Difference between revisions of "2016 AMC 10A Problems/Problem 19"
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We set coordinates for the points. Let <math>A=(0,3), B=(6,3), C=(6,0)</math> and <math>D=(0,0)</math>. Then the equation of line <math>AE</math> is <math>y = -\frac{1}{6}x + 3,</math> the equation of line <math>AF</math> is <math>y = -\frac{1}{3}x + 3,</math> and the equation of line <math>BD</math> is <math>y = \frac{1}{2}x</math>. We find that the x-coordinate of point <math>P</math> is <math>\frac 9 2</math> by solving <math> -\frac{1}{6}x + 3=\frac{1}{2}x.</math> Similarly we find that the x-coordinate of point <math>Q</math> is <math>\frac {18} 5</math> by solving <math>-\frac{1}{3}x + 3=\frac{1}{2}x.</math> It follows that <math>BP:PQ:QD=6-\frac 9 2 : \frac 9 2 - \frac {18} 5 : \frac {18} 5= \frac 3 2 : \frac 9 {10} : \frac {18} 5 = 5:3:12.</math> Hence <math>r,s,t=5,3,12</math> and <math>r+s+t=5+3+12=\boxed{\textbf{(E) } 20}.</math> ~ Solution by dolphin7 | We set coordinates for the points. Let <math>A=(0,3), B=(6,3), C=(6,0)</math> and <math>D=(0,0)</math>. Then the equation of line <math>AE</math> is <math>y = -\frac{1}{6}x + 3,</math> the equation of line <math>AF</math> is <math>y = -\frac{1}{3}x + 3,</math> and the equation of line <math>BD</math> is <math>y = \frac{1}{2}x</math>. We find that the x-coordinate of point <math>P</math> is <math>\frac 9 2</math> by solving <math> -\frac{1}{6}x + 3=\frac{1}{2}x.</math> Similarly we find that the x-coordinate of point <math>Q</math> is <math>\frac {18} 5</math> by solving <math>-\frac{1}{3}x + 3=\frac{1}{2}x.</math> It follows that <math>BP:PQ:QD=6-\frac 9 2 : \frac 9 2 - \frac {18} 5 : \frac {18} 5= \frac 3 2 : \frac 9 {10} : \frac {18} 5 = 5:3:12.</math> Hence <math>r,s,t=5,3,12</math> and <math>r+s+t=5+3+12=\boxed{\textbf{(E) } 20}.</math> ~ Solution by dolphin7 | ||
− | == | + | ==Video Solution== |
− | |||
https://www.youtube.com/watch?v=aG9JiBMd0ag | https://www.youtube.com/watch?v=aG9JiBMd0ag | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://youtu.be/us3e2jMjWnw | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==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 03:50, 19 May 2020
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. Our goal is to put the ratio in terms of . 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 are . Furthermore we find that the coordinates of are . Using the Pythagorean Theorem, we get that the length of is , and the length of is The length of . Then The ratio Then and is and , respectively. The problem tells us to find , so ~ minor LaTeX edits by dolphin7
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 .
Solution 4 (Mass Points)
Draw line segment , and call the intersection between and point . In , observe that and . Using mass points, find that . Again utilizing , observe that and . Use mass points to find that . Now, draw a line segment with points ,,, and ordered from left to right. Set the values ,, and . Setting both sides segment equal, we get . Plugging in and solving gives , ,. The question asks for , so we add to and multiply the ratio by to create integers. This creates . This sums up to
Solution 5 (Easy Coord Bash)
We set coordinates for the points. Let and . Then the equation of line is the equation of line is and the equation of line is . We find that the x-coordinate of point is by solving Similarly we find that the x-coordinate of point is by solving It follows that Hence and ~ Solution by dolphin7
Video Solution
https://www.youtube.com/watch?v=aG9JiBMd0ag
Video Solution 2
~IceMatrix
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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