2016 AMC 10B Problems/Problem 19
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 :
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