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

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- Aug 4th 2012, 04:03 AMshiny718Vertices of equilateral triangle
If (0,0), (a,11), (b, 37) are the vertices of an equilateral triangle, find the product of a and b.

- Aug 4th 2012, 04:15 AMProve ItRe: Vertices of equilateral triangle
- Aug 4th 2012, 04:25 AMshiny718Re: Vertices of equilateral triangle
Hi Prove it, I've tried evaluating the method u suggested above but I'm stuck with $\displaystyle a^2 -b^2 =1248, 2ab - b^2 =555$ and $\displaystyle a^2 -2ab =693$ Any tips on how to continue?

- Aug 4th 2012, 05:07 AMProve ItRe: Vertices of equilateral triangle
- Aug 22nd 2012, 09:26 AMthecmd999Re: Vertices of equilateral triangle
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.

- Aug 22nd 2012, 03:32 PMrichard1234Re: Vertices of equilateral triangle
I second thecmd999's post. For example, let

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

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

Dividing the second equation by the first,

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

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

Expand both sides, equate real and imaginary parts. I think I recognize this problem...was it on a past AIME? - Aug 22nd 2012, 03:34 PMthecmd999Re: Vertices of equilateral triangle
Yeah, this was 1994 #8.

- Aug 22nd 2012, 03:37 PMrichard1234Re: Vertices of equilateral triangle
Ah okay. I did AIME in HS (qualified to USAMO once). So most of the past AIME questions I've seen before.