# 2016 AIME II Problems/Problem 7

## Problem

Squares $ABCD$ and $EFGH$ have a common center and $\overline{AB} || \overline{EF}$. The area of $ABCD$ is 2016, and the area of $EFGH$ is a smaller positive integer. Square $IJKL$ is constructed so that each of its vertices lies on a side of $ABCD$ and each vertex of $EFGH$ lies on a side of $IJKL$. Find the difference between the largest and smallest positive integer values for the area of $IJKL$.

## Solution

Letting $AI=a$ and $IB=b$, we have $IJ^{2}=a^{2}+b^{2} \geq 1008$ by CS inequality. Also, since $EFGH||ABCD$, the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and $2$ adjacent vertices of a smaller square are similar. Therefore, the areas form a geometric progression, so since $2016=12^{2} \cdot 14$, we have the maximum area is $2016 \cdot \dfrac{11}{12} = 1848$ (the areas of the squares from largest to smallest are $12^{2} \cdot 14, 11 \cdot 12 \cdot 14, 11^{2} \cdot 14$ forming a geometric progression).

The minimum area is $1008$ (every square is half the area of the square whose sides its vertices touch), so the desired answer is $1848-1008=\boxed{840}$.

Solution by Shaddoll (edited by ppiittaattoo)