Question

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 $\tfrac{BC}{AD}?$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

Solution:

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

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

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 Problem 23

Solution:

Problem 23 in Other Years