Problem:
The set G is defined by the points (x,y) with integer coordinates, 3β€β£xβ£β€7, and 3β€β£yβ£β€7. How many squares of side at least 6 have their four vertices in G ?
Answer Choices:
A. 125
B. 150
C. 175
D. 200
E. 225 Solution:
Let Giβ be the subset of G contained in the i th quadrant, 1β€iβ€ 4. For a fixed i, the maximum distance among points in Giβ is 42β<6, also the distance from a point in Giβ to a point in Gjβξ =Giβ is at least 6 . Thus the required squares are exactly the squares in G with exactly one vertex in each of the Giβ. Let S=p1βp2βp3βp4β be a square with vertices piββGiβ. Let p1β²β=p1β+(β5,β5),\ p2β²β=p2β+(5,β5),p3β²β=p3β+(5,5), and p4β²β=p4β+(β5,5). Observe that p1β²β, p2β²β,p3β²β, and p4β²β are all lattice points inside the square region Gβ² defined by the points ( x,y ) with β£xβ£,β£yβ£β€2; moreover, by symmetry, Sβ²=p1β²βp2β²βp3β²βp4β²β is either a square or p1β²β=p2β²β=p3β²β=p4β²β. Reciprocally, if Sβ²=p1β²βp2β²βp3β²βp4β²β is a square in Gβ², then the points p1β=p1β²β+(5,5),p2β=p2β²β+(β5,5),p3β=p3β²β+(β5,β5), and p4β=p4β²β+(5,β5) satisfy that piββGiβ and S=p1βp2βp3βp4β is a square. The same conclusion holds if p1β²β=p2β²β=p3β²β=p4β²β. Therefore the required count consists of the number of points in Gβ² plus four times the number of squares with vertices in Gβ².
There are 52 points in Gβ² and the following number of squares with vertices in Gβ²:42 of side 1,32 of side 2,32 of side 2β (each inscribed in a unique square of side 2 ), 22 of side 3,2β 22 of side 5β (exactly two inscribed in every square of side 3), 12 of side 4,12 of side 22β, and 2β 12 of side 10β (exactly two inscribed in the square of side 4 ). Thus the answer is
