1995 AMC 12 第 28 题

Question

圆内两条平行弦的长度分别为

10

和

14

，它们之间的距离为

6

。另有一条弦与它们平行，并且位于它们正中间，其长度为

\sqrt a

。则

a

是

Two parallel chords in a circle have lengths

10

and

14,

and the distance between them is

6.

The chord parallel to these chords and midway between them is of length

\sqrt a,

where

a

is

144

156

168

176

184

答案：E

知识点：circle chords 坐标几何

难度评级：2290

小提示：

若一条弦到圆心的有向距离为

y

，其半弦长满足

h^2+y^2=r^2

If a chord is at signed distance

y

from the center, its half-length satisfies

h^2+y^2=r^2

大提示：

设长为

14

与

10

的两条弦到圆心的有向距离相差

6

，再将两式相减

Let the signed distances of the

14

- and

10

-chords differ by

6

, and subtract their equations

解答：

设圆心到长为

14

和

10

的弦的有向距离分别为

u

和

v

，且

u-v=6

。则

r^2-u^2=7^2,\qquad r^2-v^2=5^2\text{。}

因此

v^2-u^2=24

，所以

(v-u)(v+u)=24

。由于

v-u=-6

，可得

u+v=-4

，进而

u=1

，且

v=-5

。中间那条弦的有向距离为

-2

，而

r^2=49+1=50

。它的长度平方为

4(50-4)=184

，所以

a=184

。因此正确答案是 E。

Let the signed distances from the center to the

14

- and

10

-chords be

u

and

v,

with

u-v=6.

Then

r^2-u^2=7^2,\qquad r^2-v^2=5^2.

Hence

v^2-u^2=24,

so

(v-u)(v+u)=24.

Since

v-u=-6,

we get

u+v=-4,

and therefore

u=1

and

v=-5.

The midway chord is at signed distance

-2,

while

r^2=49+1=50.

Its squared length is

4(50-4)=184,

so

a=184.

Thus the correct answer is E.

1995 AMC 12 第 28 题

答案：E

难度评级：2290

小提示：

大提示：

解答：

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