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

Revision as of 18:59, 7 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

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

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 $\overline{PQ}$ or below $\overline{PQ}$ . Therefore, 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 $\overline{PQ}$ is a hypotenuse, then there are 4 possible points because the arrangement of the shorter and longer legs can be switched, and can be flipped to either side of $\overline{PQ}$ . Therefore, the answer is $2+2+4=\boxed{\textbf{(D) }8}.

Video Solution

https://youtu.be/OHR_6U686Qg

~IceMatrix

@@ Line 65: / Line 65: @@
 <math>2+2+4=\boxed{\textbf{(D) }8}</math> ~quacker88
-<math>\textbf{Completing the square}</math>
+==Solution 2==
+Note that line segment <math>\overline{PQ}</math> can either be the shorter leg, longer leg or the hypotenuse. If it is the shorter leg, there are two possible points for <math>Q</math> that can satisfy the requirements - that being above <math>\overline{PQ}</math> or below <math>\overline{PQ}</math>. Therefore, there are 2 ways for this case. Similarly, one can find that there are also 2 ways for point <math>Q</math> to lie if <math>\overline{PQ}</math> is the longer leg. If <math>\overline{PQ}</math> is a hypotenuse, then there are 4 possible points because the arrangement of the shorter and longer legs can be switched, and can be flipped to either side of <math>\overline{PQ}</math>. Therefore, the answer is $2+2+4=\boxed{\textbf{(D) }8}.
-Set the two values to be <math>x</math> and <math>y</math>. We know that <math>xy=24, x^2+y^2=64</math>. Adding <math>2</math> times equation <math>1</math> to equation <math>2</math> gives us <math>x^2+2xy+y^2=108 \implies (x+y)^2=108 \implies x+y=\sqrt{108}</math>. Plugging <math>x=\sqrt{108}-y</math> into the first equation and rearraging, we get <math>y^2-y\sqrt{108}+24</math>. The discriminant is <math>108-4(24)=12</math>, which is positive, so there are two solutions. However, we got the two solutions on the right side of <math>\overline{AB}</math>, there are two more on the left of <math>\overline{AB}</math> by symmetry. Again,
-<math>2+2+4=\boxed{\textbf{(D) }8}</math>
 ==Video Solution==

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