Difference between revisions of "2017 AMC 12A Problems/Problem 19"
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\qquad\textbf{(E)}\ \frac{13}{12}</math> | \qquad\textbf{(E)}\ \frac{13}{12}</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Analyze the first right triangle. | Analyze the first right 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>. | ||
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| + | ==Solution 2 (Pure Guessing... and Dumb Luck)== | ||
| + | |||
| + | As stated in Solution 1, the square in the first triangle <math>\triangle ABC</math> has a side length of <math>12/7</math>. Then we look at the answers. The only answer with a factor of <math>7</math> in the denominator is <math>B</math>. | ||
==See Also== | ==See Also== | ||
Revision as of 19:37, 9 February 2017
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, 
.
Solution 2 (Pure Guessing... and Dumb Luck)
As stated in Solution 1, the square in the first triangle has a side length of 
. Then we look at the answers. The only answer with a factor of 
in the denominator is 
.
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. 
