)|}^2={|(x-x_2)+i(y-y_2)|}^2$

$\Rightarrow {(x-x_1)}^2+{(y-y_1)}^2={(x-x_2)}^2+{(y-y_2)}^2$

$\Rightarrow x^2-2x{x}_1+x_1^2+y^2-2y{y}_1+y_1^2=x^2-2x{x}_2+x_2^2+y^2-2y{y}_2+y_2^2$

$\Rightarrow x_1^2+y_1^2-2x{x}_1-2y{y}_1=x_2^2+y_2^2-2x{x}_2-2y{y}_2$

$\Rightarrow -2x{x}_1-2y{y}_1+2x{x}_2+2y{y}_2=x_2^2+y_2^2-x_1^2-y_1^2$

$\therefore 2x(x_2-x_1)+2y(y_2-y_1)=x_2^2-x_1^2+y_2^2-y_1^2⟶\left(1\right)$

Given $|z-\frac{(z_1+z_2)}{2}|\le r$

$\Rightarrow |z-\frac{(z_1+z_2)}{2}|\le |z_1-z_2|$

$\Rightarrow |(x+iy)-\frac{x_1+{iy}_1+x_2+{iy}_2}{2}|\le |x_1+{iy}_1-x_2-2{iy}_2|$

$\Rightarrow |2x+2iy-x_1+{iy}_1-x_2-2{iy}_2|\le 2|x_1-x_2+i(y_1-y_2)|$

$\Rightarrow |2x-x_1-x_2+i(2y-y_1-y_2)|\le 2|x_1-x_2+i(y_1-y_2)|⟶\left(2\right)$

From $\left(1\right)$

$\Rightarrow 2x(x_2-x_1)+2y(y_2-y_1)=(x_2-x_1)(x_2+x_1)+(y_2-y_1)(y_2+y_1)$

$\Rightarrow 2x(x_2-x_1)-(x_2-x_1)(x_2+x_1)=(y_2-y_1)(y_2+y_1)-2y(y_2-y_1)$

$\therefore (x_2-x_1)[{2x-x}_2-x_1]=(y_2-y_1)[y_2{+y}_1-2y]⟶\left(3\right)$

Substitute $\left(3\right)$ in $\left(2\right)$

$\Rightarrow |(x_2-x_1)+i(y_2-y_1)|\le 2|x_1-x_2+i(y_1-y_2)|$

$\Rightarrow |x_2-i{y}_2-x_1+i{y}_1|\le 2|x_1-{iy}_1-x_2-{iy}_2|$

$\Rightarrow |z_2-z_1|\le 2|z_1-z_2|$

$\Rightarrow |z_2-z_1|\le 2x$

$=\frac{1}{2}|z_2-z_1|r$

Hence, the answer is $\frac{1}{2}|z_2-z_1|r.$