AIME 1988 · 第 11 题

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 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.