<s>[INST] If the points $(0,0),(a,11),(b,37)$ are vertices of equilateral triangle what is the value of ab? If the points $(0,0),(a,11),(b,37)$ are vertices of equilateral triangle, find the value of $ab$
 [/INST] 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,
$$\begin{align}2ab &= b^2 \pm37\sqrt{3}b\\
&=75 + 555\\
&=630\end{align}$$
$$ab = 315$$
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