2009 AMC 12B Problems/Problem 25

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Problem

The set $G$ is defined by the points $(x,y)$ with integer coordinates, $3\le|x|\le7$ , $3\le|y|\le7$ . How many squares of side at least $6$ have their four vertices in $G$ ?

$\textbf{(A)}\ 125\qquad \textbf{(B)}\ 150\qquad \textbf{(C)}\ 175\qquad \textbf{(D)}\ 200\qquad \textbf{(E)}\ 225$

Solution

We need to find a reasonably easy way to count the squares.

First, obviously the maximum distance between two points in the same quadrant is $4\sqrt 2 < 6$ , hence each square has exactly one vertex in each quadrant.

Given any square, we can circumscribe another axes-parallel square around it. In the picture below, the original square is red and the circumscribed one is blue.

Let's now consider the opposite direction. Assume that we picked the blue square, how many different red squares do share it?

Answering this question is not as simple as it may seem. Consider the picture below. It shows all three red squares that share the same blue square. In addition, the picture shows a green square that is not valid, as two of its vertices are in bad locations.

The size of the blue square can range from $6\times 6$ to $14\times 14$ , and for the intermediate sizes there is more than one valid placement. We will now examine the cases one after another. Also, we can use symmetry to reduce the number of cases.

size  upper_right  solutions  symmetries  total
   6        (3,3)          1           1      1
 
   7        (3,3)          1           4      4
 
   8        (3,3)          1           4      4
   8        (3,4)          1           4      4
   8        (4,4)          3           1      3
 
   9        (3,3)          1           4      4
   9        (3,4)          1           8      8
   9        (4,4)          3           4     12
 
  10        (3,3)          1           4      4
  10        (3,4)          1           8      8
  10        (3,5)          1           4      4
  10        (4,4)          3           4     12
  10        (4,5)          3           4     12
  10        (5,5)          5           1      5
  
  11        (4,4)          3           4     12
  11        (4,5)          3           8     24
  11        (5,5)          5           4     20
  
  12        (5,5)          5           4     20
  12        (5,6)          5           4     20
  12        (6,6)          7           1      7
  
  13        (6,6)          7           4     28
  
  14        (7,7)          9           1      9

Summing the last column, we get that the answer is $\boxed{225}$ .

2009 AMC 12B (Problems • Resources)
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2009 AMC 12B Problems/Problem 25

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