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> | + | 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>{ADQ}</math> and <math>{BFQ}</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 23:40, 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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