2021 AMC 12A Spring 第 17 题

Question

梯形

ABCD

满足

AB \parallel CD

、

BC = CD = 43

，且

AD \perp BD

。设

O

为对角线

AC

和

BD

的交点，

P

为

BD

的中点。已知

OP = 11

，长度

AD

可写成

m\sqrt n

，其中

m

和

n

是正整数，且

n

不被任何质数的平方整除。求

m + n

。

Trapezoid

ABCD

has

AB \parallel CD,

BC = CD = 43,

and

AD \perp BD.

Let

O

be the intersection of the diagonals

AC

and

BD,

and let

P

be the midpoint of

BD.

Given that

OP = 11,

the length

AD

can be written in the form

m\sqrt n,

where

m

and

n

are positive integers and

n

is not divisible by the square of any prime. What is

m + n?

65

132

157

194

215

答案：D

知识点：坐标几何梯形相似

难度评级：2080

小提示：

由于

AD \perp BD

，将

D

放在原点，让

BD

沿一条坐标轴、

AD

沿另一条坐标轴

Place

D

at the origin with

BD

along one axis and

AD

along the other, since

AD \perp BD

大提示：

利用

BC = CD

可推出

C

对应的参数为

\tfrac12

；于是

O

分割

BD

，使

OP = \tfrac{BD}{6}

Using

BC = CD

forces

C

to be the midpoint-type point with parameter

\tfrac12;

then

O

divides

BD

so that

OP = \tfrac{BD}{6}

解答：

令

D = (0,0)

，

B = (b, 0)

在一条轴上，

A = (0, a)

在另一条轴上，使

AD \perp BD

。因为

CD \parallel AB

，可写成

C = t(b, -a)

，其中

t

为某个实数。于是

CD = t\sqrt{a^2+b^2}

，且

BC^2 = b^2(1-t)^2 + t^2a^2

。由

BC = CD

得

t^2 = (1-t)^2

，所以

t = \tfrac12

。

因此

C = \left(\tfrac{b}{2}, -\tfrac{a}{2}\right)

，且

CD = 43

给出

a^2 + b^2 = 4\cdot 43^2 = 7396

。对角线

AC

与

BD

（

x

-轴）交于

O = \left(\tfrac{b}{3}, 0\right)

，而

P = \left(\tfrac{b}{2}, 0\right)

。因此

OP = \tfrac{b}{6} = 11

，所以

b = 66

。

那么

a^2 = 7396 - 66^2 = 3040

，所以

AD = a = \sqrt{3040} = 4\sqrt{190}

。取

m = 4

、

n = 190

，得

m + n = 194

。

因此，正确答案是 D。

Place

D = (0,0)

with

B = (b, 0)

on one axis and

A = (0, a)

on the other, so that

AD \perp BD.

Since

CD \parallel AB,

write

C = t(b, -a)

for some

t.

Then

CD = t\sqrt{a^2+b^2}

and

BC^2 = b^2(1-t)^2 + t^2a^2.

Setting

BC = CD

gives

t^2 = (1-t)^2,

so

t = \tfrac12.

Thus

C = \left(\tfrac{b}{2}, -\tfrac{a}{2}\right),

and

CD = 43

gives

a^2 + b^2 = 4\cdot 43^2 = 7396.

The diagonal

AC

meets

BD

(the

x

-axis) at

O = \left(\tfrac{b}{3}, 0\right),

while

P = \left(\tfrac{b}{2}, 0\right).

Hence

OP = \tfrac{b}{6} = 11,

so

b = 66.

Then

a^2 = 7396 - 66^2 = 3040,

so

AD = a = \sqrt{3040} = 4\sqrt{190}.

With

m = 4

and

n = 190,

we get

m + n = 194.

Thus, the correct answer is D.

2021 AMC 12A Spring 第 17 题

答案：D

难度评级：2080

小提示：

大提示：

解答：

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