1999 AIME 第 9 题

Question

函数

f

定义在复数集上，且

f(z) = (a + bi)z

，其中

a

和

b

为正数。这个函数满足：复平面中每个点的像到该点与到原点的距离相等。已知

|a + bi| = 8

且

b^2 = \frac{m}{n}

，其中

m

和

n

是互质的正整数。求

m + n

。

A function

f

is defined on the complex numbers by

f(z) = (a + bi)z,

where

a

and

b

are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin. Given that

|a + bi| = 8

and that

b^2 = \frac{m}{n},

where

m

and

n

are relatively prime positive integers, find

m + n.

答案

答案：259

知识点：复数代数变形

难度评级：2270

小提示：

条件表示

|f(z) - z| = |f(z)|

对每个

z

都成立；把两边都写成某个模长乘以

|z|

The condition says

|f(z) - z| = |f(z)|

for every

z;

write both sides as a modulus times

|z|

大提示：

需要

|a - 1 + bi| = |a + bi|

，这会确定

a

；然后使用

a^2 + b^2 = 64

You need

|a - 1 + bi| = |a + bi|,

which pins down

a;

then use

a^2 + b^2 = 64

解答：

条件是

|f(z) - z| = |f(z)|

对所有

z

都成立，即

|(a - 1 + bi)z| = |(a + bi)z|

。两边除以

|z|

（当

z \ne 0

时），得

|a - 1 + bi| = |a + bi|

，所以

(a - 1)^2 + b^2 = a^2 + b^2\text{，}

从而

a = \frac{1}{2}

。

由

|a + bi| = 8

可得

a^2 + b^2 = 64

，所以

b^2 = 64 - \frac{1}{4} = \frac{255}{4}

。又

\gcd(255, 4) = 1

，答案是

255 + 4 = 259

。

The condition is

|f(z) - z| = |f(z)|

for all

z,

that is,

|(a - 1 + bi)z| = |(a + bi)z|.

Dividing by

|z|

(for

z \ne 0

) gives

|a - 1 + bi| = |a + bi|,

so

(a - 1)^2 + b^2 = a^2 + b^2,

which forces

a = \frac{1}{2}.

Since

|a + bi| = 8,

we have

a^2 + b^2 = 64,

so

b^2 = 64 - \frac{1}{4} = \frac{255}{4}.

As

\gcd(255, 4) = 1,

the answer is

255 + 4 = 259.

1999 AIME 第 9 题

答案：259

难度评级：2270

小提示：

大提示：

解答：

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