2022 AMC 10A 第 23 题

Question

等腰梯形

ABCD

的平行边为

\overline{AD}

与

\overline{BC}

，其中

BC < AD

，且

AB = CD

。平面上有一点

P

，使得

PA=1

，

PB=2

，

PC=3

、

PD=4

。求

\tfrac{BC}{AD}

。

Isosceles trapezoid

ABCD

has parallel sides

\overline{AD}

and

\overline{BC},

with

BC < AD

and

AB = CD.

There is a point

P

in the plane such that

PA=1,

PB=2,

PC=3,

and

PD=4.

What is

\tfrac{BC}{AD}?

\dfrac{1}{4}

\dfrac{1}{3}

\dfrac{1}{2}

\dfrac{2}{3}

\dfrac{3}{4}

答案：B

知识点：梯形托勒密定理对称性

难度评级：2230

小提示：

把 P 关于等腰梯形的对称轴反射

Reflect P across the symmetry line of the isosceles trapezoid

大提示：

对反射得到的两个圆内接梯形使用托勒密定理

Apply Ptolemy to the two cyclic trapezoids formed by the reflection

解答：

令

P'

为

P

关于

\overline{BC}

的垂直平分线的反射点。

这样得到两个等腰梯形

CBPP'

和

DAPP'

。

由反射对称可得

\begin{gathered} P'A = PD = 4 \\ P'D = PA = 1 \\ P'C = PB = 2 \\ P'B = PC = 3 \end{gathered}\text{。}

对两个圆内接梯形使用托勒密定理，知道对角线乘积等于两组对边乘积之和。因此

\begin{gathered} PP' \cdot AD + 1 = 16 \\ PP' \cdot BC + 4 = 9 \end{gathered}\text{。}

于是

PP' \cdot AD = 15

，且

PP' \cdot BC = 5

。两式相除得到

\dfrac{BC}{AD} = \dfrac{1}{3}

。

所以正确答案是 B。

Let

P'

be the reflection of

P

across the perpendicular bisector of

\overline{BC}.

This forms two new isosceles trapezoids:

CBPP'

and

DAPP'.

Therefore, we get

\begin{gathered} P'A = PD = 4 \\ P'D = PA = 1 \\ P'C = PB = 2 \\ P'B = PC = 3. \end{gathered}

Using Ptolemy’s theorem, we know that the product of the diagonals is equal to the sum of the products of the opposite sides. Therefore:

\begin{gathered} PP' \cdot AD + 1 = 16 \\ PP' \cdot BC + 4 = 9. \end{gathered}

This gets us

PP' \cdot AD = 15

and

PP' \cdot BC = 5.

Dividing these two equations yields

\dfrac{BC}{AD} = \dfrac{1}{3}.

Thus, B is the correct answer.

2022 AMC 10A 第 23 题

答案：B

难度评级：2230

小提示：

大提示：

解答：

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