2016 AMC 10A Problems/Problem 19: Difference between revisions
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Revision as of 11:24, 5 February 2016
Problem
In rectangle 
and 
. Point 
between 
and 
, and point 
between and are such that 
. Segments 
and 
intersect 
at 
and 
, respectively. The ratio 
can be written as 
where the greatest common factor of 
and 
is 
What is 
?
Solution
![[asy] size(6cm); pair D=(0,0), C=(6,0), B=(6,3), A=(0,3); draw(A--B--C--D--cycle); draw(B--D); draw(A--(6,2)); draw(A--(6,1)); label("$A$", A, dir(135)); label("$B$", B, dir(45)); label("$C$", C, dir(-45)); label("$D$", D, dir(-135)); label("$Q$", extension(A,(6,1),B,D),dir(-90)); label("$P$", extension(A,(6,2),B,D), dir(90)); label("$F$", (6,1), dir(0)); label("$E$", (6,2), dir(0)); [/asy]](http://latex.artofproblemsolving.com/1/5/8/15893c761d311ed13254fdea73b5a3079e88884f.png)
Since 
Similarly, 
. Call the hypotonuse 
. This means that 
. Applying similar triangles to 
and 
, we see that 
. Thus 
. Therefore, 
so 
See Also
| 2016 AMC 10A (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 10 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America. 
