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: | |
+ | <math>(a+11i)(cis60)=b+37i | ||
+ | \newline(a+11i)(1/2+\sqrt{3}i/2)=b+37i</math>. | ||
+ | |||
+ | Setting the real and imaginary parts equal, we have: | ||
+ | |||
+ | <math>b=a/2-11\sqrt{3}/2 | ||
+ | \newline37=11/2+a\sqrt{3}/2</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}} |
Revision as of 20:26, 20 June 2008
Problem
The points , , and are the vertices of an equilateral triangle. Find the value of .
Solution
Using complex coordiantes is one approach. The point is then a rotation of degrees of about the origin, hence we have:
.
Setting the real and imaginary parts equal, we have:
.
Solving this system, we have: . Thus, the answer is .
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 |