2018 AMC 8 Problems/Problem 20
Contents
Problem
In a point is on with and Point is on so that and point is on so that What is the ratio of the area of to the area of
Solution 1
By similar triangles, we have . Similarly, we see that [mathjax][BEF] = \dfrac{4}{9}[ABC][/mathjax]. Using this information, we get Then, since , it follows that the [mathjax][CDEF] = \dfrac{4}{9}[ABC][/mathjax]. Thus, the answer would be .
Sidenote: [mathjax][ABC][/mathjax] denotes the area of triangle [mathjax]ABC[/mathjax]. Similarly, [mathjax][ABCD][/mathjax] denotes the area of figure [mathjax]ABCD[/mathjax].
Solution 2
Let and the height of . We can extend to form a parallelogram, which would equal . The smaller parallelogram is times . The smaller parallelogram is of the larger parallelogram, so the answer would be , since the triangle is of the parallelogram, so the answer is .
By babyzombievillager with credits to many others who helped with the solution :D
Solution 3
. We can substitute as and as , where is . Side having, distance , has parts also. And, and are and respectfully. You can consider the height of and as and respectfully. The area of is because the area formula for a triangle is or . The area of will be . So, the area of will be . The area of parallelogram will be . Parallelogram to . The answer is .
Video Solution (CREATIVE ANALYSIS!!!)
~Education, the Study of Everything
Video Solution (Meta-Solving Technique)
https://youtu.be/GmUWIXXf_uk?t=1541
~ pi_is_3.14
Video Solution 2
~savannahsolver
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=TpsuRedYOiM&t=250s
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
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