2017 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 21 Hard

A square with side length x is inscribed in a right triangle with sides of length 3, 4, and 5 so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length y is inscribed in another right triangle with sides of length 3, 4, and 5 so that one side of the square lies on the hypotenuse of the triangle. What is \dfrac{x}{y}? (2017 AMC 10A Problem, Question#21)

  • A.

    \dfrac{12}{13}

  • B.

    \dfrac{35}{37}

  • C.

    1

  • D.

    \dfrac{37}{35}

  • E.

    \dfrac{13}{12}

Answer:D

Analyze the first right triangle.

Note that \triangle ADC ant \triangle FBE are similar, so \dfrac{BF}{FE}= \dfrac{AB}{AC}.This can be written as \dfrac{4-x}{x}= \dfrac{4}{3}.

Soving, x= \dfrac{12}{7}

Now we analyze the second triangle.

Similarly, \triangle A'D'C' and \triangle RB'Qare similar, so RB'=\dfrac{3}{4}y and C'S= \dfrac{3}{4}y.

Thus, C'B'=C'S+SR+RB'= \dfrac{4}{3}y+y+ \dfrac{3}{4}y=5. Solving for y we get y= \dfrac{60}{37}. Thus \dfrac{x}{y}= (\rm D)\dfrac{37}{35}.

Related Problems
Problem 19 2017 AMC 10 A
Problem 20 2017 AMC 10 A
Problem 22 2017 AMC 10 A
Problem 23 2017 AMC 10 A
Think Academy Powered by ChatGPT