Complex Number Geometry

Question

Problem: There exist two complex numbers $c$, say $c_1$ and $c_2$, so that $2+2i$, $5+i$, and $c$ form the vertices of an equilateral triangle. Find the product $c_1c_2$.

Diagram http://cache.artofproblemsolving.com/texer/images/7efb0d210992c0ebb7708a84f468eadd198ad8f8.png

I've been struggling with this problem for a while, please help. Thanks.

EDIT: The problem has been solved but if you are still interested, see if you can solve the problem without finding $c_1$ or $c_2$.

Answer 1

If $c_1(a,b)$

We have $$(a-2)^2+(b-2)^2=(5-2)^2+(1-2)^2=(a-5)^2+(b-1)^2$$

$$(a-2)^2+(b-2)^2=(a-5)^2+(b-1)^2$$ will give us a linear relationship between $a,b$

which can be used in $$(5-2)^2+(1-2)^2=(a-5)^2+(b-1)^2$$ or in $$(a-2)^2+(b-2)^2=(5-2)^2+(1-2)^2$$ to find $a,b$

Observe that there two sets of values of $(a,b)$ corresponding to $c_1,c_2$

Answer 2

Put $z_1=2+2i,z_2=5+i$, then $z_2-z_1=3-i$. If you multiply this by $\lambda=\frac{1}{2}+\frac{\sqrt3}{2}i$, then you rotate it anticlockwise through $60^o$. So $z_1+\lambda(z_2-z_1)=c_1$. Similarly for $c_2$.

Answer 3

It's easy to see that $z_1$, $z_2$, $z_3$ are vertices of an equilateral triangle if and only if $z_1$, $z_2$, $z_3$ are the roots of a cubic equation of form $$(z-a)^3 =b$$ for some $a$, $b$ in $\mathbb{C}$. The equation above is equivalent to $z^3 - 3a z^2 + 3 a^2 z - a^3 - b =0$. By Viete's relations $$a = \frac{1}{3}( z_1 + z_2 + z_3)$$ and $$a^2 = \frac{1}{3}(z_1 z_2 + z_1 z_3 + z_2 z_3)$$ Therefore, $z_1$, $z_2$, $z_3$ are the vertices of an equilateral triangle if and only if $$(\frac{1}{3}( z_1 + z_2 + z_3))^2 = \frac{1}{3}(z_1 z_2 + z_1 z_3 + z_2 z_3)$$ or, equivalently $$ z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_1 z_3 + z_2 z_3$$ Fix $z_2$ and $z_3$, this becomes a quadratic equation in $z_1$ $$z_1^2 - (z_2+z_3)z_1 + z_2^2 + z_3^2 - z_2 z_3$$ The product of the roots is therefore $z_2^2 + z_3^2 - z_2 z_3$.