Question

If $a$ and $b$ are real number between $0$ and $1$ such that the point $(a,1),(1,b)$ and $(0,0)$ from an equilateral triangle find $a$ and $b$.

Answer 1

A] Let $z^1=a+i$

$x^2=1+bi$

$z^3=0$

$\therefore {12}^1-z{21}^2=(a-1{)}^2+(1-b{)}^2$

${12}^2-z{31}^2=1+b^2$

${12}^3-z{11}^2=a^2+1$

These are equal

$\Rightarrow 1+b^2=a^2+1$

$\Rightarrow b^2=a^2$

$\Rightarrow a=b$

$1+a^2=(a-1{)}^2+(1-b{)}^2$

$\Rightarrow 1+a^2=(a-1{)}^2+(1-b^2)$

$\Rightarrow 1+a^2=2(a-1{)}^2$

$\Rightarrow 1+a^2=2a^2-4a+2$

$\Rightarrow 0=a^2-4a+1$

$\therefore a=2\pm \sqrt{3}$

For the desired range,

$a=b=2-\sqrt{3}$