2016 AIME II Problems/Problem 7

Squares $ABCD$ and $EFGH$ have a common center at $\overline{AB} || \overline{EF}$ . 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 teh 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 $2016 \cdot \dfrac{11}{12} = 1848$ and the minimum area is $1008$ , so the desired answer is $1848-1008=\boxed{840}$ .

Solution by Shaddoll

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2016 AIME II Problems/Problem 7

Solution