1988 AIME Problems/Problem 11
Problem
Let be complex numbers. A line in the complex plane is called a mean line for the points if contains points (complex numbers) such that For the numbers , , , , and , there is a unique mean line with -intercept 3. Find the slope of this mean line.
Solution
Solution 1
Each lies on the complex line , so we can rewrite this as
Matching the real parts and the imaginary parts, we get that and . Simplifying the second summation, we find that , and substituting, the answer is .
Solution 2
We know that
And because the sum of the 5 's must cancel this out,
We write the numbers in the form and we know that
and
The line is of equation . Substituting in the polar coordinates, we have .
Summing all 5 of the equations given for each , we get
Solving for , the slope, we get
Solution 3
The mean line for must pass through the mean (the center of mass) of these points, which, if we graph these points on the complex plane, is . Since we now have two points, namely that one and , we can simply find the slope between them, which is by the good ol' slope formula.
See also
1988 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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