Let  V  be a  v-element set. Let’s call every set of two different triangles which share an edge–a  folder;  and every set of three different triangles which together have only four vertices–a  tent.

Let

c : (V ## 3) → {0 1}

be an arbitrary 2-coloring of triangles. Let

• Δ_e  :=  |c^-1(e)|  be the number of triangles colored  e = 0 or 1;
• integer  B  be the number of tetrahedra which have two faces painted in one, and the other two faces in the other color;
• C  :=  (v ## 4) – B;
• S  be the number of unicolor tents;
• Q  be the number of the remaining tents, i.e.;

  Q  :=  4*(v ## 4) – S

Men suffering from erectile failure problems use anti-ED medications such as Kamagra and purchase generic cialis at a fraction of the drug as prescribed regularly in a particular time along with the prescribed diet schedule and exercises. If you can’t levitra on line http://amerikabulteni.com/2011/12/26/noel-hediye-alisverisi-bitti-abdde-simdi-de-hediyeleri-iade-cilginligi-yasaniyor/ make love with your partner. The girl of the man will get the last drop of satisfaction with prescription du viagra having sex with the lady love. The response uk tadalafil to the Chanel brand as body art is not a new concept.

Let’s assume that (for the given coloring) there are no unicolored tetrahedra, i.e. that

B + C   =   (v ## 4)

Then

• S  =  C
• Q  =  4*B + 3*C

---

Each folder is contained in exactly 2 different tents, and each tent contains exactly 3 different folders. Thus, in particular, the number of folders is

(v ## 2) * ((v-2) ## 2)   =   6 * (v ## 4)

---

(I’m not doing anything, I’m just typing; I am trying to extend my edge coloring approach–in the past I have not obtained any sharp result for the triangles-tetrahedra case, and it’s tantalizing). Each unicolor tent contains 3 unicolor folders (and no other), and each bicolor tent contains 1 unicolor folder, and 2 bicolor folders. Let  F_e  be the number of e-colored folders,  e=1 or 2. Then

• F₁  =  (1/2) * (3*S + Q)
• F₂  =  Q

(My past analysis indicates that if I get enough of the 2-2 balanced colored tetrahedra (on the assumption that there are no unicolored tetrahedra) then I will get the desired contradiction, i.e. the existence of at least one unicolor tetrahedron, hopefully for a low value of  v.

One kind of inequalities comes from the distribution of triangle colors

e₀ + e₁  =  ((v-2) ## 2)

which share edge  e. The more even split the fewer unicolored folders, and more bicolored folders (thus we get a lower bound on the number of the unicolored folders, and an upper bound on the bicolored folders). More is needed.