If the points $(0,0),(a,11),(b,37)$ are vertices of equilateral triangle what is the value of ab?

Question

If the points $(0,0),(a,11),(b,37)$ are vertices of equilateral triangle, find the value of $ab$

Answer 1

Pretty lengthy equations (and probably bad method to solve). So here are the equations :

$$a^2+11^2=b^2+37^2=(b-a)^2+(37-11)^2$$ $$\implies b^2+a^2-2ab +676 = a^2+121\implies b^2-2ab+555=0\,\,\,\,(1)$$ $$b^2 = a^2-1278 \implies b=\sqrt{a^2-1278}\,\,\,\,\,\,(2)$$

substitute $b$ in $(1)$$$\implies (a^2-1278)+555 = 2a(\sqrt{a^2-1278})$$

Now this is ugly step :- square both sides=> $$\implies a^4 + 693^2 -2(693)a^2 = 4a^2(a^2-1278)$$

And if you assign $a^2$ to some $t$ it is in quadraric form and should be solvable.

Answer 2

I'd like to propose an exotic solution using complex numbers. Let the three vertices be represented on the Argand diagram by

$$z_0 = 0 + 0i = 0$$ $$z_1 = a + 11i$$ $$z_2 = b + 37i$$

Then we know that $z_1-z_0 = z_1$ and $z_2-z_0 = z_2$ are $\frac{\pi}{3}$ apart. That is

$$z_1 = z_2e^{\pm i\frac{\pi}{3}}$$ $$a + 11i = (b + 37i)\left(\frac{1}{2} \pm i \frac{\sqrt{3}}{2}\right)$$ $$a + 11i = \left(\frac{1}{2}b \pm\frac{37\sqrt{3}}{2}\right) + \left(\mp\frac{\sqrt{3}}{2}b + \frac{37}{2} \right)i$$

Do note that the $\pm$ and $\mp$ signs are alternate.

Next, comparing coefficients,

$$2a = b \pm 37\sqrt{3}~~,~~22 = \mp\sqrt{3}b + 37$$ $$2a = b \pm 37\sqrt{3}~~,~~\mp\sqrt{3}b = -15$$ $$2a = b \pm 37\sqrt{3}~~,~~b = \pm\frac{15}{\sqrt{3}} = \pm5\sqrt{3}$$

Hence,

$$\begin{align}2ab &= b^2 \pm37\sqrt{3}b\\ &=75 + 555\\ &=630\end{align}$$ $$ab = 315$$