2022 AMC 12A Problem 20

Below is the professionally curated solution for Problem 20 of the 2022 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 12A solutions, or check the answer key.

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Concepts:coordinate geometry trapezoid symmetry

Difficulty rating: 2110

20.

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 $\dfrac{BC}{AD}?$

$\dfrac14$

$\dfrac13$

$\dfrac12$

$\dfrac23$

$\dfrac34$

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.

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