2007 AIME I Problems/Problem 12
In isosceles triangle , is located at the origin and is located at (20,0). Point is in the first quadrant with and angle . If triangle is rotated counterclockwise about point until the image of lies on the positive -axis, the area of the region common to the original and the rotated triangle is in the form , where are integers. Find .
An image is supposed to go here. You can help us out by creating one and. Thanks.
Call the vertices of the new triangle (, the origin, is a vertex of both triangles). and intersect at a single point, . intersect at two points; the one with the higher y-coordinate will be , and the other . The intersection of the two triangles is a quadrilateral . Notice that we can find this area by subtracting .
Since and both have measures , both of their complements are , and . We know that , and since the angles of a triangle add up to , we find that .
So is a . It can be solved by drawing an altitude splitting the angle into and angles – this forms a right triangle and a isosceles right triangle. Since we know that , the base of the triangle is , the height is , and the base of the is . Thus, the total area of .
Now, we need to find , which is a right triangle. We can find its base by subtracting from . is also a triangle, so we find that . .
To solve , note that . Through algebra, we can calculate :
To finish, find . The solution is .
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