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
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
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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