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
.
Solution
Letting and
, we have
by AM-GM 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
See also
2016 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |