1994 AIME Problems/Problem 8

Problem

The points $(0,0)\,$ , $(a,11)\,$ , and $(b,37)\,$ are the vertices of an equilateral triangle. Find the value of $ab\,$ .

Solution

Consider the points on the complex plane. The point $b+37i$ is then a rotation of $60$ degrees of $a+11i$ about the origin, so:

$(a+11i)\left(\text{cis}\,60^{\circ}\right) = (a+11i)\left(\frac 12+\frac{\sqrt{3}i}2\right)=b+37i.$

Equating the real and imaginary parts, we have:

$\begin{align*}b&=\frac{a}{2}-\frac{11\sqrt{3}}{2}\\37&=\frac{11}{2}+\frac{a\sqrt{3}}{2} \end{align*}$

Solving this system, we find that $a=21\sqrt{3}, b=5\sqrt{3}$ . Thus, the answer is $\boxed{315}$ .

Note: There is another solution where the point $b+37i$ is a rotation of $-60$ degrees of $a+11i$ ; however, this triangle is just a reflection of the first triangle by the $y$ -axis, and the signs of $a$ and $b$ are flipped. However, the product $ab$ is unchanged.

1994 AIME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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1994 AIME Problems/Problem 8

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