Counting Squares

worthawholebean

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 $G$ ?

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

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rrusczyk

I went ahead and added the diagram DPatrick made.

**Posted:** Thu Feb 26, 2009 10:51 am

worthawholebean

Cool. I edited it a little because I'm crazy and want it to look as much like the diagram on the page.

**Posted:** Thu Feb 26, 2009 1:32 pm

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RunpengFAILS

According to someone that I know, the answer can be obtained by computing $(1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 4^2 + 3^2 + 2^2 + 1^2) + (1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2)$ . I understand how the stuff in the first parentheses is gotten but don't understand the second part...

Another person I know used an inccorect method that should have gotten him answer of 325 but didn't add correctly and got 225 which is the correct answer!!! Shocked

**Posted:** Thu Feb 26, 2009 2:15 pm

Math Geek

**Posted:** Thu Feb 26, 2009 4:58 pm

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math154

gf4848

Can somebody post the official solution?

**Posted:** Thu Feb 26, 2009 6:22 pm

matt276eagles

This has to be the fastest solution:

Beastly Solution

**Posted:** Thu Feb 26, 2009 7:00 pm

cipos2 · Hodge Conjecture Offline Joined: 17 Nov 2008 Posts: 71

For some reason I get 220.
The method I use is to add up all the squares from one to five, , reasoning that this corresponds to side lengths from 6-10. Then I multiply by 4 to fall short 5 from the answer. Could someone tell me if this is just serendipitous or how my method should count the next 5 squares?

**Posted:** Sat Feb 28, 2009 3:57 pm

pleurestique

**Posted:** Sun Nov 08, 2009 1:04 pm

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