The points $(0,0)$, $(a, 11)$, and $(b,37)$ are vertices of an equilateral triangle. Find the product $ab$. "The points $(0,0),\;(a, 11), \text{ and } (b,37)$ are vertices of an equilateral triangle. Find the product $ab$."
I'm not sure how to start this problem. I of course drew out an equilateral triangle with those points, but i'm not sure what information I can draw from them. We know that the sides all have equal length obviously. I'm not sure what to do.

I guess we obtain the following equations:
$a^2+11^2 = b^2+37^2$
$a^2+11^2 = (a-b)^2 + (37-11)^2$
These simplify to:
$a^2-b^2 = 1248$
$2ab = b^2 + 555$
We can solve the second equation for $a$, and obtain $a=\frac{b^2+555}{2b}$. Substituting this into the first equation, we obtain:
$\left(\frac{b^2+555}{2b}\right)^2 - b^2 = 1248$
Multiplying by $4b^2$, this becomes:
$(b^2+555)^2 - 4b^4 = 4992b^2$, 
or:
$3b^4+3882b^2-308025=0$
Now we can use the quadratic formula to obtain:
$b^2 = \frac{-3882\pm\sqrt{3882^2+4\cdot 3\cdot 308025}}{6}$
We need a positive answer, so we choose the plus sign, and get:
$b^2= 75 \\
b = 5\sqrt{3} \\
a^2 = 1323 \\
a=21\sqrt{3}$ 
Finally, we want the product $ab$, which equals $5\sqrt{3}\cdot 21\sqrt{3} = 315$
