Difference between revisions of "2017 AMC 10A Problems/Problem 21"

Revision as of 19:44, 24 July 2024

Problem

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 $\dfrac{x}{y}$ ?

$\textbf{(A) } \dfrac{12}{13} \qquad \textbf{(B) } \dfrac{35}{37} \qquad \textbf{(C) } 1 \qquad \textbf{(D) } \dfrac{37}{35} \qquad \textbf{(E) } \dfrac{13}{12}$

Solution 1

Analyze the first right triangle.

$[asy] pair A,B,C; pair D, e, F; A = (0,0); B = (4,0); C = (0,3); D = (0, 12/7); e = (12/7 , 12/7); F = (12/7, 0); draw(A--B--C--cycle); draw(D--e--F); label("$x$", D/2, W); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$D$", D, W); label("$E$", e, NE); label("$F$", F, S); [/asy]$

Note that $\triangle ABC$ and $\triangle FBE$ are similar, so $\frac{BF}{FE} = \frac{AB}{AC}$ . This can be written as $\frac{4-x}{x}=\frac{4}{3}$ . Solving, $x = \frac{12}{7}$ .

Now we analyze the second triangle.

$[asy] pair A,B,C; pair q, R, S, T; A = (0,0); B = (4,0); C = (0,3); q = (1.297, 0); R = (2.27 , 1.297); S = (0.973, 2.27); T = (0, 0.973); draw(A--B--C--cycle); draw(q--R--S--T--cycle); label("$y$", (q+R)/2, NW); label("$A'$", A, SW); label("$B'$", B, SE); label("$C'$", C, N); label("$Q$", (q-(0,0.3))); label("$R$", R, NE); label("$S$", S, NE); label("$T$", T, W); [/asy]$

Similarly, $\triangle A'B'C'$ and $\triangle RB'Q$ are similar, so $RB' = \frac{4}{3}y$ , and $C'S = \frac{3}{4}y$ . Thus, $C'B' = C'S + SR + RB' = \frac{4}{3}y + y + \frac{3}{4}y = 5$ . Solving for $y$ , we get $y = \frac{60}{37}$ . Thus, $\frac{x}{y} = \boxed{\textbf{(D)}\:\frac{37}{35}}$ .

Solution 2 (Alternate solution in finding x)

Set the right-angle vertex of the triangle as $(0,0)$ . Notice that the hypotenuse of the triangle, as depicted in solution one, can be described by $y = 3 - \frac{3}{4}x$ , while $AE$ can be describe by $y=x$ . Hence, we may solve for $x$ by solving $3- \frac{3}{4}x = x$ , which yields $\frac{12}{7}$ .

Proceed by finding the value of y via the method described in solution 1, and we will get $y = \frac{60}{37}$ . Thus, $\frac{x}{y} = \boxed{\textbf{(D)}\:\frac{37}{35}}$ .

Note

In general, if the legs were $a$ and $b$ , we have that $x= \frac{ab}{a+b}, y=\frac{ab\sqrt{a^2+b^2}}{a^2+ab+b^2}.$ This can be verified by plugging in $a=3$ and $b=4$ . ~anduran

Video Solutions

https://youtu.be/THeq4ZiZxIA -Video Solution by TheBeautyOfMath

https://youtu.be/MF2QFOInbYc -Video Solution by Richard Rusczyk

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.

@@ Line 70: / Line 70: @@
 ==Note==
 In general, if the legs were <math>a</math> and <math>b</math>, we have that
-<cmath>x= \frac{ab}{a+b}, y=\frac{ab\sqrt{a^2+b^2}}{a^2+ab+b^2},</cmath>
+<cmath>x= \frac{ab}{a+b}, y=\frac{ab\sqrt{a^2+b^2}}{a^2+ab+b^2}.</cmath>
-Which can be verified by plugging in <math>a=3</math> and <math>b=4</math>.
+This can be verified by plugging in <math>a=3</math> and <math>b=4</math>.
 ~anduran

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