2017 AMC 10A Problems/Problem 21
Contents
Problem
A square with side length is inscribed in a right triangle with sides of length , , and so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length is inscribed in another right triangle with sides of length , , and so that one side of the square lies on the hypotenuse of the triangle. What is ?
Solution 1
Analyze the first right triangle.
Note that and are similar, so . This can be written as . Solving, .
Now we analyze the second triangle.
Similarly, and are similar, so , and . Thus, . Solving for , we get . Thus, .
Solution 2 (Alternate solution in finding x)
Set the right-angle vertex of the triangle as . Notice that the hypotenuse of the triangle, as depicted in solution one, can be described by , while can be describe by . Hence, we may solve for by solving , which yields .
Proceed by finding the value of y via the method described in solution 1, and we will get . Thus, .
Note
In general, if the legs were and , we have that This can be verified by plugging in and . ~anduran
Video Solutions
https://youtu.be/THeq4ZiZxIA -Video Solution by TheBeautyOfMath
https://youtu.be/MF2QFOInbYc -Video Solution by Richard Rusczyk
See Also
2017 AMC 10A (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 AMC 10 Problems and Solutions |
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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 12 Problems and Solutions |
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