2022 AMC 12A Problem 20

Problem 20 of 25HarderGeometry

Isosceles trapezoid

ABCD

has parallel sides

\overline{AD}

and

\overline{BC},

with

BC\lt 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

\dfrac{BC}{AD}?

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Solution

Place the trapezoid symmetric about the

y

-axis:

A=(-p,0),

D=(p,0),

B=(-q,h),

C=(q,h),

with

P=(x,y).

Then

PA^2-PD^2=4px

=1-16=-15

and

PB^2-PC^2=4qx

=4-9=-5.

Dividing gives

\dfrac{p}{q}=3.

Since

AD=2p

and

BC=2q,

we get

\dfrac{BC}{AD}=\dfrac{q}{p}=\dfrac13.

Thus, the correct answer is B.

More practice

Concepts: coordinate geometry · trapezoid · symmetry

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.