Question

If $a$ and $b$ are real numbers between 0 and 1 such that the points $z_1=a+i,z_2=1+bi$ and $z_3=0$ form an equilateral triangle, then

• A. $a=b=2-\sqrt{3}$ should look like this
• B. $a=2-\sqrt{3},b=\sqrt{3}-1$
• C. $a=\sqrt{3}-1,b=2-\sqrt{3}$
• D. none of these

Answer 1

Answer: (A)

Solution: By the hypothesis $0<a,b<1$
and $\left|z_1-z_2\right|=\left|z_2-z_3\right|=\left|z_3-z_1\right|$
$\Rightarrow |(a-1)+i(1-b)|=|1+ib|=|a+i|$
$\Rightarrow (a-1{)}^2+(1-b{)}^2=1+b^2=a^2+1$
$\Rightarrow a^2-2a+1-2b=0\text{ and }{b}^2=a^2$
As $0<a,b<1$ and $a^2=b^2$, we get $a=b$.
$\therefore a^2-2a+1-2a=0\Rightarrow a^2-4a+1=0$
$\Rightarrow a=2\pm \sqrt{3}\text{ As }0<a<1,a=2-\sqrt{3}.$