Problem:
Squares ABCD and EFGH have a common center and AB∥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 possible integer values for the area of IJKL.
Solution:
Without loss of generality, let E and F be the vertices of EFGH that are nearest A and B, respectively, and let I and J lie on AB and BC, respectively. Because â–³EIF is similar to â–³JBI, it follows that
implying that AB,IJ, and EF, in that order, form a decreasing geometric sequence. Hence the three squares have areas AB2=2016,IJ2=2016r, and EF2=2016r2 for some 0<r<1. For all areas to be integers, r must be rational, and when r2 is written as a fraction in lowest terms, its denominator must divide 2016=25â‹…32â‹…7. Thus the only possible denominators for r written in lowest terms are 2,3,4,6, and 12. Note that if x=BI, then
which implies that IJ≥2​1​AB. Similarly, EF≥2​1​IJ, implying that EF≥21​AB. Thus the only possible values of r are 21​,32​,43​,65​,127​, and 1211​. Therefore the difference between the largest and smallest possible values of the area of IJKL is (1211​−21​)(2016)=840​.
