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

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==Solution==
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==Problem==
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Points <math>P</math> and <math>Q</math> lie in a plane with <math>PQ=8</math>. How many locations for point <math>R</math> in this plane are there such that the triangle with vertices <math>P</math>, <math>Q</math>, and <math>R</math> is a right triangle with area <math>12</math> square units?
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<math>\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\  6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12</math>
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==Solution 1==
  
 
There are <math>3</math> options here:
 
There are <math>3</math> options here:
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<asy>
 
<asy>
 
 
pair  A, B, C, D, X, Y;
 
pair  A, B, C, D, X, Y;
 
A = (0,0);
 
A = (0,0);
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X = (0,4);
 
X = (0,4);
 
Y = (1.5,6.64575131106);
 
Y = (1.5,6.64575131106);
 
  
 
draw(A--B--C--A);
 
draw(A--B--C--A);
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Label AB= Label("$8$", position=MidPoint);
 
Label AB= Label("$8$", position=MidPoint);
 
 
</asy>
 
</asy>
  
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We know that the minimum value of <math>\overline{BC}^2+\overline{AC}^2=64</math> is when <math>\overline{BC} = \overline{AC} = \sqrt{24}</math>. In this case, the equation becomes <math>24+24=48</math>, which is LESS than <math>64</math>.  
 
We know that the minimum value of <math>\overline{BC}^2+\overline{AC}^2=64</math> is when <math>\overline{BC} = \overline{AC} = \sqrt{24}</math>. In this case, the equation becomes <math>24+24=48</math>, which is LESS than <math>64</math>.  
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<math>\overline{BC}=1, \overline{AC} =24</math>. The equation becomes <math>1+576=577</math>, which is obviously greater than <math>64</math>. We can conclude that there are values for <math>\overline{BC}</math> and <math>\overline{AC}</math> in between that satisfy the Pythagorean Theorem.
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And since  <math>\overline{BC} \neq \overline{AC}</math>, the triangle is not isoceles, meaning we could reflect it over <math>\overline{AB}</math> and/or the line perpendicular to <math>\overline{AB}</math> for a total of <math>4</math> triangles this case.
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Therefore, the answer is <math>2+2+4=\boxed{\textbf{(D) }8}</math>.
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==Solution 2==
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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 or below <math>\overline{PQ}</math>. As such, there are <math>2</math> ways for this case. Similarly, one can find that there are also <math>2</math> ways for point <math>Q</math> to lie if <math>\overline{PQ}</math> is the longer leg. If it is a hypotenuse, then there are <math>4</math> 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 <math>2+2+4=\boxed{\textbf{(D) }8}</math>.
  
Another possibility is if <math>\overline{BC}=1, \overline{AC} =24</math>. The equation becomes <math>1+576=577</math>, which is obviously greater than <math>64</math>. We can conclude that there are values for <math>\overline{BC}</math> and <math>\overline{AC}</math> in between that satisfy the Pythagorean Theorem.
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==Video Solution==
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https://youtu.be/OHR_6U686Qg
  
And since  <math>\overline{BC} \neq \overline{AC}</math>, the triangle is not isoceles, meaning we could reflect it over <math>\overline{AB}</math> and/or the line perpendicular to <math>\overline{AB}</math> for a total of <math>4</math> triangles this case.
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~IceMatrix
 
<math>2+2+4=\boxed{\textbf{D) }8}</math> ~quacker88
 
  
<math>\textbf{Completing the square}</math>
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https://youtu.be/cUzK5DqKaRY
  
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,
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~savannahsolver
<math>2+2+4=\boxed{\textbf{D) }8}</math>
 
  
 
== See Also ==
 
== See Also ==
  
{{AMC10 box|year=2020|ab=B|before=Problem 7|num-a=9}}
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{{AMC10 box|year=2020|ab=B|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:39, 7 July 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.

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

Video Solution

https://youtu.be/OHR_6U686Qg

~IceMatrix

https://youtu.be/cUzK5DqKaRY

~savannahsolver

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
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

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