Vertices of equilateral triangle

**shiny718** · August 4th 2012, 04:03 AM

If (0,0), (a,11), (b, 37) are the vertices of an equilateral triangle, find the product of a and b.

**Prove It** · August 4th 2012, 04:15 AM

Originally Posted by shiny718

If (0,0), (a,11), (b, 37) are the vertices of an equilateral triangle, find the product of a and b.

Each side of the triangle is equal, so that means $\displaystyle \begin{align*} \sqrt{(a - 0)^2 + (11 - 0)^2} = \sqrt{(b - a)^2 + (37 - 11)^2} = \sqrt{(b - 0)^2 + (37 - 0)^2} \end{align*}$ .

See if you can evaluate a and b from here.

**shiny718** · August 4th 2012, 04:25 AM

Hi Prove it, I've tried evaluating the method u suggested above but I'm stuck with $a^2 -b^2 =1248, 2ab - b^2 =555$ and $a^2 -2ab =693$ Any tips on how to continue?

**Prove It** · August 4th 2012, 05:07 AM

Originally Posted by shiny718

Hi Prove it, I've tried evaluating the method u suggested above but I'm stuck with $a^2 -b^2 =1248, 2ab - b^2 =555$ and $a^2 -2ab =693$ Any tips on how to continue?

From the last equation, write b in terms of a, then substitute into another equation.

**thecmd999** · August 22nd 2012, 09:26 AM

This method will work, but in a competition situation, you'll have to be very good at bashing. A much simpler way to do this problem is to consider (b, 37) as a rotation of 60 degrees of (a, 11). This method is much easier and requires almost no bash. You can either use trigonometric sum identities, rotations on the complex plane, or rotation mapping of matrices; they are all equivalent.

**richard1234** · August 22nd 2012, 03:32 PM

I second thecmd999's post. For example, let

$b + 37i = re^{i \theta}$

$a + 11i = re^{i \theta + \frac{\pi}{3}$

Dividing the second equation by the first,

$\frac{a+11i}{b+37i} = e^{i \frac{\pi}{3}} = \frac{1 + \sqrt{3}i}{2}$

$(b+37i)(1 + \sqrt{3}i) = 2(a+11i)$

Expand both sides, equate real and imaginary parts. I think I recognize this problem...was it on a past AIME?

**thecmd999** · August 22nd 2012, 03:34 PM

Yeah, this was 1994 #8.

**richard1234** · August 22nd 2012, 03:37 PM

Ah okay. I did AIME in HS (qualified to USAMO once). So most of the past AIME questions I've seen before.

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