2016 AMC 10A Problems/Problem 19
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
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
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 | |
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