Difference between revisions of "2017 AMC 12A Problems/Problem 19"
Eatingstuff (talk | contribs) 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, | + | A = (0,1); |
B = (4,0); | B = (4,0); | ||
− | C = ( | + | C = (2,3); |
D = (0, 12/7); | D = (0, 12/7); | ||
Line 27: | Line 27: | ||
draw(D--e--F); | draw(D--e--F); | ||
− | label("$ | + | label("$q$", D/2, W); |
label("$A$", A, SW); | label("$A$", A, SW); | ||
− | label("$ | + | label("$3$", B, SE); |
− | label("$ | + | label("$e$", 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>. 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 | + | Now we analyze the third triangle. |
Line 67: | Line 67: | ||
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 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.
Note that and are similar, so . Th1s can be wr1tten 4s . Solving, .
Now we analyze the third triangle.
Similarly, and are similar, so , and . Thus, . Solving for , we get . Thus, .
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.