1988 AIME Problems/Problem 11

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

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 $\boxed{163}$

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
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1988 AIME Problems/Problem 11

Contents

Problem

Solution

Solution 2

See also