AMC12 2022 A

AMC12 2022 A · Q20

AMC12 2022 A · Q20. It mainly tests Circle theorems.

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 $\tfrac{BC}{AD}?$

等腰梯形 $ABCD$ 有平行边 $\overline{AD}$ 和 $\overline{BC}$，且 $BC < AD$，$AB = CD$。平面上存在点 $P$ 使得 $PA=1, PB=2, PC=3,$ 和 $PD=4$。求 $\tfrac{BC}{AD}$？

(A) \frac{1}{4} \frac{1}{4}

(B) \frac{1}{3} \frac{1}{3}

(D) \frac{2}{3} \frac{2}{3}

(E) \frac{3}{4} \frac{3}{4}

Answer

Correct choice: (B)

正确答案：（B）

Solution

Consider the reflection $P^{\prime}$ of $P$ over the perpendicular bisector of $\overline{BC}$, creating two new isosceles trapezoids $DAPP^{\prime}$ and $CBPP^{\prime}$. Under this reflection, $P^{\prime}A=PD=4$, $P^{\prime}D=PA=1$, $P^{\prime}C=PB=2$, and $P^{\prime}B=PC=3$. Since $DAPP'$ and $CBPP'$ are isosceles trapezoids, they are cyclic. Using Ptolemy's theorem on $DAPP'$, we get that $(PP')(AD) + (PA)(P'D) = (AP')(PD)$, so \[PP' \cdot AD + 1 \cdot 1 = 4 \cdot 4.\] Then, using Ptolemy's theorem again on $CBPP'$, we get that $(BC)(PP') + (BP)(CP') = (BP')(CP)$, so \[PP' \cdot BC + 2 \cdot 2 = 3 \cdot 3.\] Thus, $PP^{\prime}\cdot AD=15$ and $PP^{\prime}\cdot BC=5$; dividing these two equations and taking the reciprocal yields $\frac{BC}{AD}=\boxed{\textbf{(B) }\frac{1}{3}}$. (diagram by cinnamon_e)

考虑 $P$ 关于 $\overline{BC}$ 垂直平分线的反射 $P^{\prime}$，从而创建两个新的等腰梯形 $DAPP^{\prime}$ 和 $CBPP^{\prime}$。在此反射下，$P^{\prime}A=PD=4$，$P^{\prime}D=PA=1$，$P^{\prime}C=PB=2$，$P^{\prime}B=PC=3$。由于 $DAPP'$ 和 $CBPP'$ 是等腰梯形，故是循环四边形。对 $DAPP'$ 使用托勒密定理，得到 $(PP')(AD) + (PA)(P'D) = (AP')(PD)$，即 \[PP' \cdot AD + 1 \cdot 1 = 4 \cdot 4.\] 然后，对 $CBPP'$ 再次使用托勒密定理，得到 $(BC)(PP') + (BP)(CP') = (BP')(CP)$，即 \[PP' \cdot BC + 2 \cdot 2 = 3 \cdot 3.\] 因此，$PP^{\prime}\cdot AD=15$ 和 $PP^{\prime}\cdot BC=5$；将这两个方程相除并取倒数，得 $\frac{BC}{AD}=\boxed{\textbf{(B) }\frac{1}{3}}$。（图由 cinnamon_e 绘制）

Topics

GE.006 Circle theorems