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Difference between revisions of "2019 AMC 8 Problems/Problem 21"

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==Solution 1==
 
==Solution 1==
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First we need to find the coordinates where the graphs intersect.  
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First, we need to find the coordinates where the graphs intersect.  
  
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<math>y=5</math>, and <math>y=x+1</math> intersect at <math>(4,5)</math>,  
+
We want the points x and y to be the same. Thus, we set <math>5=x+1,</math> and get <math>x=4.</math> Plugging this into the equation, <math>y=1-x,</math>
 +
<math>y=5</math>, and <math>y=1+x</math> intersect at <math>(4,5)</math>, we call this line x.
  
−
<math>y=5</math>, and <math>y=1-x</math> intersect at <math>(-4,5)</math>,
+
Doing the same thing, we get <math>x=-4.</math> Thus, <math>y=5</math>. Also,
 +
<math>y=5</math> and <math>y=1-x</math> intersect at <math>(-4,5)</math>, and we call this line y.
  
−
<math>y=1-x</math> and <math>y=1+x</math> intersect at <math>(0,1)</math>.  
+
It's apparent the only solution to <math>1-x=1+x</math> is <math>0.</math> Thus, <math>y=1.</math>
 +
<math>y=1-x</math> and <math>y=1+x</math> intersect at <math>(0,1)</math>, we call this line z.
  
−
Using the [[Shoelace Theorem]] we get: <cmath>\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}</cmath> <math>=</math> So our answer is <math>\boxed{\textbf{(E)}\ 16}</math>.
+
Using the [[Shoelace Theorem]] we get: <cmath>\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}</cmath> <math>=</math> So, our answer is <math>\boxed{\textbf{(E)}\ 16.}</math>
  
−
~heeeeeeheeeee
+
We might also see that the lines <math>y</math> and <math>x</math> are mirror images of each other. This is because, when rewritten, their slopes can be multiplied by <math>-1</math> to get the other. As the base is horizontal, this is a isosceles triangle with base 8, as the intersection points have distance 8. The height is <math>5-1=4,</math> so <math>\frac{4\cdot 8}{2} = \boxed{\textbf{(E)} 16.}</math>
  
−
~more edits by BakedPotato66
+
Warning: Do not use the distance formula for the base then use Heron's formula. It will take you half of the time you have left!
  
 
==Solution 2==
 
==Solution 2==
−
Graphing the lines, using the intersection points we found in Solution 1, we can see that the height of the triangle is 4, and the base is 8. Using the formula for the area of a triangle, we get <math>\frac{4\cdot8}{2}</math> which is equal to <math>\boxed{\textbf{(E)}\ 16}</math>.  
+
Graphing the lines, using the intersection points we found in Solution 1, we can see that the height of the triangle is 4, and the base is 8. Using the formula for the area of a triangle, we get <math>\frac{4\cdot8}{2}</math> which is equal to <math>\boxed{\textbf{(E)}\ 16}</math>.
  
−
~SmileKat32
+
==Solution 3==
  
−
~more edits by BakedPotato66
+
<math>y = x + 1</math> and <math>y = -x + 1</math> have <math>y</math>-intercepts at <math>(0, 1)</math> and slopes of <math>1</math> and <math>-1</math>, respectively. Since the product of these slopes is <math>-1</math>, the two lines are perpendicular. From <math>y = 5</math>, we see that <math>(-4, 5)</math> and <math>(4, 5)</math> are the other two intersection points, and they are <math>8</math> units apart. By symmetry, this triangle is a <math>45-45-90</math> triangle, so the legs are <math>4\sqrt{2}</math> each and the area is <math>\frac{(4\sqrt{2})^2}{2} = \boxed{\textbf{(E)}\ 16}</math>.
  
 
==Video Solutions==
 
==Video Solutions==
 +
==Video Solution by Marshmallow==
 +
https://youtu.be/DUhONk6cPy4
 +
 +
==Video Solution by Math-X (First understand the problem + diagram included!!!)==
 +
https://youtu.be/IgpayYB48C4?si=i9Nnwi2KpuHkL82l&t=6214
 +
 +
~Math-X
 +
 +
 +
https://www.youtube.com/watch?v=mz3DY1rc5ao
 +
 +
- Happytwin
 +
 +
Associated Video - https://www.youtube.com/watch?v=ie3tlSNyiaY
 +
 
https://www.youtube.com/watch?v=9nlX9VCisQc
 
https://www.youtube.com/watch?v=9nlX9VCisQc
  
−
https://www.youtube.com/watch?v=mz3DY1rc5ao
+
https://www.youtube.com/watch?v=Z27G0xy5AgA&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=3
 +
 
 +
~ MathEx
 +
 
 +
https://youtu.be/-5C6ACg5Hl0
 +
 
 +
~savannahsolver
 +
 
 +
https://www.youtube.com/watch?v=SBGcumVOroI&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=23
 +
 
 +
==Video Solutions==
 +
 
 +
https://www.youtube.com/watch?v=x4cF3o3Fzj8&t=123s
 +
 
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/j3QSD5eDpzU?t=2607
 +
 
 +
~ pi_is_3.14
 +
 
 +
==Video Solution (CREATIVE THINKING!!!)==
 +
https://youtu.be/1RLoLeJjWko
 +
 
 +
~Education, the Study of Everything
  
−
https://www.youtube.com/watch?v=Z27G0xy5AgA&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=3 ~ MathEx
+
==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)==
 +
https://youtu.be/Xm4ZGND9WoY
  
−
https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25
+
~Hayabusa1
  
 
==See Also==
 
==See Also==

Revision as of 06:23, 17 January 2024

Problem 21

What is the area of the triangle formed by the lines $y=5$, $y=1+x$, and $y=1-x$?

$\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16$

Solution 1

First, we need to find the coordinates where the graphs intersect.

We want the points x and y to be the same. Thus, we set $5=x+1,$ and get $x=4.$ Plugging this into the equation, $y=1-x,$ $y=5$, and $y=1+x$ intersect at $(4,5)$, we call this line x.

Doing the same thing, we get $x=-4.$ Thus, $y=5$. Also, $y=5$ and $y=1-x$ intersect at $(-4,5)$, and we call this line y.

It's apparent the only solution to $1-x=1+x$ is $0.$ Thus, $y=1.$ $y=1-x$ and $y=1+x$ intersect at $(0,1)$, we call this line z.

Using the Shoelace Theorem we get: \[\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}\] $=$ So, our answer is $\boxed{\textbf{(E)}\ 16.}$

We might also see that the lines $y$ and $x$ are mirror images of each other. This is because, when rewritten, their slopes can be multiplied by $-1$ to get the other. As the base is horizontal, this is a isosceles triangle with base 8, as the intersection points have distance 8. The height is $5-1=4,$ so $\frac{4\cdot 8}{2} = \boxed{\textbf{(E)} 16.}$

Warning: Do not use the distance formula for the base then use Heron's formula. It will take you half of the time you have left!

Solution 2

Graphing the lines, using the intersection points we found in Solution 1, we can see that the height of the triangle is 4, and the base is 8. Using the formula for the area of a triangle, we get $\frac{4\cdot8}{2}$ which is equal to $\boxed{\textbf{(E)}\ 16}$.

Solution 3

$y = x + 1$ and $y = -x + 1$ have $y$-intercepts at $(0, 1)$ and slopes of $1$ and $-1$, respectively. Since the product of these slopes is $-1$, the two lines are perpendicular. From $y = 5$, we see that $(-4, 5)$ and $(4, 5)$ are the other two intersection points, and they are $8$ units apart. By symmetry, this triangle is a $45-45-90$ triangle, so the legs are $4\sqrt{2}$ each and the area is $\frac{(4\sqrt{2})^2}{2} = \boxed{\textbf{(E)}\ 16}$.

Video Solutions

Video Solution by Marshmallow

https://youtu.be/DUhONk6cPy4

Video Solution by Math-X (First understand the problem + diagram included!!!)

https://youtu.be/IgpayYB48C4?si=i9Nnwi2KpuHkL82l&t=6214

~Math-X


https://www.youtube.com/watch?v=mz3DY1rc5ao

- Happytwin

Associated Video - https://www.youtube.com/watch?v=ie3tlSNyiaY

https://www.youtube.com/watch?v=9nlX9VCisQc

https://www.youtube.com/watch?v=Z27G0xy5AgA&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=3

~ MathEx

https://youtu.be/-5C6ACg5Hl0

~savannahsolver

https://www.youtube.com/watch?v=SBGcumVOroI&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=23

Video Solutions

https://www.youtube.com/watch?v=x4cF3o3Fzj8&t=123s

Video Solution by OmegaLearn

https://youtu.be/j3QSD5eDpzU?t=2607

~ pi_is_3.14

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/1RLoLeJjWko

~Education, the Study of Everything

Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~Hayabusa1

See Also

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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 AJHSME/AMC 8 Problems and Solutions

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

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