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>. This means that <math>{DQ}=\frac{3\cdot BD}{5}</math>. As <math>\triangle ADP</math> and <math>\triangle BEP</math> are similar, we see that <math>\frac{PD}{PB}=\frac{3}{1}</math>. Thus <math>PB=\frac{BD}{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> | + | Use similar triangles. 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>. This means that <math>{DQ}=\frac{3\cdot BD}{5}</math>. As <math>\triangle ADP</math> and <math>\triangle BEP</math> are similar, we see that <math>\frac{PD}{PB}=\frac{3}{1}</math>. Thus <math>PB=\frac{BD}{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> |
==Solution 2== | ==Solution 2== |
Revision as of 23:12, 13 February 2017
Contents
[hide]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 1
Use similar triangles. Since Similarly, . This means that . As and are similar, we see that . Thus . Therefore, so
Solution 2
Coordinate Bash: We can set coordinates for the points. and . The line 's equation is , line 's equation is , and line 's equation is . Adding the equations of lines and , we find that the coordinates of is . Furthermore we find that the coordinates is . Using the Pythagorean Theorem, the length of is , and the length of = The length of . Then The ratio Then and is and , respectively. The problem tells us to find , so
An alternate solution is to perform the same operations, but only solve for the x-coordinates. By similar triangles, the ratios will be the same.
Solution 3
Extend to meet at point . Since and , by similar triangles and . It follows that . Now, using similar triangles and , . WLOG let . Solving for gives and . So our desired ratio is and .
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 |
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