Difference between revisions of "2020 AMC 10B Problems/Problem 8"

Revision as of 18:25, 19 February 2020

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.

(Diagram temporarily removed due to asymptote error)

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.

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

Video Solution

https://youtu.be/OHR_6U686Qg

~IceMatrix

@@ Line 19: / Line 19: @@
 . The new point is the right angle.
-(Diagram temprarily removed due to asymptote error)
+(Diagram temporarily removed due to asymptote error)
 The diagram looks something like this. We know that the altitude to base <math>\overline{AB}</math> must be <math>3</math> since the area is <math>12</math>. From here, we must see if there are valid triangles that satisfy the necessary requirements.

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

Page

Toolbox

Search