# 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.

• You are wrong, the lengths are calculated as $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ – enzotib Jul 13 '14 at 14:31
• Edited appropriately. – nyaameow Jul 13 '14 at 14:34

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, $$\boxed{ab=315}$$

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 $5\sqrt{3}\cdot 21\sqrt{3} = 315$

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 $\cos(\theta+\pi/3)$?

• What does you mean by $d$? – nyaameow Jul 13 '14 at 15:03
• Sorry, $d$ is the length of the triangle's side. – Empy2 Jul 13 '14 at 15:11

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.

• I am sorry, But how are we using pythogoras theorem on an equilateral triangle. Everyone on this question is using it. I think I am missing some thing. Can you please elaborate? – MonK Jul 14 '14 at 18:47
• We are using the Pythagorean theorem to calculate the distance between two points in $\mathbb{R}^2$, nothing more than that. – Jack D'Aurizio Jul 21 '14 at 1:21