Problem:
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?
Answer Choices:
A. 2
B. 4
C. 6
D. 8
E. 12
Solution:
Let the brackets denote areas. We are given that
[PQR]=21⋅PQ⋅hR=12
Since PQ=8, it follows that hR=3. We construct a circle with diameter PQ. All such locations for R are shown below:
We apply casework to the right angle of △PQR:
1. If ∠P=90∘, then R∈{R1,R5} by the tangent.
2. If ∠Q=90∘, then R∈{R4,R8} by the tangent.
3. If ∠R=90∘, then R∈{R2,R3,R6,R7} by the Inscribed Angle Theorem.
Together, there are (D)8 such locations for R.
Remarks:
The reflections of R1,R2,R3,R4 about PQ are R5,R6,R7,R8, respectively
The reflections of R1,R2,R5,R6 about the perpendicular bisector of PQ are R4,R3,R8,R7, respectively.
OR
Let the brackets denote areas. We are given that
[PQR]=21⋅PQ⋅hR=12
Since PQ=8, it follows that hR=3. Without the loss of generality, let P=(−4,0) and Q=(4,0). We conclude that the y-coordinate of R must be ±3.
We apply casework to the right angle of △PQR:
1.∠P=90∘. The x-coordinate of R must be −4, so we have R=(−4,±3). In this case, there are 2 such locations for R.
2.∠Q=90∘. The x-coordinate of R must be 4, so we have R=(4,±3). In this case, there are 2 such locations for R.
3.∠R=90∘. For R=(x,3), the Pythagorean Theorem PR2+QR2=PQ2 gives
[(x+4)2+32]+[(x−4)2+32]=82
Solving this equation, we have x=±7, or R=(±7,3). For R=(x,−3), we have R=(±7,−3) by a similar process. In this case, there are 4 such locations for R.
Together, there are 2+2+4=(D)8 such locations for R.
