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1994 AIME Problems/Problem 8

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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}.

See also

1994 AIME (ProblemsResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.
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