2001 AIME II 第 4 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

4.

设

R = (8, 6)

。方程为

8y = 15x

和

10y = 3x

的直线分别含有点

P

和

Q

，且

R

是

\overline{PQ}

的中点。长度

PQ

等于

\frac{m}{n}

，其中

m

和

n

是互质正整数。求

m + n

。

Let

R = (8, 6).

The lines whose equations are

8y = 15x

and

10y = 3x

contain points

P

and

Q,

respectively, such that

R

is the midpoint of

\overline{PQ}.

The length

PQ

equals

\frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m + n.

答案

答案：67

知识点：坐标几何中点方程组距离公式

难度评级：2170

小提示：

写成

P = (8t, 15t)

、

Q = (10u, 3u)

，这样两个直线条件会自动满足。

Write

P = (8t, 15t)

and

Q = (10u, 3u)

so that both line conditions hold automatically

大提示：

中点条件给出

8t + 10u = 16

和

15t + 3u = 12

；解出

t

和

u

，再用距离公式。

The midpoint condition gives

8t + 10u = 16

and

15t + 3u = 12;

solve for

t

and

u,

then use the distance formula

解答：

两条直线上的点可写成

P = (8t, 15t)

和

Q = (10u, 3u)

。因为

R = (8, 6)

是

\overline{PQ}

的中点，

8t + 10u = 16

且

15t + 3u = 12\text{。}

第二个方程给出

u = 4 - 5t

；代入第一个方程，得

8t + 40 - 50t = 16

，所以

t = \frac{4}{7}

，且

u = \frac{8}{7}

。于是

P = \left(\frac{32}{7}, \frac{60}{7}\right)

，

Q = \left(\frac{80}{7}, \frac{24}{7}\right)

。

因而

PQ = \sqrt{\left(\frac{48}{7}\right)^2 + \left(\frac{36}{7}\right)^2}

= \frac{12}{7}\sqrt{4^2 + 3^2}

= \frac{60}{7}

，所以

m + n = 60 + 7 = 67

。

Points on the two lines can be written

P = (8t, 15t)

and

Q = (10u, 3u).

Since

R = (8, 6)

is the midpoint of

\overline{PQ},

8t + 10u = 16

and

15t + 3u = 12.

The second equation gives

u = 4 - 5t;

substituting into the first,

8t + 40 - 50t = 16,

so

t = \frac{4}{7}

and

u = \frac{8}{7}.

Thus

P = \left(\frac{32}{7}, \frac{60}{7}\right)

and

Q = \left(\frac{80}{7}, \frac{24}{7}\right).

Then

PQ = \sqrt{\left(\frac{48}{7}\right)^2 + \left(\frac{36}{7}\right)^2}

= \frac{12}{7}\sqrt{4^2 + 3^2}

= \frac{60}{7},

so

m + n = 60 + 7 = 67.