Difference between revisions of "2017 AMC 12A Problems/Problem 19"
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pair A,B,C; | pair A,B,C; | ||
pair D, e, F; | pair D, e, F; | ||
| − | A = (0, | + | A = (0,0); |
B = (4,0); | B = (4,0); | ||
| − | C = ( | + | C = (0,3); |
D = (0, 12/7); | D = (0, 12/7); | ||
| Line 27: | Line 27: | ||
draw(D--e--F); | draw(D--e--F); | ||
| − | label("$ | + | label("$x$", D/2, W); |
label("$A$", A, SW); | label("$A$", A, SW); | ||
| − | label("$ | + | label("$B$", B, SE); |
| − | label("$ | + | label("$C$", C, N); |
label("$D$", D, W); | label("$D$", D, W); | ||
label("$E$", e, NE); | label("$E$", e, NE); | ||
| Line 36: | Line 36: | ||
</asy> | </asy> | ||
| − | Note that <math>\triangle ABC</math> and <math>\triangle FBE</math> are similar, so <math>\frac{BF}{FE} = \frac{AB}{AC}</math>. | + | 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 | + | Now we analyze the second triangle. |
| Line 66: | Line 66: | ||
</asy> | </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}</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} \implies \boxed{\textbf{D}}</math>. |
| + | |||
| + | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | ||
| + | https://youtu.be/BbEgHNtuY8c | ||
| + | |||
| + | ~Education, the Study of Everything | ||
==See Also== | ==See Also== | ||
Latest revision as of 13:55, 10 June 2023
Problem
A square with side length 
is inscribed in a right triangle with sides of length 
, 
, and 
so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length 
is inscribed in another right triangle with sides of length , , and so that one side of the square lies on the hypotenuse of the triangle. What is 
?
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]](http://latex.artofproblemsolving.com/a/2/d/a2d8d4056c8f797b4635c3aa467fd8f36bb9281c.png)
Note that 
and 
are similar, so 
. This can be written as 
. Solving, 
.
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]](http://latex.artofproblemsolving.com/a/b/c/abc0a7730ccae9580fa62d17f8130c558ceec4bc.png)
Similarly, 
and 
are similar, so 
, and 
. Thus, 
. Solving for , we get 
. Thus, 
.
Video Solution (HOW TO THINK CREATIVELY!!!)
~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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
