Problem:
The set G is defined by the points (x,y) with integer coordinates, 3≤∣x∣≤7, and 3≤∣y∣≤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 Gi​ be the subset of G contained in the i th quadrant, 1≤i≤ 4. For a fixed i, the maximum distance among points in Gi​ is 42​<6, also the distance from a point in Gi​ to a point in Gjâ€‹î€ =Gi​ is at least 6 . Thus the required squares are exactly the squares in G with exactly one vertex in each of the Gi​. Let S=p1​p2​p3​p4​ be a square with vertices pi​∈Gi​. Let p1′​=p1​+(−5,−5),\ p2′​=p2​+(5,−5),p3′​=p3​+(5,5), and p4′​=p4​+(−5,5). Observe that p1′​, p2′​,p3′​, and p4′​ are all lattice points inside the square region G′ defined by the points ( x,y ) with ∣x∣,∣y∣≤2; moreover, by symmetry, S′=p1′​p2′​p3′​p4′​ is either a square or p1′​=p2′​=p3′​=p4′​. Reciprocally, if S′=p1′​p2′​p3′​p4′​ is a square in G′, then the points p1​=p1′​+(5,5),p2​=p2′​+(−5,5),p3​=p3′​+(−5,−5), and p4​=p4′​+(5,−5) satisfy that pi​∈Gi​ and S=p1​p2​p3​p4​ is a square. The same conclusion holds if p1′​=p2′​=p3′​=p4′​. Therefore the required count consists of the number of points in G′ plus four times the number of squares with vertices in G′.
There are 52 points in G′ and the following number of squares with vertices in G′:42 of side 1,32 of side 2,32 of side 2​ (each inscribed in a unique square of side 2 ), 22 of side 3,2⋅22 of side 5​ (exactly two inscribed in every square of side 3), 12 of side 4,12 of side 22​, and 2⋅12 of side 10​ (exactly two inscribed in the square of side 4 ). Thus the answer is
