Difference between revisions of "2019 AMC 8 Problems/Problem 21"
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==Video Solutions== | ==Video Solutions== | ||
| + | ==Video Solution by Marshmallow== | ||
| + | https://youtu.be/DUhONk6cPy4 | ||
==Video Solution by Math-X (First understand the problem + diagram included!!!)== | ==Video Solution by Math-X (First understand the problem + diagram included!!!)== | ||
Revision as of 06:23, 17 January 2024
Contents
- 1 Problem 21
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Video Solutions
- 6 Video Solution by Marshmallow
- 7 Video Solution by Math-X (First understand the problem + diagram included!!!)
- 8 Video Solutions
- 9 Video Solution by OmegaLearn
- 10 Video Solution (CREATIVE THINKING!!!)
- 11 Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
- 12 See Also
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: ![\[\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}\]](http://latex.artofproblemsolving.com/4/e/8/4e88c659a71f5b5560729f07308d87bb87438a06.png)
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
Video Solution by Marshmallow
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
~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!!!)
~Education, the Study of Everything
Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
~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. 
