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 â–³PQR:
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 â–³PQR:
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.
