2020 AMC 10B Problem 8

Take $PQ$ as the base. Area $12 = \tfrac12 \cdot 8 \cdot h$ gives $h = 3$, so $R$ must lie on one of the two lines parallel to $PQ$ at distance $3$ from it.

Now decide where the right angle is.

Right angle at $P$. Then $PR \perp PQ$, so $R$ is on the perpendicular to $PQ$ at $P$, at distance $3$: two points (one on each side). Likewise right angle at $Q$ gives two more points. Subtotal: $4$.

Right angle at $R$. By Thales' theorem, $R$ lies on the circle with diameter $PQ$, radius $4$, centered at the midpoint of $PQ$. Each parallel line at distance $3 < 4$ from the center crosses this circle in two points, so the two lines give $4$ points.

None of these coincide: the first four points are $\sqrt{4^2 + 3^2} = 5$ from the midpoint of $PQ$, so they are not on the circle of radius $4$. Total $4 + 4 = 8$.

The answer is $\boxed{\textbf{(D)}\ 8}$.

Two constraints, two loci: the area condition pins $R$ to a pair of parallel lines, and "right triangle" pins $R$ to one of three curves (two perpendicular lines and the Thales circle). Counting the intersections case by case is systematic and avoids missing configurations. Always compare the offset ($3$) to the radius ($4$) to confirm the circle is actually hit.

Coordinates: $P = (0,0)$, $Q = (8,0)$, $R = (x, \pm 3)$. Right angle at $P$: $x = 0$; at $Q$: $x = 8$; at $R$: $\vec{RP}\cdot\vec{RQ} = x(x-8) + 9 = 0$, so $x = 4 \pm \sqrt{7}$. That is $4$ values of $x$ times $2$ signs, giving $8$.

• Counting only the four positions where the right angle is at P or Q, and missing the four on the circle with diameter PQ.
• Forgetting that $R$ can be on either side of line $PQ$, which halves the count to $4$.