Difference between revisions of "2016 AMC 10B Problems/Problem 19"
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First, we will define point <math>D</math> as the origin. Then, we will find the equations of the following three lines: <math>AG</math>, <math>AC</math>, and <math>EF</math>. The slopes of these lines are <math>-\frac{3}{5}</math>, <math>-\frac{4}{5}</math>, and <math>2</math>, respectively. Next, we will find the equations of <math>AG</math>, <math>AC</math>, and <math>EF</math>. They are as follows: | First, we will define point <math>D</math> as the origin. Then, we will find the equations of the following three lines: <math>AG</math>, <math>AC</math>, and <math>EF</math>. The slopes of these lines are <math>-\frac{3}{5}</math>, <math>-\frac{4}{5}</math>, and <math>2</math>, respectively. Next, we will find the equations of <math>AG</math>, <math>AC</math>, and <math>EF</math>. They are as follows: | ||
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<cmath>=\sqrt{\frac{400}{8281}+\frac{1600}{8281}}=\sqrt{\frac{2000}{8281}}=\frac{20\sqrt{5}}{91}</cmath> Also using the distance formula, <cmath>EF=\sqrt{\left(4-2\right)^{2}+\left(4-0\right)^{2}}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}</cmath> Finally, <cmath>\frac{PQ}{EF}=\frac{\frac{20\sqrt{5}}{91}}{2\sqrt{5}}=\frac{10}{91} \Longrightarrow \boxed{\textbf{(D)}~\frac{10}{91}}</cmath> | <cmath>=\sqrt{\frac{400}{8281}+\frac{1600}{8281}}=\sqrt{\frac{2000}{8281}}=\frac{20\sqrt{5}}{91}</cmath> Also using the distance formula, <cmath>EF=\sqrt{\left(4-2\right)^{2}+\left(4-0\right)^{2}}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}</cmath> Finally, <cmath>\frac{PQ}{EF}=\frac{\frac{20\sqrt{5}}{91}}{2\sqrt{5}}=\frac{10}{91} \Longrightarrow \boxed{\textbf{(D)}~\frac{10}{91}}</cmath> | ||
~JH. L | ~JH. L | ||
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+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/GrCtzL0S-Uo?t=911 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=18|num-a=20}} | {{AMC10 box|year=2016|ab=B|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:16, 1 November 2024
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 (Coordbash)
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 ratio of area of to .
Because and share the same base , the ratio is equal to the ratio of the altitude of to to that of to , which is equal to :
Solution 5 (Area)
I will calculate using the ratio of area of to that of .
Because and share the same base , the ratio is equal to the ratio of altitude of to to that of to , which is equal to :
Solution 6 (Coordinate Bash, not as efficient as Solution 1 but it works)
We set the points , , , , , , and . The equation of is , the equation of is , and the equation of is . Solving the system of equations for and to find point , and . So the coordinate of point P is . Next find point Q by solving the system of equations for and to get . Using the distance formula, Also using the distance formula, Finally, ~JH. L
Video Solution by OmegaLearn
https://youtu.be/GrCtzL0S-Uo?t=911
~ pi_is_3.14
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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