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

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== Solution ==
 
== Solution ==
{{solution}}
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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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<math>(a+11i)(cis60)=b+37i
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\newline(a+11i)(1/2+\sqrt{3}i/2)=b+37i</math>.
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Setting the real and imaginary parts equal, we have:
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<math>b=a/2-11\sqrt{3}/2
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\newline37=11/2+a\sqrt{3}/2</math>.
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Solving this system, we have:
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<math>a=21\sqrt{3}, b=5\sqrt{3}</math>.
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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 19:26, 20 June 2008

Problem

The points $(0,0)\,$, $(a,11)\,$, and $(b,37)\,$ are the vertices of an equilateral triangle. Find the value of $ab\,$.

Solution

Using complex coordiantes is one approach. The point $b+37i$ is then a rotation of $60$ degrees of $a+11i$ about the origin, hence we have:

$(a+11i)(cis60)=b+37i \newline(a+11i)(1/2+\sqrt{3}i/2)=b+37i$.

Setting the real and imaginary parts equal, we have:

$b=a/2-11\sqrt{3}/2 \newline37=11/2+a\sqrt{3}/2$.

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

See also

1994 AIME (ProblemsAnswer KeyResources)
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
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