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>. | ||
− | ==Solution 2== | + | ===Solution 2=== |
We know that | We know that | ||
Revision as of 15:09, 16 May 2020
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
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.