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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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===Solution 1===
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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}^5 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>.
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===Solution 2===
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We know that
 +
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<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>
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 +
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
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<math>504 =  3m + 15</math>
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Solving for <math>m</math>, the slope, we get <math>\boxed{163}</math>
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===Solution 3===
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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 ==
−
* [[1988 AIME Problems]]
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{{AIME box|year=1988|num-b=10|num-a=12}}
 +
 
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[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 23:22, 5 September 2020

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

Solution 1

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

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

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

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

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

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 $m \cdot 3 = 489 \Longrightarrow m = 163$.

Solution 2

We know that

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

And because the sum of the 5 $z$'s must cancel this out,

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

We write the numbers in the form $a + bi$ and we know that

$\sum_{k=1}^5 a_k = 3$ and $\sum_{k=1}^5 b_k = 504$

The line is of equation $y=mx+3$. Substituting in the polar coordinates, we have $b_k = ma_k + 3$.

Summing all 5 of the equations given for each $k$, we get

$504 = 3m + 15$

Solving for $m$, the slope, we get $\boxed{163}$

Solution 3

The mean line for $w_1, . . ., w_5$ must pass through the mean (the center of mass) of these points, which, if we graph them on the complex plane, is $(\frac{3}{5}, \frac{504i}{5})$. Since we now have two points, namely that one and $(0, 3i)$, we can simply find the slope between them, which is $\boxed{163}$ 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

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