# 2016 AIME II Problems/Problem 7

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## 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 AM-GM 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 $$1848-1008=\boxed{840}$$ $[asy] pair A,B,C,D,E,F,G,H,I,J,K,L; A=(0,0); B=(2016,0); C=(2016,2016); D=(0,2016); I=(1008,0); J=(2016,1008); K=(1008,2016); L=(0,1008); E=(504,504); F=(1512,504); G=(1512,1512); H=(504,1512); draw(A--B--C--D--A); draw(I--J--K--L--I); draw(E--F--G--H--E); label("A",A,SW); label("B",B,SE); label("C",C,NE); label("D",D,NW); label("E",E,SW); label("F",F,SE); label("G",G,NE); label("H",H,NW); label("I",I,S); label("J",J,NE); label("K",K,N); label("L",L,NW); [/asy]$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 