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2017 AMC 10A Problem 21

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2017 AMC 10A Problem 21AMC 10 Step by Step
Name:                             Date: September 13, 2026
Similar & Congruent Triangles · difficulty 4 of 5 · about 7 min · position P21-25

A square with side length xx is inscribed in a right triangle with sides of length 33 , 44 , and 55 so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length yy is inscribed in another right triangle with sides of length 33 , 44 , and 55 so that one side of the square lies on the hypotenuse of the triangle. What is xy\tfrac{x}{y} ?

  1. A)

    1213\dfrac{12}{13}

  2. B)

    3537\dfrac{35}{37}

  3. C)

    11

  4. D)

    3735\dfrac{37}{35}

  5. E)

    1312\dfrac{13}{12}

Answer

D

Key insight

Each square cuts off triangles similar to the 3-4-5; the corner square gives x = 12/7, and the hypotenuse splits as 4y/3 + y + 3y/4 = 5.

Solution

The square at the right angle. Put the right angle at the origin with legs along the axes, the leg of length 44 on the xx-axis and 33 on the yy-axis. The hypotenuse is the line X4+Y3=1\frac{X}{4} + \frac{Y}{3} = 1, and the square's far corner (x,x)(x, x) lies on it:
x4+x3=1⟹7x12=1⟹x=127.\frac{x}{4} + \frac{x}{3} = 1 \quad\Longrightarrow\quad \frac{7x}{12} = 1 \quad\Longrightarrow\quad x = \frac{12}{7}.

The square on the hypotenuse. The square's two upper corners touch the legs, cutting off two small right triangles at the ends of the hypotenuse plus a triangle on top. The small triangle at the vertex whose angle is opposite the side 33 has one leg yy (a side of the square, perpendicular to the hypotenuse) and the other leg along the hypotenuse; it is similar to the 33-44-55 triangle with legs in ratio 3:43 : 4, so its hypotenuse-leg is 43y\tfrac{4}{3}y. At the other vertex the roles swap and the hypotenuse-leg is 34y\tfrac{3}{4}y. These two pieces plus the square's side make up the hypotenuse:
43y+y+34y=5⟹3712y=5⟹y=6037.\frac{4}{3}y + y + \frac{3}{4}y = 5 \quad\Longrightarrow\quad \frac{37}{12}y = 5 \quad\Longrightarrow\quad y = \frac{60}{37}.

Ratio.
xy=12/760/37=12⋅377⋅60=3735.\frac{x}{y} = \frac{12/7}{60/37} = \frac{12 \cdot 37}{7 \cdot 60} = \frac{37}{35}.

The answer is (D) 3735\boxed{\textbf{(D)}\ \frac{37}{35}}.

Why this works

Every right triangle cut off from a right triangle by a line parallel to one side, or by a perpendicular to the hypotenuse, is similar to the original. A square inscribed in a right triangle always creates such pieces, so both configurations reduce to writing one segment (a leg or the hypotenuse) as a sum of lengths proportional to the square's side.

Alternative approach

For the hypotenuse square, use the top triangle instead: it is similar to the whole triangle with base yy (the square's top side) and height h−yh - y, where h=125h = \tfrac{12}{5} is the altitude to the hypotenuse. So yh−y=5h\frac{y}{h - y} = \frac{5}{h}, giving y=5h5+h=1237/5=6037y = \frac{5h}{5 + h} = \frac{12}{37/5} = \frac{60}{37}.

The trap

Mixing up which small triangle has which leg ratio (3:4 versus 4:3) when splitting the hypotenuse, or inverting the final ratio to get 35/37.

Common mistakes

  • Mixing up which small triangle has which leg ratio (3:4 versus 4:3) when splitting the hypotenuse, or inverting the final ratio to get 35/37.
  • Assuming the two squares have the same side and answering 11.

Techniques

Set up the equation/formula and compute; no special trick needed · Cut the figure into known shapes (triangles, rectangles, sectors)