Difference between revisions of "2016 AMC 10A Problems/Problem 19"
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As <math>\triangle APD \sim \triangle XPB,</math> <math>\frac{DP}{PD}=\frac{AD}{BX}=\frac{3}{2}.</math> Similarly, <math>\frac{DQ}{BQ}=3.</math> From this, it is not hard to find <math>r+s+t=\boxed{\textbf{(E) }20.}</math> | As <math>\triangle APD \sim \triangle XPB,</math> <math>\frac{DP}{PD}=\frac{AD}{BX}=\frac{3}{2}.</math> Similarly, <math>\frac{DQ}{BQ}=3.</math> From this, it is not hard to find <math>r+s+t=\boxed{\textbf{(E) }20.}</math> | ||
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=A|num-b=18|num-a=20}} | {{AMC10 box|year=2016|ab=A|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:05, 4 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("$P$", extension(A,(6,1),B,D),dir(-90)); label("$Q$", extension(A,(6,2),B,D), dir(90)); label("$X$", (6,1), dir(0)); label("$Y$", (6,2), dir(0)); [/asy]](http://latex.artofproblemsolving.com/9/7/3/973a3fd654ca6fa9a2f9c5933b4f1548674af2e1.png)
As 
Similarly, 
From this, it is not hard to find 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
