2016 AMC 10B Problems/Problem 19
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 (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 2 (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 3 (Similar Triangles)
[asy]
pair A1=(2,0),A2=(4,4);
pair B1=(0,4),B2=(5,1); pair C1=(5,0),C2=(0,4); pair H = (20/3,0); draw(A1--A2); draw(B1--B2); draw(C1--C2); draw(B1--H); draw((0,0)--H); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); label("",(3.07692307692,2.15384615384),N); label("",(20/7,12/7),W); label("",(0,4), NW); label("",(5,4), NE); label("",(5,0),SE); label("",(0,0),SW); label("",(2,0),S); label("",(5,1),E); label("",(4,4),N); label("",H,E);
dot(A1); dot(A2); dot(B1); dot(B2); dot(C1); dot(C2); dot(H); dot((0,0)); dot((5,4)); [\asy]
Extend to intersect at . Letting , we have that
Then, notice that and . Thus, we see that and Thus, we see that
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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