Difference between revisions of "1994 AIME Problems/Problem 8"

Revision as of 14:01, 19 July 2009

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

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 $\boxed{315}$ .

@@ Line 9: / Line 9: @@
 Equating the real and imaginary parts, we have:
-<cmath>\begin{align*}b&=a/2-11\sqrt{3}/2 \\37&=11/2+a\sqrt{3}/2 \end{align*}</cmath>
+<cmath>\begin{align*}b&=\frac{a}{2}-\frac{11\sqrt{3}}{2}\\37&=\frac{11}{2}+\frac{a\sqrt{3}}{2} \end{align*}</cmath>
 Solving this system, we find that <math>a=21\sqrt{3}, b=5\sqrt{3}</math>. Thus, the answer is <math>\boxed{315}</math>.

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