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Difference between revisions of "1988 AIME Problems/Problem 11"

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== Problem ==
 
== Problem ==
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Let <math>w_1, w_2, \dots, w_n</math> be [[complex number]]s.  A line <math>L</math> in the [[complex plane]] is called a mean [[line]] for the [[point]]s <math>w_1, w_2, \dots, w_n</math> if <math>L</math> contains points (complex numbers) <math>z_1, z_2, \dots, z_n</math> such that
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<cmath>
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\sum_{k = 1}^n (z_k - w_k) = 0.
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</cmath>
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For the numbers <math>w_1 = 32 + 170i</math>, <math>w_2 = - 7 + 64i</math>, <math>w_3 = - 9 + 200i</math>, <math>w_4 = 1 + 27i</math>, and <math>w_5 = - 14 + 43i</math>, there is a unique mean line with <math>y</math>-intercept 3.  Find the [[slope]] of this mean line.
  
 
== Solution ==
 
== Solution ==
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<math>\sum_{k=1}^5 z_k - \sum_{k=1}^5 w_k = 0</math>
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<math>\sum_{k=1}^5 z_k = 3 + 504i</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
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<math>\sum_{k=1}^5 z_k = \sum_{k=1}^5 x_k + \sum_{k=1}^n y_ki</math>
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<math>3 + 504i = \sum_{k=1}^5 x_k + i \sum_{k=1}^5 (mx_k + 3)</math>
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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>.
  
 
== See also ==
 
== See also ==
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* [[1988 AIME Problems]]
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{{AIME box|year=1988|num-b=10|num-a=12}}
  
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{{AIME box|year=1988|num-b=10|num-a=12}}
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[[Category:Intermediate Complex Numbers Problems]]
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[[Category:Intermediate Geometry Problems]]

Revision as of 16:10, 28 September 2007

Problem

Let $w_1, w_2, \dots, w_n$ be complex numbers. A line $L$ in the complex plane is called a mean line for the points $w_1, w_2, \dots, w_n$ if $L$ contains points (complex numbers) $z_1, z_2, \dots, z_n$ such that \[\sum_{k = 1}^n (z_k - w_k) = 0.\] For the numbers $w_1 = 32 + 170i$, $w_2 = - 7 + 64i$, $w_3 = - 9 + 200i$, $w_4 = 1 + 27i$, and $w_5 = - 14 + 43i$, there is a unique mean line with $y$-intercept 3. Find the slope of this mean line.

Solution

$\sum_{k=1}^5 z_k - \sum_{k=1}^5 w_k = 0$

$\sum_{k=1}^5 z_k = 3 + 504i$

Each $z_k = x_k + y_ki$ lies on the complex line $y = mx + 3$, so we can rewrite this as

$\sum_{k=1}^5 z_k = \sum_{k=1}^5 x_k + \sum_{k=1}^n y_ki$

$3 + 504i = \sum_{k=1}^5 x_k + i \sum_{k=1}^5 (mx_k + 3)$

Matching the real parts and the imaginary parts, we get that $\sum_{k=1}^5 x_k = 3$ and $\sum_{k=1}^5 (mx_k + 3) = 504$. Simplifying the second summation, we find that $m\sum_{k=1}^5 x_k = 504 - 3 \cdot 5 = 489$, and substituting, the answer is $m \cdot 3 = 489 \Longrightarrow m = 163$.

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
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