Third point of a triangle in the complex plane

Question

I have an equilateral triangle with two points equal to $(2+2i)$ and $(5+i)$. I want to find the third point(s) (there are $2$ of these). I have that the side length of the triangle is $\sqrt{10}$.

Answer 1

A different Hint: if $A,B\in\mathbb{C}$ and $\omega=e^{\frac{\pi i}{3}}=\frac{1+\sqrt{-3}}{2}$, the points you are looking for are given by: $$ A+(B-A)\omega, \qquad A+(B-A)\omega^{-1}.$$ This happens because the multiplication by $\omega$ ($\omega^{-1}$) acts like a $60^\circ$ ($-60^\circ$) rotation.

Answer 2

Hint: You already know that $\big|z-(5+i)\big|=\big|z-(2+2i)\big|=\ldots$ (you are to fulfil the gap), where $z=x+yi$ the complex number(s) you are looking for. Plus, you have an equilateral triangle, which means all sides are equal!

Answer 3

Yet another hint: Why not work in $\Bbb R^2$ with vectors? You have the points $A = (2,2)$ and $B = (5,1)$, and the length $\ell = \sqrt{10}$. Find the perpendicular bisector of $\overline{AB}$ and go $h = \frac{\ell\sqrt{3}}{2} = \frac{\sqrt{30}}{2}$ in both senses.

Answer 4

(I'm purposely not trying to be clever here and doing all calculations in my head.)

The distance $d$ between the two known points satisfies $d^2 =(2-5)^2+(2-1)^2 =10 $.

If the point is $(x, y)$, then $(x-2)^2+(y-2)^2 =(x-5)^2+(y-1)^2 =10 $, or $x^2-4x+4+y^2-4y+4 =x^2-10x+25+y^2-2y+1 =10 $, or $x^2-4x+y^2-4y+8 =x^2-10x+y^2-2y+26 =10 $.

Subtracting the first two, $6x-2y-18 = 0$, or $y = 3x-9$.

Substituting this in the first equation, $10 =x^2-4x+(3x-9)^2-4(3x-9)+8 =x^2-4x+9x^2-54x+81-12x+36+8 =10x^2-70x+125 $ or $10x^2-70x+115=0$ or $2x^2-14x+23=0$

Using the good old quadratic equation formuls, $D^2 = b^2-4ac = 14^2-4\cdot 2 \cdot 23 =196-184 =12 $ so $D = 2\sqrt{3}$.

The roots are therefore $x =\dfrac{-b \pm D}{2 a } =\dfrac{14\pm 2\sqrt{3}}{2\cdot 2} =\dfrac{7\pm \sqrt{3}}{2} $ and $y =3x-9 =\dfrac{21\pm 3\sqrt{3}-18}{2} =\dfrac{3\pm 3\sqrt{3}}{2} $.

Note: I find it somewhat embarrassing and dishearting how many errors I made while doing this all in my head.