Greatest Problem Ever! (You can't do it)

How many non-intersecting chords of length N can fit into a circle of radius R?

P.S. If you don't know what this means, don't even try.
P.P.S. Show your work or you'll get no credit

P.P.P.S: This really isn't that great, I just wanted to grab your attention.
P.P.P.P.S. This came to me in a flash of inspiration

who care? 😉

Originally posted by Codfish
How many non-intersecting chords of length N can fit into a circle of radius R?

P.S. If you don't know what this means, don't even try.
P.P.S. Show your work or you'll get no credit

At first glance I would say the answer is the following:

1, if N = 2R.

The nearest whole number that is less than or equal to Pi/[InverseSine(N/2R)], if N < 2R.

Is this what you got?

Edit: To get the above answer, I am simply dividing 2*Pi by the angle ACB, where A and B are the terminal points of some chord of length N, and C is the center of the circle.

INFINITE!! 😀

Hint: If the circle is a unit circle, then you can fit 2chords between N=2 and N=Sqrrt3, and 3 chords between sqrtt3 and sqrrt2.

Originally posted by Codfish
Hint: If the circle is a unit circle, then you can fit 2chords between N=2 and N=Sqrrt3, and 3 chords between sqrtt3 and sqrrt2.

That is consistent with the answer I already gave.

Sorry lemon Jello, I missed your post. (Somehow😳)
That is indeed what I got. Please note, though, that the maximum side length for each maximum #of chords is .................................................................
Never mind, its too obvious for me to say it out loud without being blinded
:'(

it would be infinite, as your chords could be infinitely thin.

Originally posted by laur3tta
it would be infinite, as your chords could be infinitely thin.

No, because they are all of the same length and can't intersect

Originally posted by Codfish
No, because they are all of the same length and can't intersect

By intersect do you mean its not allowed for even the ends of the chords to touch? I think even if that were the case you still could get an infinite #, they can asymtotically approach each other not touching, but infinitely close. Don't see a limiting factor here.
For every chord you can construct there would be a smaller one that could be fitted in the same space, ad infinitum.

i have to agree with sonhouse, i can't find anything wrong with the argument. please prove it wrong!

If I inscribe a regular N-gon in a circle of radius R then surely all of the N sides are chords. There is, however, no limit on N. So I would have to conclude the answer is infinity. Please explain to me how I am wrong

Wait, unless N (the length of the chord) is fixed. in which case it is not infinity, there would have to be a relation between N and R. I'm an idiot, i just got the same answer as LemmonJello. Sorry. So in response to sonhouse draw a picture and you'll see it is not infinite, N has to be fixed.

Originally posted by frew
Wait, unless N (the length of the chord) is fixed. in which case it is not infinity, there would have to be a relation between N and R. I'm an idiot, i just got the same answer as LemmonJello. Sorry. So in response to sonhouse draw a picture and you'll see it is not infinite, N has to be fixed.

Yes, N is a fixed length. Sorry for the confusion