Difference between revisions of "2016 AMC 10B Problems/Problem 19"
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<cmath>[AEG]=\frac{1}{2} \cdot 4\cdot 3=6</cmath> | <cmath>[AEG]=\frac{1}{2} \cdot 4\cdot 3=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> | ||
− | Because <math>\triangle AEG</math> and <math>\triangle AFG</math> share the same base <math>AG</math>, the ratio <math>\frac{[AEG]}{[AFG]}</math> is equal to the ratio of the | + | Because <math>\triangle AEG</math> and <math>\triangle AFG</math> share the same base <math>AG</math>, the ratio <math>\frac{[AEG]}{[AFG]}</math> is equal to the ratio of the altitude of <math>\triangle AEG</math> to <math>AG</math> to that of <math>\triangle AFG</math> to <math>AG</math>, which is equal to <math>\frac{EQ}{QF}</math>: |
<cmath>\frac{[AEG]}{[AFG]}=\frac{EQ}{QF}=\frac{6}{7}</cmath> | <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> |
Revision as of 11:33, 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 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 :
~isabelchen
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 altitudes of to to that of 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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