Question

Points $P$ and $Q$ lie in a plane with $PQ=8.$ How many locations for point $R$ in this plane are there such that the triangle with vertices $P,$ $Q,$ and $R$ is a right triangle with area $12$ square units?

$2$

$4$

$6$

$8$

$12$

Solution:

Place $P=(-4,0)$ and $Q=(4,0)$ . Since the area is $12$ and $PQ=8$ , the distance from $R$ to line $PQ$ is $3$ , so $R=(x,\pm 3)$ .

If the right angle is at $P$ , then $R=(-4,\pm 3)$ , giving $2$ points. If it is at $Q$ , then $R=(4,\pm 3)$ , giving $2$ more points.

If the right angle is at $R$ , then $PQ$ is the hypotenuse, so $64=PR^2+QR^2=(x+4)^2+9+(x-4)^2+9=2x^2+50.$ Thus $x^2=7$ , giving $R=(\pm\sqrt7,\pm 3)$ , another $4$ points.

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

Thus, the correct answer is D .

2020 AMC 10B Problem 8

Solution:

Problem 8 in Other Years