2022 AMC 10A Problem 23

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}

Reflect P across the symmetry line of the isosceles trapezoid

Apply Ptolemy to the two cyclic trapezoids formed by the reflection

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.

Problem 23 in other years