2016 AIME II Problems/Problem 7
Problem
Squares and have a common center and . 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 .
Solution
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 (the areas of the squares from largest to smallest are forming a geometric progression).
The minimum area is (every square is half the area of the square whose sides its vertices touch), so the desired answer is .
Solution by Shaddoll (edited by ppiittaattoo)
See also
2016 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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