Complex Equilateral triangle

Geometry Level 2

Let there be an equilateral triangle on the complex plane with vertices $z_1,z_2,$ and $z_3$ . Let the circumcenter of the triangle be $z_0$ .

If $z_0\ne 0$ , find the value of

$\frac{(z_1)^2+(z_2)^2+(z_3)^2}{(z_0)^2}.$

The answer is 3.

2 solutions

Ashish Menon
Dec 21, 2016

The circumcenter of an equilateral triangle is given by $z_0 = \dfrac{z_1 + z_2 + z_3}{3}$ .
Also in an equilateral triangle, ${z_1}^2 + {z_2}^2 + {z_3}^2 = z_1z_2 + z_2z_3 + z_3z_1$ .

So, $\dfrac{{z_1}^2 + {z_2}^2 + {z_3}^2}{{z_0}^2}\\ \\ = \dfrac{z_1z_2 + z_2z_3 + z_3z_1}{ {z_1}^2 + {z_2}^2 + {z_3}^2 +2(z_1z_2 + z_2z_3 + z_3z_1)}\\ \\ = \color{#69047E}{\boxed{3}}$ .

Bakul Choudhary
Aug 15, 2015

Let us assume $z_1$ on origin and $z_2$ on positive real axis.

Then, $z_3$ will correspond to, $\left| z_3 \right| \left(\cos \frac{\pi}{3} +i\sin\frac{\pi}{3}\right) = \left| z_3 \right| \left( \frac{1}{2} +i\frac{\sqrt 3}{2}\right)$

Let $a = \left|z_2\right|$ ,then $z_1 = 0$ , $z_2 = a$ , $z_3 = \frac{a}{2} \left( 1 + i \sqrt 3\right)$

So, $z_1 ^2 + z_2 ^2 + z_3 ^2 = \frac{a^2}{2} \left( 1 + i \sqrt 3 \right)$

As, orthocenter, incenter, median and circumcenter of an equilateral triangle coincide, $a_0 = \frac{z_1 + z_2 + z_3}{3}$ or, $a_0 = \frac{a}{6} \left( 3 + i \sqrt 3 \right)$ and, $a_0 ^2 = \frac{a^2}{6} \left( 1 + i \sqrt 3 \right)$

Therefore, $\frac{z_1 ^2 + z_2 ^2 + z_3 ^2}{z_0 ^2} = \frac{\frac{a^2}{2} \left( 1 + i \sqrt 3 \right)}{\frac{a^2}{6} \left( 1 + i \sqrt 3 \right)}$ or, $\frac{z_1 ^2 + z_2 ^2 + z_3 ^2}{z_0 ^2} = \boxed{3}$

Here's a more general solution:

Let $\omega_1$ , $\omega_2$ , and $\omega_3$ be the $3^\text{rd}$ roots of unity. Then the equilateral triangle's vertices can be defined by:

$z_1=\omega_1 r e^{i\theta}+z_0$

$z_2=\omega_2 r e^{i\theta}+z_0$

$z_3=\omega_3 r e^{i\theta}+z_0$

Where $\theta$ is the relative rotation of the vertices from the $3^\text{rd}$ roots of unity and $r$ is the radius of the circle that the triangle is inscribed in.

$(z_1)^2=\omega_1^2 r^2 e^{2i\theta}+2z_0\omega_1 r e^{i\theta} + z_0^2$

$(z_2)^2=\omega_2^2 r^2 e^{2i\theta}+2z_0\omega_2 r e^{i\theta} + z_0^2$

$(z_3)^2=\omega_3^2 r^2 e^{2i\theta}+2z_0\omega_3 r e^{i\theta} + z_0^2$

$(z_1)^2+(z_2)^2+(z_3)^2=r^2 e^{2i\theta}(\omega_1^2+\omega_2^2+\omega_3^2)+2z_0 r e^{i\theta}(\omega_1+\omega_2+\omega_3)+3z_0^2$

By the properties of roots of unity , $\omega_1^2+\omega_2^2+\omega_3^2=0$ and $\omega_1+\omega_2+\omega_3=0$ .

Thus,

$\frac{(z_1)^2+(z_2)^2+(z_3)^2}{z_0^2}=\frac{3z_0^2}{z_0^2}=\boxed{3}$

Andy Hayes - 4 years, 9 months ago

Can u explain how did u get that expression for z?

toshali mohapatra - 4 years, 6 months ago

When the 3rd roots of unity are graphed in the complex plane, they form an equilateral triangle. This equilateral triangle can be transformed into any equilateral triangle in the complex plane by performing a dilation, rotation, and translation.

$r$ is the scale factor (It's not known how big the triangle is. This variable is chosen to represent the distance from the circumcenter to a vertex). Multiplying the 3rd root of unity by $r$ accomplishes the dilation.
$\theta$ is the angle of rotation (It's not known how much the triangle is rotated by. This variable is chosen to represent how much the triangle is rotated from the 3rd roots of unity). Multiplying the previous result by $e^{i\theta}$ accomplishes the rotation.
$z_0$ is the circumcenter of the triangle, so this is chosen as the amount of translation. Adding $z_0$ to the previous result accomplishes the translation.

These three transformations gives the coordinates of the vertices of the triangle in terms of the 3rd roots of unity. The key to this problem is that, although we don't know the values of $r,$ $\theta,$ and $z_0,$ they are the same for each transformation.