Difference between revisions of "2016 AMC 10A Problems/Problem 19"
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− | Since <math>\triangle APD \sim \triangle EPB,</math> <math>\frac{DP}{PB}=\frac{AD}{BE}=3.</math> Similarly, <math>\frac{DQ}{QB}=\frac{3}{2}</math>. Call the hypotonuse <math>l</math>. This means that <math>{DQ}=\frac{3l}{5}</math>. Applying similar triangles to <math>{ | + | Since <math>\triangle APD \sim \triangle EPB,</math> <math>\frac{DP}{PB}=\frac{AD}{BE}=3.</math> Similarly, <math>\frac{DQ}{QB}=\frac{3}{2}</math>. Call the hypotonuse <math>l</math>. This means that <math>{DQ}=\frac{3l}{5}</math>. Applying similar triangles to <math>{ADP}</math> and <math>{BEP}</math>, we see that <math>\frac{PD/DB}=\frac{3}{1}</math>. Thus <math>DB=\frac{1}{4}</math> Therefore, <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 23:41, 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
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 |
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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