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

(Solution)
(Solution)
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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 find that <math>a=21\sqrt{3}, b=5\sqrt{3}</math>. Thus, the answer is <math>\boxed{315}</math>.
  
''Note'': There is another solution where the point <math>b+37i</math> is a rotation of <math>-60</math> degrees of <math>a+11i</math>; however, this gives <math>a=-21\sqrt 3</math> and <math>b=-5\sqrt 3</math>, which gives the same final answer, <math>315</math>.
+
'''Note''': There is another solution where the point <math>b+37i</math> is a rotation of <math>-60</math> degrees of <math>a+11i</math>; however, this triangle is just a reflection of the first triangle by the <math>y</math>-axis, and the signs of <math>a</math> and <math>b</math> are switched. However, the product <math>ab</math> is unchanged.
  
 
== See also ==
 
== See also ==

Revision as of 20:14, 27 February 2011

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 switched. However, the product $ab$ is unchanged.

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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