Difference between revisions of "1988 AIME Problems/Problem 11"
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Matching the real parts and the imaginary parts, we get that <math>\sum_{k=1}^5 x_k = 3</math> and <math>\sum_{k=1}^5 (mx_k + 3) = 504</math>. Simplifying the second summation, we find that <math>m\sum_{k=1}^5 x_k = 504 - 3 \cdot 5 = 489</math>, and substituting, the answer is <math>m \cdot 3 = 489 \Longrightarrow m = 163</math>. | Matching the real parts and the imaginary parts, we get that <math>\sum_{k=1}^5 x_k = 3</math> and <math>\sum_{k=1}^5 (mx_k + 3) = 504</math>. Simplifying the second summation, we find that <math>m\sum_{k=1}^5 x_k = 504 - 3 \cdot 5 = 489</math>, and substituting, the answer is <math>m \cdot 3 = 489 \Longrightarrow m = 163</math>. | ||
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+ | |||
+ | ==Solution 2== | ||
+ | We know that | ||
+ | |||
+ | <math>\sum_{k=1}^5 w_k = 3 + 504i</math> | ||
+ | |||
+ | And because the sum of the 5 <math>z</math>'s must cancel this out, | ||
+ | |||
+ | <math>\sum_{k=1}^5 z_k = 3 + 504i</math> | ||
+ | |||
+ | We write the numbers in the form <math>a + bi</math> and we know that | ||
+ | |||
+ | <math>\sum_{k=1}^5 a_k = 3</math> and <math>\sum_{k=1}^5 b_k = 504</math> | ||
+ | |||
+ | The line is of equation <math>y=mx+3</math>. Substituting in the polar coordinates, we have <math>b_k = ma_k + 3</math>. | ||
+ | |||
+ | Summing all 5 of the equations given for each <math>k</math>, we get | ||
+ | |||
+ | <math>504 = 3m + 15</math> | ||
+ | |||
+ | Solving for <math>m</math>, the slope, we get <math>\boxed{163}</math> | ||
== See also == | == See also == |
Revision as of 18:45, 22 February 2014
Contents
[hide]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
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
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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.