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12 hours ago history edited Toby Speight CC BY-SA 4.0
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15 hours ago answer added Neil timeline score: 2
17 hours ago answer added Level River St timeline score: 3
20 hours ago comment added Arnauld Yes, I think so!
20 hours ago comment added Glory2Ukraine @Arnauld do you think it is correct now?
20 hours ago history edited Glory2Ukraine CC BY-SA 4.0
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20 hours ago comment added Glory2Ukraine @Arnauld I will add the requirement for the chords to have unequal their corresponding parts. This indeed aligns with your answers as with well as the idea of this challenge. I think it would be enough to state that the chords need to have different lengthes.
20 hours ago comment added Arnauld According to your last edit, we should only exclude chords that are diameters. But something like (m,n,r)=(1,7,5) (with CS=BS) now looks valid. For X=100, I believe this would lead to 74 solutions instead of 37.
21 hours ago history edited Glory2Ukraine CC BY-SA 4.0
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yesterday comment added Ajax1234 @LevelRiverSt Thank you very much, that makes sense
yesterday history edited Glory2Ukraine CC BY-SA 4.0
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yesterday comment added Level River St @Ajax1234 All the additional solutions you find: 2, 14, 10 , 1,7,5 etc. have one thing in common: the x and y coordinates of the intersection are equal, and both chords are split equally. These solutions are in some sense trivial. I understood "a diagonal cannot be a solution m+n<2r" to mean "a diameter cannot be a solution" but it seems the intent may indeed have been to exclude solutions where the intersection falls on a diagonal as well as ones where a chord is a diameter. I will wait for OP response but that's the only way I can explain the test cases.
yesterday comment added Ajax1234 Can you clarify how 2, 14, 10 is not a solution?
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yesterday answer added Ajax1234 timeline score: 4
yesterday answer added Arnauld timeline score: 4
yesterday history edited Glory2Ukraine
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yesterday history asked Glory2Ukraine CC BY-SA 4.0