Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "==Problem== A set <math>S</math> is constructed as follows. To begin, <math>S = {{0,10}}</math>. Repeatedly, as long as possible, if <math>x</math> is an integer root of some...")
 
(→‎Solution 1)
 
(15 intermediate revisions by 9 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
−
A set <math>S</math> is constructed as follows. To begin, <math>S = {{0,10}}</math>. Repeatedly, as long as possible, if <math>x</math> is an integer root of some polynomial <math>a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{1}x + a_0</math> for some <math>n\geq{1}</math>, all of whose coefficients <math>a_i</math> are elements of <math>S</math>, then <math>x</math> is put into <math>S</math>.
+
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>\frac{x}{y}</math>?
  
−
<math> \textbf{(A)}\ 4
+
<math> \textbf{(A)}\ \frac{12}{13}
−
\qquad \textbf{(B)}\ 5
+
\qquad \textbf{(B)}\ \frac{35}{37}
−
\qquad\textbf{(C)}\ 7
+
\qquad\textbf{(C)}\ 1
−
\qquad\textbf{(D)}\ 9
+
\qquad\textbf{(D)}\ \frac{37}{35}
−
\qquad\textbf{(E)}\ 11</math>
+
\qquad\textbf{(E)}\ \frac{13}{12}</math>
 +
 
 +
==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 <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>.
 +
 
 +
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, <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} \implies \boxed{\textbf{D}}</math>.
 +
 
 +
==Video Solution (HOW TO THINK CREATIVELY!!!)==
 +
https://youtu.be/BbEgHNtuY8c
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2017|ab=A|num-b=20|num-a=22}}
 +
{{AMC12 box|year=2017|ab=A|num-b=18|num-a=20}}
 +
{{MAA Notice}}

Latest revision as of 13:55, 10 June 2023

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 = (0,0); 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, $\frac{x}{y} = \frac{37}{35} \implies \boxed{\textbf{D}}$.

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/BbEgHNtuY8c

~Education, the Study of Everything

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12A_Problems/Problem_19&oldid=194140"