2020 AMC 10B Problem 8

Attempt Problem 8 of the 2020 AMC 10B 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 2020 AMC 10B solutions, or check the answer key.

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8.

Points PP and QQ lie in a plane with PQ=8.PQ=8. How many locations for point RR in this plane are there such that the triangle with vertices P,P, Q,Q, and RR is a right triangle with area 1212 square units?

22

44

66

88

1212

Answer: D
Concepts:right triangletriangle areacoordinate geometrycasework
Difficulty rating: 1420
Solution:

Place P=(4,0)P=(-4,0) and Q=(4,0).Q=(4,0). Since the area is 1212 and PQ=8,PQ=8, the distance from RR to line PQPQ is 3,3, so R=(x,±3).R=(x,\pm 3).

If the right angle is at P,P, then R=(4,±3),R=(-4,\pm 3), giving 22 points. If it is at Q,Q, then R=(4,±3),R=(4,\pm 3), giving 22 more points.

If the right angle is at R,R, then PQPQ is the hypotenuse, so 64=PR2+QR2=(x+4)2+9+(x4)2+9=2x2+50. \begin{aligned} &64=PR^2+QR^2 \\ &\quad =(x+4)^2+9 \\ &\quad {}+(x-4)^2+9 \\ &\quad =2x^2+50. \end{aligned} Thus x2=7,x^2=7, giving R=(±7,±3),R=(\pm\sqrt7,\pm 3), another 44 points.

The total is 2+2+4=8.2+2+4=8.

Thus, the correct answer is D .

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