2020 AMC 10B Problems/Problem 8

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?

$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12$

Solution

There are $3$ options here:

1. $\textbf{P}$ is the right angle.

It's clear that there are $2$ points that fit this, one that's directly to the right of $P$ and one that's directly to the left. We don't need to find the length, we just need to know that it is possible, which it is.

2. $\textbf{Q}$ is the right angle.

Using the exact same reasoning, there are also $2$ solutions for this one.

3. The new point is the right angle.

$[asy] pair A, B, C, D, X, Y; A = (0,0); B = (0,8); C = (3,6.64575131106); D = (0,6.64575131106); X = (0,4); Y = (1.5,6.64575131106); draw(A--B--C--A); draw(C--D); label("$8$", X, W); label("$3$", Y, S); dot("$A$", A, S); dot("$B$", B, N); dot("$C$", C, E); draw(rightanglemark(A, C, B)); draw(rightanglemark(A, D, C)); Label AB= Label("$8$", position=MidPoint); [/asy]$

The diagram looks something like this. We know that the altitude to base $\overline{AB}$ must be $3$ since the area is $12$ . From here, we must see if there are valid triangles that satisfy the necessary requirements.

First of all, $\frac{\overline{BC}\cdot\overline{AC}}{2}=12 \implies \overline{BC}\cdot\overline{AC}=24$ because of the area.

Next, $\overline{BC}^2+\overline{AC}^2=64$ from the Pythagorean Theorem.

From here, we must look to see if there are valid solutions. There are multiple ways to do this:

$\textbf{Recognizing min \& max:}$

We know that the minimum value of $\overline{BC}^2+\overline{AC}^2=64$ is when $\overline{BC} = \overline{AC} = \sqrt{24}$ . In this case, the equation becomes $24+24=48$ , which is LESS than $64$ .

Another possibility is if $\overline{BC}=1, \overline{AC} =24$ . The equation becomes $1+576=577$ , which is obviously greater than $64$ . We can conclude that there are values for $\overline{BC}$ and $\overline{AC}$ in between that satisfy the Pythagorean Theorem.

And since $\overline{BC} \neq \overline{AC}$ , the triangle is not isoceles, meaning we could reflect it over $\overline{AB}$ and/or the line perpendicular to $\overline{AB}$ for a total of $4$ triangles this case.

$2+2+4=\boxed{\textbf{(D) }8}$ ~quacker88

$\textbf{Completing the square}$

Set the two values to be $x$ and $y$ . We know that $xy=24, x^2+y^2=64$ . Adding $2$ times equation $1$ to equation $2$ gives us $x^2+2xy+y^2=108 \implies (x+y)^2=108 \implies x+y=\sqrt{108}$ . Plugging $x=\sqrt{108}-y$ into the first equation and rearraging, we get $y^2-y\sqrt{108}+24$ . The discriminant is $108-4(24)=12$ , which is positive, so there are two solutions. However, we got the two solutions on the right side of $\overline{AB}$ , there are two more on the left of $\overline{AB}$ by symmetry. Again, $2+2+4=\boxed{\textbf{(D) }8}$

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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2020 AMC 10B Problems/Problem 8

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