2014 AMC 10A Problems/Problem 16
Contents
Problem
In rectangle , , , and points , , and are midpoints of , , and , respectively. Point is the midpoint of . What is the area of the shaded region?
Solution 1
Denote . Then . Let the intersection of and be , and the intersection of and be . Then we want to find the coordinates of so we can find . From our points, the slope of is , and its -intercept is just . Thus the equation for is . We can also quickly find that the equation of is . Setting the equations equal, we have . Because of symmetry, we can see that the distance from to is also , so . Now the area of the kite is simply the product of the two diagonals over . Since the length , our answer is .
Solution 2
Let the area of the shaded region be . Let the other two vertices of the kite be and with closer to than . Note that . The area of is and the area of is . We will solve for the areas of and in terms of x by noting that the area of each triangle is the length of the perpendicular from to and to respectively. Because the area of = based on the area of a kite formula, for diagonals of length and , . So each perpendicular is length . So taking our numbers and plugging them into gives us Solving this equation for gives us
Solution 3
From the diagram in Solution 1, let be the height of and be the height of . It is clear that their sum is as they are parallel to . Let be the ratio of the sides of the similar triangles and , which are similar because is parallel to and the triangles share angle . Then , as 2 is the height of . Since and are similar for the same reasons as and , the height of will be equal to the base, like in , making . However, is also the base of , so where so . Subbing into gives a system of linear equations, and . Solving yields and , and since the area of the kite is simply the product of the two diagonals over and , our answer is .
Solution 4
Let the unmarked vertices of the shaded area be labeled and , with being closer to than . Noting that kite can be split into triangles and .
Lemma: The distance from line segment to is half the distance from to
Proof: Drop perpendiculars of triangles and to line , and let the point of intersection be . Note that and are similar to and , respectively. Now, the ratio of to is , which shows that the ratio of to is , because of similar triangles as described above. Similarly, the ratio of to is . Since these two triangles contain the same base, , the ratio of .
Because kite is orthodiagonal, we multiply
~Lemma proof by sakshamsethi
Solution 5 (Similarity)
The area of the shaded area is the area of minus the two triangles on the side. Extend so that it hits point . Call the intersection of and point . Drop altitudes from down to and ; call the intersection points and respectively. Thus the two triangles on the side have area . Since there are two, their total area is . The area of is . The shaded region is which is .
~JH. L :)
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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.