# 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 1

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 $64$. $\overline{BC}=1, \overline{AC} =24$. 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.

Therefore, the answer is $2+2+4=\boxed{\textbf{(D) }8}$.

## Solution 2

Note that line segment $\overline{PQ}$ can either be the shorter leg, longer leg or the hypotenuse. If it is the shorter leg, there are two possible points for $Q$ that can satisfy the requirements - that being above or below $\overline{PQ}$. As such, there are $2$ ways for this case. Similarly, one can find that there are also $2$ ways for point $Q$ to lie if $\overline{PQ}$ is the longer leg. If it is a hypotenuse, then there are $4$ possible points because the arrangement of the shorter and longer legs can be switched, and can be either above or below the line segment. Therefore, the answer is $2+2+4=\boxed{\textbf{(D) }8}$.

## Solution 3 (Algebra)

Let a and b be the distances of R from P and Q, respectively. By the Pythagorean Theorem, we have $a^2 + b^2 = 64$. Since the area of this triangle is 12, we get $a * b = 12 * 2 = 24$. Thus $b = 24/a$. Now substitute this into the other equation to get $a^2 + (24/a)^2 = 64$. Multiplying by $a^2$ on both sides, we get $a^4 + 24 = 64*a^2$. Now let $y = a^2$. Substituting and rearranging, we get $y^2 - 64*y + 24 = 0$. We reduced this problem to a quadratic equation! By the quadratic formula, our solutions are $y = 32 \pm 10*\sqrt{10}$. Now substitute back $y = a^2$ to get $a = \pm \sqrt{32 \pm 10*\sqrt{10}}$. All 4 of these solutions are rational and will work. But our answer is actually $4 * 2 = 8$ as we have only calculated the number of places where R could go above line segment PQ. We need to multiply our 4 by 2 to account for all the places R could go below segment PQ. ~mewto

## Solution 4 (Illustration)

Let the brackets denote areas. We know that $$[PQR]=\frac12\cdot PQ\cdot h_R=12.$$ Since $PQ=8,$ it follows that $h_R=3.$

All such locations for $R$ are shown below:

For $\triangle PQR,$ note that

1. If $\angle P=90^\circ,$ then $R\in\{R_1,R_5\}$ by construction.
2. If $\angle Q=90^\circ,$ then $R\in\{R_4,R_8\}$ by construction.
3. If $\angle R=90^\circ,$ then $R\in\{R_2,R_3,R_6,R_7\}$ by the Inscribed Angle Theorem.

Together, there are $\boxed{\textbf{(D)}\ 8}$ such locations for $R.$

Remarks

1. The reflections of $R_1,R_2,R_3,R_4$ about $\overleftrightarrow{PQ}$ are $R_5,R_6,R_7,R_8,$ respectively.
2. The reflections of $R_1,R_2,R_5,R_6$ about the perpendicular bisector of $\overline{PQ}$ are $R_4,R_3,R_8,R_7,$ respectively.

~MRENTHUSIASM

## Video Solution

~IceMatrix

~savannahsolver

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 