Difference between revisions of "2016 AMC 10B Problems/Problem 19"
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label("$E$",(4,4),N);</asy> | label("$E$",(4,4),N);</asy> | ||
− | <cmath>\triangle AEP \sim \triangle CFP, \frac{AE}{CF}=\frac{EP}{FP}, \frac{ | + | <cmath>\triangle AEP \sim \triangle CFP, \frac{AE}{CF}=\frac{EP}{FP}, \frac{EP}{FP}=\frac{4}{3}, \frac{EP}{EF}=\frac{4}{7}</cmath> |
+ | |||
<cmath>[AEG]=4\cdot 3\cdot \frac{1}{2}=6</cmath> | <cmath>[AEG]=4\cdot 3\cdot \frac{1}{2}=6</cmath> | ||
<cmath>[AFG]=[ABCD]-[ADF]-[CFG]-[ABG]=20-4-\frac{3}{2}-\frac{15}{2}=7</cmath> | <cmath>[AFG]=[ABCD]-[ADF]-[CFG]-[ABG]=20-4-\frac{3}{2}-\frac{15}{2}=7</cmath> | ||
− | <cmath>\frac{EQ}{QF}=\frac{6}{7}</cmath> | + | Because the ratio of the altitudes of <math>\triangle AEG</math> and <math>\triangle AFG</math> are equal to <math>\frac{EQ}{QF}</math>: |
+ | <cmath>\frac{[AEG]}{[AFG]}=\frac{EQ}{QF}=\frac{6}{7}</cmath> | ||
<cmath>\frac{EQ}{EF}=\frac{6}{13}</cmath> | <cmath>\frac{EQ}{EF}=\frac{6}{13}</cmath> | ||
− | |||
− | <cmath>PQ= | + | <cmath>\frac{PQ}{EF}=\frac{4}{7}-\frac{6}{13}</cmath> |
− | <cmath>PQ=\frac{10}{91} | + | <cmath>\frac{PQ}{EF}=\frac{10}{91}</cmath> |
<cmath>\frac{PQ}{EF}=\boxed{\textbf{(D)}~\frac{10}{91}}</cmath> | <cmath>\frac{PQ}{EF}=\boxed{\textbf{(D)}~\frac{10}{91}}</cmath> | ||
Revision as of 09:41, 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 the ratio of the altitudes of and are 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.