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

(cleanup)
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Equating the real and imaginary parts, we have:
 
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>.
 
Solving this system, we find that <math>a=21\sqrt{3}, b=5\sqrt{3}</math>. Thus, the answer is <math>\boxed{315}</math>.

Revision as of 14:01, 19 July 2009

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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