Difference between revisions of "2019 AMC 8 Problems/Problem 21"
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First we need to find the coordinates where the graphs intersect. | First we need to find the coordinates where the graphs intersect. | ||
− | <math>y=5</math>, and <math>y=x | + | 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>, 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> |
− | + | We might also see that the lines <math>y</math> and <math>z</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> | |
− | + | 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>. |
− | + | ==Solution 3== | |
− | + | <math>y = x + 1</math> and <math>y = -x + 1</math> have <math>y</math>-intercepts at <math>(1, 0)</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== | ||
+ | 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= | + | https://www.youtube.com/watch?v=Z27G0xy5AgA&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=3 ~ MathEx |
− | https:// | + | https://youtu.be/RvtOX17DemY |
− | + | ~savannahsolver | |
==See Also== | ==See Also== |
Latest revision as of 19:48, 16 January 2021
Problem 21
What is the area of the triangle formed by the lines , , and ?
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 and get Plugging this into the equation, , and intersect at , we call this line x.
Doing the same thing, we get Thus also. , and intersect at , we call this line y.
It's apparent the only solution to is Thus, and intersect at , we call this line z.
Using the Shoelace Theorem we get: So our answer is
We might also see that the lines and are mirror images of each other. This is because, when rewritten, their slopes can be multiplied by 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 so
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 which is equal to .
Solution 3
and have -intercepts at and slopes of and , respectively. Since the product of these slopes is , the two lines are perpendicular. From , we see that and are the other two intersection points, and they are units apart. By symmetry, this triangle is a triangle, so the legs are each and the area is .
Video Solutions
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
~savannahsolver
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.