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.
 A: Use complex numbers. Notice that:
$$b+37i=(a+11i)e^{i\pi/3} \Rightarrow b+37i=\frac{1}{2}(1+\sqrt{3}i)(a+11i)$$
Comparing the imaginary parts:
$$37=\frac{1}{2}(\sqrt{3}a+11) \Rightarrow a=21\sqrt{3}$$
Comparing the real part:
$$b=\frac{1}{2}(a-11\sqrt{3})\Rightarrow b=5\sqrt{3}$$
Hence,
$$\boxed{ab=315}$$
A: 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 $5\sqrt{3}\cdot 21\sqrt{3} = 315$
A: Let the angle between $(a,11)$ and the $x$-axis be $\theta$.
$$\begin{align}d\cos\theta&=a\\d\sin\theta&=11\\d\cos(\theta+\pi/3)&=b\\d\sin(\theta+\pi/3)&=37\end{align}$$
Do you know how to expand $\cos(\theta+\pi/3)$?
A: By Pythagoras, the squares of the three side lengtsh are $a^2+11^2$, $b^2+37^2$ and $(a-b)^2+26^2$. Equate these.
