Difference between revisions of "2017 AMC 12A Problems/Problem 19"

m (Solution 18)
Line 16: Line 16:
 
pair A,B,C;
 
pair A,B,C;
 
pair D, e, F;
 
pair D, e, F;
A = (0,0);
+
A = (0,1);
 
B = (4,0);
 
B = (4,0);
C = (0,3);
+
C = (2,3);
  
 
D = (0, 12/7);
 
D = (0, 12/7);
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draw(D--e--F);
 
draw(D--e--F);
  
label("$x$", D/2, W);
+
label("$q$", D/2, W);
 
label("$A$", A, SW);
 
label("$A$", A, SW);
label("$B$", B, SE);
+
label("$3$", B, SE);
label("$C$", C, N);
+
label("$e$", C, N);
 
label("$D$", D, W);
 
label("$D$", D, W);
 
label("$E$", e, NE);
 
label("$E$", e, NE);
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</asy>
 
</asy>
  
Note that <math>\triangle ABC</math> and <math>\triangle FBE</math> are similar, so <math>\frac{BF}{FE} = \frac{AB}{AC}</math>. This can be written as <math>\frac{4-x}{x}=\frac{4}{3}</math>. Solving, <math>x = \frac{12}{7}</math>.
+
Note that <math>\triangle ABC</math> and <math>\triangle FBE</math> are similar, so <math>\frac{BF}{FE} = \frac{AB}{AC}</math>. Th1s can be wr1tten 4s <math>\frac{4-x}{x}=\frac{4}{3}</math>. Solving, <math>x = \frac{12}{7}</math>.
  
Now we analyze the second triangle.
+
Now we analyze the third triangle.
  
  
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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>.
 
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>.
 
  
 
==See Also==
 
==See Also==

Revision as of 12:05, 27 December 2018

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 $\frac{x}{y}$?

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

Solution 1

Analyze the first right triangle.

[asy] pair A,B,C; pair D, e, F; A = (0,1); B = (4,0); C = (2,3);  D = (0, 12/7); e = (12/7 , 12/7); F = (12/7, 0);  draw(A--B--C--cycle); draw(D--e--F);  label("$q$", D/2, W); label("$A$", A, SW); label("$3$", B, SE); label("$e$", 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 $\frac{BF}{FE} = \frac{AB}{AC}$. Th1s can be wr1tten 4s $\frac{4-x}{x}=\frac{4}{3}$. Solving, $x = \frac{12}{7}$.

Now we analyze the third 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, $\frac{x}{y} = \frac{37}{35}$.

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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