Difference between revisions of "2016 AMC 10B Problems/Problem 19"
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<cmath>\frac{PQ}{QE}=\frac{[APG]}{[AEG]}=\frac{\frac{10}{7}}{6}=\frac{10}{42}=\frac{5}{21}</cmath> | <cmath>\frac{PQ}{QE}=\frac{[APG]}{[AEG]}=\frac{\frac{10}{7}}{6}=\frac{10}{42}=\frac{5}{21}</cmath> | ||
<cmath>\frac{PQ}{PE}=\frac{5}{21+5}=\frac{5}{26}</cmath> | <cmath>\frac{PQ}{PE}=\frac{5}{21+5}=\frac{5}{26}</cmath> | ||
− | <cmath>\frac{ | + | <cmath>\frac{PE}{PF}=\frac{AE}{CF}=\frac{4}{3}</cmath> |
− | |||
<cmath>\frac{PE}{EF}=\frac{4}{7}</cmath> | <cmath>\frac{PE}{EF}=\frac{4}{7}</cmath> | ||
− | <cmath>\frac{ | + | <cmath>\frac{PQ}{PE} \cdot \frac{PE}{EF} = \frac{5}{26} \cdot \frac{4}{7} = \frac{10}{91}</cmath> |
− | <cmath>\frac{PQ}{EF | + | <cmath>\frac{PQ}{EF}=\boxed{\textbf{(D)}~\frac{10}{91}}</cmath> |
~isabelchen | ~isabelchen |
Revision as of 11:22, 8 October 2021
Contents
Problem
Rectangle has and . Point lies on so that , point lies on so that , and point lies on so that . Segments and intersect at and , respectively. What is the value of ?
Solution 1 (Coordinate Geometry)
First, we will define point as the origin. Then, we will find the equations of the following three lines: , , and . The slopes of these lines are , , and , respectively. Next, we will find the equations of , , and . They are as follows: After drawing in altitudes to from , , and , we see that because of similar triangles, and so we only need to find the x-coordinates of and . Finding the intersections of and , and and gives the x-coordinates of and to be and . This means that . Now we can find
Solution 2 (Similar Triangles)
Extend to intersect at . Letting , we have that
Then, notice that and . Thus, we see that and Thus, we see that
Solution 3 (Answer Choices)
Since the opposite sides of a rectangle are parallel and due to vertical angles, . Furthermore, the ratio between the side lengths of the two triangles is . Labeling and , we see that turns out to be equal to . Since the denominator of must now be a multiple of 7, the only possible solution in the answer choices is .
Solution 4 (Area)
I will calculate using similar triangle, and using area of and .
Because and share the same base , the ratio is equal to the ratio of the altitudes of and to , which is equal to :
~isabelchen
Solution 5 (Area)
I will calculate using the ratio of the area of and .
Because and share the same base , the ratio is equal to the ratio of the altitudes of and to , which is equal to :
~isabelchen
See Also
2016 AMC 10B (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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