Comments on Accumulated Evidence: And a triangulated triangle

my solution needs a clarification. if you start fr...

2008-02-24T06:19:00.000-08:00

my solution needs a clarification. if you start from the green vertex, the other green vertex may lie on the opposite side -of the outer triangle. in that case my argument is wrong. in any case it is possible to chose -indifferently- another outer vertex. it cannot be directly linked to the opposite side, so the proof run naming red green, green blue, blue red, R* G*, G* B* and B* R*, for any *.

Hi, i think that it is possible to make your proof...

2008-02-23T06:42:00.000-08:00

Hi, i think that it is possible to make your proof work with a couple of small tweaks, just reasoning inductively on the triangulations that are homeomorphic to a disc, and since the solution that i had in my mind is different i find your idea quite interesting.
I remark that i already supposed the triangulation to be finite (should a complicated case such as an infinite triangulation be contemplated in a problem, i will make it explicit). However, just out of curiosity, it can be proved that the triangulation of a compact space must always be finite (if you think about it, in you infinte 'counterexample' the point 0 does not belong to any triangle) because a triangulation is usually defined as a homeomorphism (note, not just a continuous bijective map) from a simplicial complex to the space, and in a simplicial complex the set of all the simplices that have a vertex v as 'face' (ie the triangles, edges, tetrahedrons, etc such that their closure contains v, plus of course v itself) are an open neighborhood N_v of v. Each such N_v contains exacty one vertex of the triangulation (that is v), and if i cover the triangle with all such N_v for all the vertices of the triangulation i have an open covering that is minimal, in the sense that no smaller subcovering can be found. On the other side, the triangle is compact, and thus a finite subcovering can always be found. So, the vertices must be in a finite number.

great! another one. no more geometry problems :)

2008-02-16T22:09:00.000-08:00

great! another one. no more geometry problems :)

please excuse me for my bad english language.i hav...

2008-02-16T22:04:00.000-08:00

please excuse me for my bad english language.
i have a solution, if you want find it by yourself don't read more.
solution -i hope :)
for the first comment at this post i can suppose that the number of vertices in the triangulation is finite. let be r_0, g_0, b_0 the vertices of the greatest triangle respectively of colour red, green, blue. the proof proceeds by induction in the number of triangles in triangulation.
if there is only a triangle obviously the thesys is true.
now let suppose the thesys for all triangulation made of less than n triangles is true.
if there is a directly connected to g_0 green vertices, let g_1 be its name, then g_1 is different from r_0 and b_0 and it is possible to delete g_0 and put in its location g_1, mantaining arcs exiting from g_1 and so obtaining a new great triangle with satisfies hipothesys and triangulated by less than n triangles. so, by inductive hipothesys, there exists a little triangle whit a red vertex, a green one and a blue one. that triangle is as subtriangle -or its deformation- of the starting triangle.
now let suppose there are no directly connected to g_0 green vertices, so the vertices on g_0b_0 side have to be blue. let b_1 be its name. remark that possibly b_0 = b_1. there is only a directly linked to both g_0 and b_1 vertex. let x_2 be its name, where x = b if vertex color is blue, otherwise x = r. let x_k be defined. if there exists only a directly linked to both g_0 and x_k vertex and k is not 1 then x_k = r_k lies on the g_0r_0 side. otherwise -yet k different from 1- there exists only a directly linked to both g_0 and x_k vertex different from x_k-1. let x_k+1 be its name. because the number of vertices is finite for any directly linked to g_0 vertex x there exists a k strictly positive integer s. t. x = x_k. let A be the set of directly linked to g_0 vertices. let be < the equivalence relation on A so defined: x_i < x_j iff i < j. A is well ordered by <, so because a red element exists is well defined the <-minimum of the set of red element of A. let r_k its name. so g_0, b_k-1 and r_k satisfy the thesys.

in the previous comment i meant finite, not finish...

2008-02-16T21:13:00.000-08:00

in the previous comment i meant finite, not finished. sorry.

please excuse me for my bad english language.first...

2008-02-16T21:05:00.000-08:00

please excuse me for my bad english language.
first of all a consideration.
the number of vertices in the triangulation have to be finished. as a counterexample consider the triangle in C of vertices the third roots of unity, say a, a^2, a^3. let s_1, s_2, s_3 be respectively the segments that connect a, a^2, a^3 to the origin. for every n strictly positive integer and for i = 1, 2, or 3, let x_i,n be the point of intersection of s_i and the circonference centered in the origin and of ray 1/n. for example x_1,1 = a, x_2,1 = a^2, x_3,1 = a^3. for all n let x_1,n be red, let x_2,n be green and let x_3,n be blue. for all n > 2 let x_1,n be connected to x_2,n, x_3,n, x_1,n-1, x_2,n-1 and x_3,n-1. for all n > 2 let x_2,n be connected to x_3,n, x_2,n-1 and x_3,n-1. for all n > 2 let x_3,n be connected to x_3,n-1. this triangulation satisfies the hypothesis but doesn't thesys.