Difference between revisions of "2016 AMC 10B Problems/Problem 19"
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+ | ==Solution 6 (Coordinate Bash, not as efficient as Solution 1 but it works)== | ||
+ | We set the points | ||
+ | <math>D(0, 0)</math>, <math>A(0, 4)</math>, <math>E(4, 4)</math>, <math>B(5, 4)</math>, <math>G(5, 1)</math>, <math>C(5, 0)</math>, and <math>F(2, 0)</math>. | ||
+ | The equation of <math>\overline{AC}</math> is <math>y=-\frac{4}{5}x+4</math>, the equation of <math>\overline{AG}</math> is <math>y=-\frac{3}{5}x+4</math>, and the equation of <math>\overline{EF}</math> is <math>y=2x-4</math>. Solving the system of equations for <math>\overline{AC}</math> and <math>\overline{EF}</math> to find point <math>P</math>, <math>y=-\frac{4}{5}x+4=2x-4 \longrightarrow \frac{14}{5}x=8 \longrightarrow x=\frac{20}{7}</math> and <math>y=2x-4=\frac{12}{7}</math>. So the coordinate of point P is <math>P(\frac{20}{7}, \frac{12}{7})</math>. Next find point Q by solving the system of equations for <math>\overline{AG}</math> and <math>\overline{EF}</math> to get <math>Q(\frac{40}{13}, \frac{28}{13})</math>. Using the distance formula, <cmath>PQ=\sqrt{\left(\frac{40}{13}-\frac{20}{7}\right)^{2}+\left(\frac{28}{13}-\frac{12}{7}\right)^{2}}=\sqrt{\left(\frac{20}{91}\right)^{2}+\left(\frac{40}{91}\right)^{2}}</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 | ||
==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}} |
Revision as of 20:09, 18 June 2022
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 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
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.