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

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== Solution ==
 
== Solution ==
Using complex coordiantes is one approach. The point <math>b+37i</math> is then a rotation of <math>60</math> degrees of <math>a+11i</math> about the origin, hence we have:
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Consider the points on the [[complex plane]]. The point <math>b+37i</math> is then a rotation of <math>60</math> degrees of <math>a+11i</math> about the origin, so:
  
<math>(a+11i)(cis60)=b+37i
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<cmath>(a+11i)\left(\text{cis}\,60^{\circ}\right) = (a+11i)\left(\frac 12+\frac{\sqrt{3}i}2\right)=b+37i.</cmath>
\newline(a+11i)(1/2+\sqrt{3}i/2)=b+37i</math>.
 
  
Setting the real and imaginary parts equal, we have:
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Equating the real and imaginary parts, we have:
  
<math>b=a/2-11\sqrt{3}/2
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<cmath>\begin{align*}b&=a/2-11\sqrt{3}/2 \\37&=11/2+a\sqrt{3}/2 \end{align*}</cmath>
\newline37=11/2+a\sqrt{3}/2</math>.
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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 have:
 
<math>a=21\sqrt{3}, b=5\sqrt{3}</math>.
 
Thus, the answer is <math>315</math>.
 
 
== See also ==
 
== See also ==
 
{{AIME box|year=1994|num-b=7|num-a=9}}
 
{{AIME box|year=1994|num-b=7|num-a=9}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 18:06, 16 October 2008

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&=a/2-11\sqrt{3}/2 \\37&=11/2+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
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All AIME Problems and Solutions
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