2016 AIME II Problems/Problem 7
Squares and have a common center at . The area of is 2016, and the area of is a smaller positive integer. Square is constructed so that each of its vertices lies on a side of and each vertex of lies on a side of . Find the difference between the largest and smallest positive integer values for the area of .
Letting and , we have by CS inequality. Also, since , the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and adjacent vertices of a smaller square are similar. Therefore, the areas form a geometric progression, so since , we have the maximum area is and the minimum area is , so the desired answer is .
Solution by Shaddoll