"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.

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

$(b^2+555)^2 - 4b^4 = 4992b^2$,



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

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

  • $\begingroup$ 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? $\endgroup$
    – MonK
    Jul 14 '14 at 18:47
  • 1
    $\begingroup$ We are using the Pythagorean theorem to calculate the distance between two points in $\mathbb{R}^2$, nothing more than that. $\endgroup$ Jul 21 '14 at 1:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.