Difference between revisions of "2016 AMC 10A Problems/Problem 19"
Line 18: | Line 18: | ||
label("$C$", C, dir(-45)); | label("$C$", C, dir(-45)); | ||
label("$D$", D, dir(-135)); | label("$D$", D, dir(-135)); | ||
− | label("$ | + | label("$Q$", extension(A,(6,1),B,D),dir(-90)); |
− | label("$ | + | label("$P$", extension(A,(6,2),B,D), dir(90)); |
− | label("$ | + | label("$F$", (6,1), dir(0)); |
− | label("$ | + | label("$E$", (6,2), dir(0)); |
</asy> | </asy> | ||
− | As <math>\triangle APD \sim \triangle XPB,</math> <math>\frac{DP}{ | + | As <math>\triangle APD \sim \triangle XPB,</math> <math>\frac{DP}{PB}=\frac{AD}{BE}=3.</math> Similarly, <math>\frac{DQ}{QB}=\frac{3}{2}.</math> From this, it is not hard to find <math>r:s:t=\frac{1}{4}:\frac{2}{5}-\frac{1}{4}:\frac{3}{5}=5:3:12,</math> so <math>r+s+t=\boxed{\textbf{(E) }20.}</math> |
==See Also== | ==See Also== |
Revision as of 13:16, 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
As
Similarly,
From this, it is not hard to find
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 |
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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 |
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