2014 AMC 10A Problems/Problem 16
In rectangle , , , and points , , and are midpoints of , , and , respectively. Point is the midpoint of . What is the area of the shaded region?
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 .
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
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 .
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 :)
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