Most interesting problem

Hey Brilliantinians, here I'm posting the most beautiful problem I've ever seen. The idea is that everyone can share the most interesting/beautiful/awesome problem he/she has ever faced to. The problem needs not to come from Brilliant. Solutions to the problems are also accepted in the comments

Here's mine. Let p3p\geq 3 be a prime number. We divide every side of a triangle in pp equal parts and every point of division is connected to the opposite vertex. Compute the maximum number of pairwise disjoint parts in which the triangle is divided in funciton of pp.

Enjoy!!!

#Combinatorics #NumberTheory #Awesome #Interesting

Note by Jordi Bosch
6 years, 5 months ago

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Comments

Well, if pp is an odd prime, then the maximum such number must be 3p23p+13{ p }^{ 2 }-3p+1

so that for p=3,5,7,11,13,19,...p={3,5,7,11,13,19,...}, we have 19,61,127,331,469,817,...{19,61,127,331,469,817,...}

We first note that the triangle is divided into p2{ p }^{ 2 } disjoint parts by the first 22 sets of rays from 22 vertices. Then each ray from the last vertex intersects those rays 2(p1)2(p-1) times, thereby increasing the number of disjoint parts by 2(p1)+12(p-1)+1, and there are p1p-1 such rays from that last vertex. Hence, the total is

p2+(p1)(2(p1)+1)=3p23p+1{ p }^{ 2 }+(p-1)(2(p-1)+1)=3{ p }^{ 2 }-3p+1

This derivation depends on the fact that no point is shared by 33 such rays, hence the primes only.

This graphic shows the case where p=5p=5, so one can see how this is done. Affine transforms of this triangle leaves the number of parts unaffected.

Michael Mendrin - 5 years, 12 months ago
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