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

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

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

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

解答：

其他年份的第 23 题