Lovely problem! To generate all possible values of n given some k... Consider n lines with k intersections per line (0 k=0, a contradiction) => n = mk/(m-1) To ensure that n is a positive integer, either m-1=1 (m=2) or m-1 divides k (for m>2 m-1 does not divide m). But m-1 divides k includes the case when m-1=1 since 1 divides k Therefore the possible values for n are: n = mk/(m-1) where m-1 is a factor of k Eg when k=10: => m-1 = 1, 2, 5, 10 => m = 2, 3, 6, 11 => n = 20, 15, 12, 11
Generalization: Let an n-star be n lines intersecting one anchor point; each line in n-star intesects n-1 other lines in that n-star. Note that any set of straight lines as given in that task, can be transformed into p n-stars, whose anchors are not on any other n-star, by translating lines in parallel. Then p n-stars, whose anchors are not on any other n-star, have k:=p*(n-1) intesection points and in total N:=p*n=k+p lines. So, if k is a multiple of p, then N=k+p lines can be the number of such set (given in the task in the video).
Lovely problem! To generate all possible values of n given some k...
Consider n lines with k intersections per line (0 k=0, a contradiction)
=> n = mk/(m-1)
To ensure that n is a positive integer, either m-1=1 (m=2) or m-1 divides k (for m>2 m-1 does not divide m). But m-1 divides k includes the case when m-1=1 since 1 divides k
Therefore the possible values for n are:
n = mk/(m-1) where m-1 is a factor of k
Eg when k=10:
=> m-1 = 1, 2, 5, 10
=> m = 2, 3, 6, 11
=> n = 20, 15, 12, 11
Generalization:
Let an n-star be n lines intersecting one anchor point; each line in n-star intesects n-1 other lines in that n-star.
Note that any set of straight lines as given in that task, can be transformed into p n-stars, whose anchors are not on any other n-star, by translating lines in parallel.
Then p n-stars, whose anchors are not on any other n-star, have k:=p*(n-1) intesection points and in total N:=p*n=k+p lines.
So, if k is a multiple of p, then N=k+p lines can be the number of such set (given in the task in the video).
Interesting question and answer. Never would have thought about it, without the video. Thanks.
@@tom-kz9pb no worries! Glad you enjoyed it
I did figure it out with basically the same thought process.
@@up1663 nice! Great minds think alike!
Me too
Oh I remember doing a question like this before
yup my answer was correct!
@@Orillians amazing!
@@JPiMaths yes sir solved in 40 seconds :)>