Difference between revisions of "2017 AMC 10A Problems/Problem 21"
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+ | https://youtu.be/DJ1105lcJJM?si=Z24jb7mCjzjLPdKh | ||
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+ | ~ Pi Academy | ||
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+ | ==Other video solutions== | ||
https://youtu.be/THeq4ZiZxIA | https://youtu.be/THeq4ZiZxIA | ||
-Video Solution by TheBeautyOfMath | -Video Solution by TheBeautyOfMath |
Latest revision as of 15:20, 11 October 2024
Contents
[hide]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 Solution by Pi Academy
https://youtu.be/DJ1105lcJJM?si=Z24jb7mCjzjLPdKh
~ Pi Academy
Other 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.