2013 AMC 12A Problems/Problem 23
is a square of side length . Point is on such that . The square region bounded by is rotated counterclockwise with center , sweeping out a region whose area is , where , , and are positive integers and . What is ?
We first note that diagonal is of length . It must be that divides the diagonal into two segments in the ratio to . It is not difficult to visualize that when the square is rotated, the initial and final squares overlap in a rectangular region of dimensions by . The area of the overall region (of the initial and final squares) is therefore twice the area of the original square minus the overlap, or .
The area also includes circular segments. Two are quarter-circles centered at of radii (the segment bounded by and ) and (that bounded by and ). Assuming is the bottom-left vertex and is the bottom-right one, it is clear that the third segment is formed as swings out to the right of the original square [recall that the square is rotated counterclockwise], while the fourth is formed when overshoots the final square's left edge. To find these areas, consider the perpendicular from to . Call the point of intersection . From the previous paragraph, it is clear that and . This means , and swings back inside edge at a point unit above (since it left the edge unit below). The triangle of the circular sector is therefore an equilateral triangle of side length , and so the angle of the segment is . Imagining the process in reverse, it is clear that the situation is the same with point .
The area of the segments can be found by subtracting the area of the triangle from that of the sector; it follows that the two quarter-segments have areas and . The other two segments both have area .
The total area is therefore
Since , , and , the answer is .
Video Solution by Richard Rusczyk
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