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

Revision as of 18:06, 16 October 2008

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:

$\begin{align*}b&=a/2-11\sqrt{3}/2 \\37&=11/2+a\sqrt{3}/2 \end{align*}$

Solving this system, we find that $a=21\sqrt{3}, b=5\sqrt{3}$ . Thus, the answer is $\boxed{315}$ .

@@ Line 3: / Line 3: @@
 == 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:
+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
+<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:
+Equating the real and imaginary parts, we have:
-<math>b=a/2-11\sqrt{3}/2
+<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>.
+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 ==
 {{AIME box|year=1994|num-b=7|num-a=9}}
+[[Category:Intermediate Geometry Problems]]

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