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

(Created page with "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 squ...")
 
 
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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>?
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==Problem==
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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>\dfrac{x}{y}</math>?
  
 
<math>\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}</math>
 
<math>\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}</math>
  
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==Solution==
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==Solution 1==
  
 
Analyze the first right triangle.
 
Analyze the first right triangle.
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</asy>
 
</asy>
  
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Similary, <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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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} = \boxed{\textbf{(D)}\:\frac{37}{35}}</math>.
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==Solution 2 (Alternate solution in finding x)==
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Set the right-angle vertex of the triangle as <math>(0,0)</math>. Notice that the hypotenuse of the triangle, as depicted in solution one, can be described by <math>y = 3 - \frac{3}{4}x</math>, while <math>AE</math> can be describe by <math>y=x</math>. Hence, we may solve for <math>x</math> by solving <math> 3- \frac{3}{4}x = x</math>, which yields <math>\frac{12}{7}</math>.
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Proceed by finding the value of y via the method described in solution 1, and we will get <math>y = \frac{60}{37}</math>. Thus, <math>\frac{x}{y} = \boxed{\textbf{(D)}\:\frac{37}{35}}</math>.
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==Note==
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In general, if the legs were <math>a</math> and <math>b</math>, we have that
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<cmath>x= \frac{ab}{a+b}, y=\frac{ab\sqrt{a^2+b^2}}{a^2+ab+b^2}.</cmath>
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This can be verified by plugging in <math>a=3</math> and <math>b=4</math>.
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~anduran
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==Video Solution by Pi Academy==
 +
 
 +
https://youtu.be/DJ1105lcJJM?si=Z24jb7mCjzjLPdKh
 +
 
 +
~ Pi Academy
 +
 
 +
==Other video solutions==
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https://youtu.be/THeq4ZiZxIA
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-Video Solution by TheBeautyOfMath
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 +
https://youtu.be/MF2QFOInbYc
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-Video Solution by Richard Rusczyk
 +
 
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==See Also==
 +
{{AMC10 box|year=2017|ab=A|num-b=20|num-a=22}}
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{{AMC12 box|year=2017|ab=A|num-b=18|num-a=20}}
 +
{{MAA Notice}}
 +
 
 +
[[Category:Introductory Geometry Problems]]

Latest revision as of 15:20, 11 October 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 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

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