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If the points $(0,0),(a,11),(b,37)$ are vertices of equilateral triangle, find the value of $ab$

Mythomorphic
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  • Well, what's the formula for the distance between two points? What's the distance between (0,0) and (a,11)? What's the distance between (a,11) and (b,37)? What's the distance between (b,37) and (0,0)? If all three of those distance are the same what chooses do you have for a and b to be? – fleablood Feb 25 '17 at 06:26
  • I did made the equations but i can't complete –  Feb 25 '17 at 06:57

2 Answers2

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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,

$$\begin{align}2ab &= b^2 \pm37\sqrt{3}b\\ &=75 + 555\\ &=630\end{align}$$ $$ab = 315$$

Yiyuan Lee
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Pretty lengthy equations (and probably bad method to solve). So here are the equations :

$$a^2+11^2=b^2+37^2=(b-a)^2+(37-11)^2$$ $$\implies b^2+a^2-2ab +676 = a^2+121\implies b^2-2ab+555=0\,\,\,\,(1)$$ $$b^2 = a^2-1278 \implies b=\sqrt{a^2-1278}\,\,\,\,\,\,(2)$$

substitute $b$ in $(1)$$$\implies (a^2-1278)+555 = 2a(\sqrt{a^2-1278})$$

Now this is ugly step :- square both sides=> $$\implies a^4 + 693^2 -2(693)a^2 = 4a^2(a^2-1278)$$

And if you assign $a^2$ to some $t$ it is in quadraric form and should be solvable.

user5954246
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