How many sqares can you draw in given pattern, when each corner of a sqare has to lie in the exact center of one of the 20 colored circles?

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There are an infinite number of squares whose sides pass through exactly four circles. In fact, there are an infinite number of squares passing through the top four circles alone.

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Each frame of the above gif represents one possible square you can draw through those circles. In addition to the 90 squares represented in these 90 frames, there are infinitely many more. For example, "the square that lies halfway between frame 1 and frame 2".

Edit: If we add the additional constraint that the corners of the square must lie on a circle, then there are 17, as originally determined by kaine.

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  • $\begingroup$ Note that this answer applies to "how many squares are there whose lines pass through these circles?", but not "how many squares are there whose corners lie on these circles?". I wasn't sure how to interpret the question, so I went with the former. $\endgroup$ – Kevin Aug 13 '14 at 13:21
  • $\begingroup$ wrong answer, it seems you didn't get the question,look at kaine's answer. $\endgroup$ – Masoud Mohammadi Aug 13 '14 at 13:52
  • $\begingroup$ @Riddle, What's wrong with it? All of my squares satisfy the criteria "your square shouldn't have more than 4 or less than 4 circles. and your lines should pass the 4 circles." Please clarify your question. $\endgroup$ – Kevin Aug 13 '14 at 14:13
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    $\begingroup$ +1. The infinite answer should be kept. That is the reason how people should formulate their question very precisely :) $\endgroup$ – justhalf Aug 14 '14 at 5:41
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    $\begingroup$ Even with "The Corners lie on four circles" you still have infinite squares by tilting and moving. The question should be "Which corners are at the exact centers of one of the 20 colored circles" $\endgroup$ – Falco Aug 18 '14 at 9:21

9 squares that are $1$x$1$ (trivial)

4 squares that are $\sqrt{2}$x$\sqrt{2}$ (Diamonds centered at the 4 central circles)

2 squares that are $\sqrt{5}$x$\sqrt{5}$ (Each starts at a yellow circle. Move like a knight: example 2 right 1 up, 2 up 1 left, 2 left 1 down, 2 down 1 right)

2 squares that are $\sqrt{13}$x$\sqrt{13}$ (Each starts at the orange circles. Move 3 right 2 up or 3 left 2 up and then keep going)

Are there more than that?

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    $\begingroup$ I can only see nine $1 \times 1$ squares. There are also four $\sqrt{8} \times \sqrt{8}$ squares $\endgroup$ – r3mainer Aug 13 '14 at 13:21
  • $\begingroup$ Thank you, miscounted. Don't the $\sqrt 8$ squares go through additional circles? 8 is made from 2 square numbers by 4+4 and the 4 squares with those dimensions I found go through 3 more circles each. $\endgroup$ – kaine Aug 13 '14 at 13:26
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    $\begingroup$ can you show √5 and √13 in picture? $\endgroup$ – Masoud Mohammadi Aug 13 '14 at 13:54
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    $\begingroup$ @kaine According to the current wording of the puzzle there is nothing wrong with the squares intersecting additional circles. $\endgroup$ – aaaaaaaaaaaa Aug 27 '14 at 17:25

I found 21

  • 9 variations of black square
  • 2 variations of orange square
  • 4 variations of brown square
  • 4 variations of gray square
  • 2 variations of red square

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    $\begingroup$ actually, i told in question that each square must be exactly on 4 circles ,but one of your square is on 8 circles, but my question was edited. so your answer is correct, but other answeres are correct too, because they answered before the edit happens. $\endgroup$ – Masoud Mohammadi Aug 28 '14 at 17:31

My answer is 21. It break down as follows:

9: Square from dots right next to each other

4: Squares with 1 dot in the middle

2: Squares with central 4 dots in the middle

2: Squares with previous 2 squares in the middle (The above can be more easily visualized from watching Kevin's answer)

4: From the 4 squares with 1 dot in the middle, extend out 1 more dot on each corner to make another square.

The total of 9 + 4 + 2 + 2 + 4 = 21.

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    $\begingroup$ Given the recent change to the question, this answer appears to be the most correct now (addition of 4 squares that previously were considered incorrect). $\endgroup$ – Joel Rondeau Aug 28 '14 at 15:59

I have found 21 Squares, too.

As Kevin illustrated in his beautiful answer there are 17 squares. PLUS there are 4 more squares.

all the squares are as follows:

Dimensions How many

  • Nine × [1x1]
  • Four × [√2×√2]
  • Two × [√5×√5]
  • Four × [√8×√8]
  • Two × [√13×√13]
  • TOTAL of 21 squares

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