2015 AIME I Problems/Problem 7
In the diagram below, is a square. Point is the midpoint of . Points and lie on , and and lie on and , respectively, so that is a square. Points and lie on , and and lie on and , respectively, so that is a square. The area of is 99. Find the area of .
Let us find the proportion of the side length of and . Let the side length of and the side length of .
Now, examine . We know , and triangles and are similar to since they are triangles. Thus, we can rewrite in terms of the side length of .
Now examine . We can express this length in terms of since . By using similar triangles as in the first part, we have
Now, it is trivial to see that
We begin by denoting the length , giving us and . Since angles and are complementary, we have that (and similarly the rest of the triangles are triangles). We let the sidelength of be , giving us:
Solving for in terms of yields
We now use the given that , implying that . We also draw the perpendicular from to and label the point of intersection as in the diagram at the top
This gives that and
Since = , we get
So our final answer is .
This is a relatively quick solution but a fakesolve. We see that with a ruler, cm and cm. Thus if corresponds with an area of , then ('s area) would correspond with - aops5234
|2015 AIME I (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|