Difference between revisions of "2021 Fall AMC 10B Problems/Problem 13"
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~Hefei417, or 陆畅 Sunny from China | ~Hefei417, or 陆畅 Sunny from China | ||
− | == Solution 3 (With two | + | == Solution 3 (With two different endings)== |
This solution is based on this figure: [[:Image:2021_AMC_10B_(Nov)_Problem_13,_sol.png]] | This solution is based on this figure: [[:Image:2021_AMC_10B_(Nov)_Problem_13,_sol.png]] | ||
Revision as of 21:01, 31 December 2021
Contents
[hide]Problem
A square with side length is inscribed in an isosceles triangle with one side of the square along the base of the triangle. A square with side length has two vertices on the other square and the other two on sides of the triangle, as shown. What is the area of the triangle?
Solution 1
Let's split the triangle down the middle and label it:
We see that by AA similarity. because cuts the side length of the square in half; similarly, . Let : then by side ratios,
.
Now the height of the triangle is . By side ratios, .
The area of the triangle is
~KingRavi
Solution 2
By similarity, the height is and the base is . Thus the area is , or .
~Hefei417, or 陆畅 Sunny from China
Solution 3 (With two different endings)
This solution is based on this figure: Image:2021_AMC_10B_(Nov)_Problem_13,_sol.png
Denote by the midpoint of .
Because , , , we have .
We observe . Hence, . Hence, . By symmetry, .
Therefore, .
Because is the midpoint of , .
We observe . Hence, . Hence, .
Therefore, .
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Alternatively, we can find the height in a slightly different way.
Following from our finding that the base of the large triangle , we can label the length of the altitude of as . Notice that . Hence, . Substituting and simplifying, . Therefore, the area of the triangle is .
~mahaler
Video Solution by Interstigation
https://www.youtube.com/watch?v=mq4e-s9ENas
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.