Difference between revisions of "1988 AIME Problems/Problem 11"
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== Solution == | == Solution == | ||
+ | ===Solution 1=== | ||
<math>\sum_{k=1}^5 z_k - \sum_{k=1}^5 w_k = 0</math> | <math>\sum_{k=1}^5 z_k - \sum_{k=1}^5 w_k = 0</math> | ||
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Each <math>z_k = x_k + y_ki</math> lies on the complex line <math>y = mx + 3</math>, so we can rewrite this as | Each <math>z_k = x_k + y_ki</math> lies on the complex line <math>y = mx + 3</math>, so we can rewrite this as | ||
− | <math>\sum_{k=1}^5 z_k = \sum_{k=1}^5 x_k + \sum_{k=1}^ | + | <math>\sum_{k=1}^5 z_k = \sum_{k=1}^5 x_k + \sum_{k=1}^5 y_ki</math> |
<math>3 + 504i = \sum_{k=1}^5 x_k + i \sum_{k=1}^5 (mx_k + 3)</math> | <math>3 + 504i = \sum_{k=1}^5 x_k + i \sum_{k=1}^5 (mx_k + 3)</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>. | 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=== | ||
+ | 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> | ||
+ | |||
+ | ===Solution 3=== | ||
+ | |||
+ | The mean line for <math>w_1, . . ., w_5</math> must pass through the mean (the center of mass) of these points, which, if we graph them on the complex plane, is <math>(\frac{3}{5}, \frac{504i}{5})</math>. Since we now have two points, namely that one and <math>(0, 3i)</math>, we can simply find the slope between them, which is <math>\boxed{163}</math> by the good ol' slope formula. | ||
== See also == | == See also == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 00:22, 6 September 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
Solution 3
The mean line for must pass through the mean (the center of mass) of these points, which, if we graph them 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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.