2016 AMC 10A Problems/Problem 19
Contents
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 (Similar Triangles)
Use similar triangles. Our goal is to put the ratio in terms of . Since Therefore, . Similarly, . This means that . Therefore, so
Solution 2 (Mass points and Similar Triangles - Easy)
This problem breaks down into finding and . We can find the first using Mass Points, and the second using similar triangles.
Draw point on such that . Then, by similar triangles . Again, by similar triangles and , . Now we begin Mass Points.
We will consider the triangle with center , so that balances and , and balances and . Assign a mass of to . Then, so . By mass points addition, since balances and . Also, so so . Then, .
To calculate , extend past to point such that lies on . Then is similar to so . Also, is similar to so
Now, we wish to get . Observe that . So, so (since has sum ), . now, we may combine the two and get so .
~Firebolt360
Solution 3(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 4
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 5 (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 6 (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
Video Solution 3 by OmegaLearn
https://youtu.be/4_x1sgcQCp4?t=3406
~ pi_is_3.14
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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