Decomposing a square into acute triangles 
The square $ABCD$ is to be decomposed into $n$ triangles (non overlapping) all of whose angles are acute. Find the smallest integer $n$ for which there exists a solution to this problem.

Solution

My questions
Ignore the picture and last sentence because that deals with another part of this question that I didn't include. What I don't get is what they mean by the last sentence: "Since each of the vertices inside the square belongs to at least five triangles, and last most two conatin both, it follows that $n \geq 8$". I get the at least $5$ triangles part (that was given to in the beginning of the solution), but how does this fact and the other fact which I don't understand "and at most two contain both" (what do they mean by both and are they talking about vertices?) lead to $n \geq 8$.
 A: The proof shows that there are at least two vertices in the interior of the square. If we assumed for a second that there are exactly 2(if there are more we will get a number more than 8) then we can note that at most two triangles are incident with both interior points. Thus if we count the triangles incident with both interior points, we see that there are $2(5)-2=8$ triangles minimally(of course this is on the assumption that there are exactly two interior vertices. The case of more than 2 is left out since it obviously will give you more triangles.)
Edited:
We have in (1) that $$3n\geq8+3b+5i$$ and from counting the sum of the angles we get $$n=2+b+2i$$ if we plug this into (1) we get $$3(2+b+2i)\geq8+3b+5i$$ if we simplify this down we get $$i\geq 2$$.
As for where $2(5)-2=8$ comes from, let us count the triangles around the two interior points. We know there are at least five triangles around each interior point, so that gives us $2(5)$ triangles(two interior points five triangles each). We should be worried though because we may have double counted some triangles, namely any triangles that are incident with both interior points. How many of those could there be? Two(do you see this?) so if we did over count, we over counted by at most 2 so we can definitely say there are at least $2(5)-2=8$. 
Let me know if you need further clarification.
A: There are at least 2 interior vertices, since i≥2. Each interior vertex belongs to a minimum of 5 triangles. By "at most two contain both" the proof means that there can only be 2 triangles that share both interior vertices. If we add the 5 triangles that interior point 1 belongs to with the 5 triangles that interior point 2 belongs 2 we get 10 triangles. However, since two of these triangles share both vertices they were counted twice, once within the 5 triangles containing interior point 1 and once within the 5 triangles containing interior point 2. Therefore we can subtract 2 out from the 10 total triangles (to avoid duplication) to get the minimum of n≥5+5-2=8.
