Difference between revisions of "1994 AIME Problems/Problem 8"
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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 | + | '''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 
, 
, and 
are the vertices of an equilateral triangle. Find the value of 
.
Solution
Consider the points on the complex plane. The point 
is then a rotation of 
degrees of 
about the origin, so:
![\[(a+11i)\left(\text{cis}\,60^{\circ}\right) = (a+11i)\left(\frac 12+\frac{\sqrt{3}i}2\right)=b+37i.\]](http://latex.artofproblemsolving.com/f/3/b/f3b0745d80c034402737de449a7bba856fafef5b.png)
Equating the real and imaginary parts, we have:
Solving this system, we find that 
. Thus, the answer is 
.
Note: There is another solution where the point is a rotation of 
degrees of ; however, this triangle is just a reflection of the first triangle by the 
-axis, and the signs of 
and 
are switched. However, the product 
is unchanged.
See also
| 1994 AIME (Problems • Answer Key • Resources) | ||
| 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 | ||
