AMC10 2022 A

AMC10 2022 A · Q23

AMC10 2022 A · Q23. It mainly tests Geometry misc.

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$\overline{BC}$，其中$BC < AD$且$AB = CD$。平面上存在一点$P$使得$PA=1, PB=2, PC=3,$和$PD=4$。$\tfrac{BC}{AD}$是多少？

(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，$P^{\prime}A=PD=4$，$P^{\prime}D=PA=1$，$P^{\prime}C=PB=2$，且$P^{\prime}B=PC=3$。由于$DAPP'$和$CBPP'$是等腰梯形，它们是循环四边形。使用$DAPP'$上的托勒密定理，我们得到$(PP')(AD) + (PA)(P'D) = (AP')(PD)$，所以 \[PP' \cdot AD + 1 \cdot 1 = 4 \cdot 4.\] 然后，再次使用$CBPP'$上的托勒密定理，我们得到$(BC)(PP') + (BP)(CP') = (BP')(CP)$，所以 \[PP' \cdot BC + 2 \cdot 2 = 3 \cdot 3.\] 因此，$PP^{\prime}\cdot AD=15$且$PP^{\prime}\cdot BC=5$；将这两个方程相除并取倒数得到 $\frac{BC}{AD}=\boxed{\textbf{(B) }\frac{1}{3}}$。 (图由 cinnamon_e 绘制)

Topics

GE.020 Geometry misc