Difference between revisions of "2017 AMC 10A Problems/Problem 21"
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A square with side length <math>x</math> is inscribed in a right triangle with sides of length <math>3</math>, <math>4</math>, and <math>5</math> so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length <math>y</math> is inscribed in another right triangle with sides of length <math>3</math>, <math>4</math>, and <math>5</math> so that one side of the square lies on the hypotenuse of the triangle. What is <math>\tfrac{x}{y}</math>? | A square with side length <math>x</math> is inscribed in a right triangle with sides of length <math>3</math>, <math>4</math>, and <math>5</math> so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length <math>y</math> is inscribed in another right triangle with sides of length <math>3</math>, <math>4</math>, and <math>5</math> so that one side of the square lies on the hypotenuse of the triangle. What is <math>\tfrac{x}{y}</math>? | ||
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− | + | Similarly, <math>\triangle A'B'C'</math> and <math>\triangle RB'Q</math> are similar, so <math>RB' = \frac{4}{3}y</math>, and <math>C'S = \frac{3}{4}y</math>. Thus, <math>C'B' = C'S + SR + RB' = \frac{4}{3}y + y + \frac{3}{4}y = 5</math>. Solving for <math>y</math>, we get <math>y = \frac{60}{37}</math>. Thus, <math>\frac{x}{y} = \frac{37}{35}</math>. | |
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+ | ==See Also== | ||
+ | {{AMC10 box|year=2017|ab=A|num-b=20|num-a=22}} | ||
+ | {{MAA Notice}} |
Revision as of 17:03, 8 February 2017
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
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, .
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 |
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