2019 AMC 8 Problems/Problem 21
Contents
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 , and 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
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
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.