Difference between revisions of "2016 AMC 10A Problems/Problem 19"
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==Solution 5 (Cheap Solution)== | ==Solution 5 (Cheap Solution)== | ||
− | Use your ruler ( | + | Use your ruler (you should probably recommended you bring ruler and protractor to AMC10 tests) and accurately draw the diagram as one in solution 1, then measure the length of the segments, you should get a ratio of <math>r:s:t</math> being <math>\frac{5}{3}:1:\frac{12}{3}</math>, multiplying each side by <math>3</math> the result is <math>r+s+t = 5+3+12 = \boxed{\textbf{(E) }20}</math> |
==See Also== | ==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 20:14, 17 August 2018
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
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 .
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 (Cheap Solution)
Use your ruler (you should probably recommended you bring ruler and protractor to AMC10 tests) and accurately draw the diagram as one in solution 1, then measure the length of the segments, you should get a ratio of being , multiplying each side by the result is
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.