Circles Puzzle

[wblipp]

Originally Posted by LaurV

How many integer point can you get on a circle of big/whatever/almost_infinite (grrr!) radius? Or say, rational points on a circle of radius 1.

Closely related to the problem of how many ways can an integer be written as the sum of two squares. A well studied problem with a connection to the factors of the integer.

[axn]

Originally Posted by wblipp

The first circle gets B, C, I, N, R, Q. K, F

The second circle gets H, I, O, T, X, Y, Q, L

I & Q ?

[fivemack]

Originally Posted by wblipp

NO

A  B  C  D  E

F  G  H  I  J

K  L  M  N  O

P  Q  R  S  T

U  V  W  X  Y

The first circle gets B, C, I, N, R, Q. K, F

The second circle gets H, I, O, T, X, Y, Q, L

Two problems. One: the second circle gets HIOTXWQL. Two: The strings BCINRQKF and HIOTXWQL have I and Q in common..

[ewmayer]

Originally Posted by fivemack

Two problems. One: the second circle gets HIOTXWQL. Two: The strings BCINRQKF and HIOTXWQL have I and Q in common..

But the problem-as-posed said nothing about circles having no points in common, thus moot. One could of course have some fun with various side-branch problems, e.g. "Sum the #circles passing through each point and find the solution(s) which minimize/maximize this quantity."

[LaurV]

Originally Posted by ewmayer

But the problem-as-posed said nothing about circles having no points in common, thus moot.

No, it is not. You said you didn't read posts except the OP, so you are excused. But read again post 2 and 3 (I didn't expect any reply to axn's post 3, actually, hehe).

[wblipp]

Originally Posted by axn

I & Q ?

If I squeeze my eyes shut and say "I believe" three times, will that make it true?

[ewmayer]

Originally Posted by LaurV

No, it is not. You said you didn't read posts except the OP, so you are excused. But read again post 2 and 3 (I didn't expect any reply to axn's post 3, actually, hehe).

I meant the problem as stated in the link supplied by the OP - is there another formal problem statement which hijacked the thread? Here from the OP link:

There is a square grid containing twenty-five points, all equally spaced and arranged in a 5 × 5 layout. Over this grid you can draw circles that pass through points (a couple of examples are shown on the left).

Here’s the question: Is it possible to use just five circles and make sure that at least one circle goes through every point on the grid?

Note the prominent 'at least one circle' - moreover the 'examples shown on the left' include circles which both share a grid point. The OPer does not modify this in any way. In post #2 wblipp gives a solution, and in #3 axn raises an objection which is not based on the actual problem statement, thus moot.

[axn]

Originally Posted by ewmayer

Note the prominent 'at least one circle' - moreover the 'examples shown on the left' include circles which both share a grid point. The OPer does not modify this in any way. In post #2 wblipp gives a solution, and in #3 axn raises an objection which is not based on the actual problem statement, thus moot.

wblipp: 25-(8+8)=9=3x3. QED.
me: 25-(8+8-2)=11 > 3x3. So, not QED. At least, not yet.

[LaurV]

Originally Posted by ewmayer

thus moot.

Think again.

[ewmayer]

Originally Posted by axn

wblipp: 25-(8+8)=9=3x3. QED.
me: 25-(8+8-2)=11 > 3x3. So, not QED. At least, not yet.

Ah, now the fog lifts - I didn't actually sketch out wblipp's 'solution', just took his word for it. I assumed his 'redfaced' post was due to his claim that the first 2 circles of his 'solution' had no points in common; you and LaurV were trying to point me to the 'not a solution' aspect. Gotcha. So I was the only one here to give a solution to the 5x5? To paraphrase South Park's Cartman, you guys are weak!

[ No wonder this place has become a magnet for cranks trying to fob off crap 'Mersenne prime finding' software which is 'really fast, if you throw at least $25000 of GPUs at it and don't ask what it's really doing with all those FLOPs.' We're clearly a very credulous bunch. :) ]

[LaurV]

Originally Posted by ewmayer

So I was the only one here to give a solution to the 5x5? To paraphrase South Park's Cartman, you guys are weak!

Haha,