2022 AMC 12A Problem 20

Attempt Problem 20 of the 2022 AMC 12A below, then check your answer against the professionally curated solution 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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20.

Isosceles trapezoid ABCDABCD has parallel sides AD\overline{AD} and BC,\overline{BC}, with BC<ADBC\lt AD and AB=CD.AB=CD. There is a point PP in the plane such that PA=1,PA=1, PB=2,PB=2, PC=3,PC=3, and PD=4.PD=4. What is BCAD?\dfrac{BC}{AD}?

14\dfrac14

13\dfrac13

12\dfrac12

23\dfrac23

34\dfrac34

Answer: B
Concepts:coordinate geometrytrapezoidsymmetry
Difficulty rating: 2110
Solution:

Place the trapezoid symmetric about the yy-axis: A=(p,0),A=(-p,0), D=(p,0),D=(p,0), B=(q,h),B=(-q,h), C=(q,h),C=(q,h), with P=(x,y).P=(x,y).

Then PA2PD2=4pxPA^2-PD^2=4px =116=15=1-16=-15 and PB2PC2=4qxPB^2-PC^2=4qx =49=5.=4-9=-5. Dividing gives pq=3.\dfrac{p}{q}=3.

Since AD=2pAD=2p and BC=2q,BC=2q, we get BCAD=qp=13.\dfrac{BC}{AD}=\dfrac{q}{p}=\dfrac13.

Thus, the correct answer is B.

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