# Dissecting an obtuse triangle into acute triangles.
Take $T$ be any obtuse triangle (namely one of the angle is $> 90^{\degree}$). Can you dissect it into only acute triangles? An acute triangle is one where each angle is strictly $< 90\degree$, so right triangles do not count.
We claim that no dissection of $T$ into $6$ or less acute triangles is possible.
Recall Euler polyhedron formula on the plane, a planar graph satisfies $$
V - E + F = 1
$$where $V =$ total number of vertices, $E =$ total number of edges, and $F =$ total number of faces.
Suppose we have dissected $T$ in a way so that we have added $a$ many interior vertices and $b$ many boundary vertices to $T$, so $$
V = 3 + a + b
$$
Since there are $3+b$ boundary vertices, there are $3+b$ many boundary edges. So there are $E - 3 - b$ many interior edges.
Now, each face is a triangle with 3 sides, so the quantity $3F$ double counts the number of interior edges and once each the boundary edges. So $$
3 F = 2 (E-3-b) + 3 + b = 2E-3-b,
$$in other words, $$
3F + 3 + b = 2E.
$$
So substituting into polyhedral formula, we get $$
F = 1+2a+b.
$$
Note, the total degree of the graph is $2E$, and substituting $F$ in, we get $$
2E = 6a+4b+ 6
$$
So the condition $F \le 6$ forces $0\le 2 a + b \le 5$. This shows we must have $0 \le a \le 2$. This gives the following possible triplets $(a,b,F)$:
Note also, if a vertex is an interior vertex, then it must have degree $5$ or more, other wise it would have an obtuse angle. If a vertex is on the boundary but it is not one of the original three corner vertices, then it must have a degree of 4 or more to avoid having an obtuse angle.
So, given $a$ many interior vertices and $b$ many non-corner boundary vertices, and each of the corner vertices has degrees at least 2, and the obtuse corner at least 3. So we need degree at least $5a + 4b + 7$.
Now, if we make a table and think about which vertices can connect to which, we will arrive at contradictions.
When $a = 0$ then we have $b\le 5$. The total degree is then $$
6 + 4b
$$while we need $$
4b+7
$$hence this is impossible.
When $a = 1$, then $b \le 3$. In this case the total degree is $$
12+4b
$$while we need $$
12 + 4b.
$$
This means the interior vertex must have degree 5 and the $b$ boundary vertices each have degree 4. Since only the obtuse corner connects to another vertex, along with $b$ others, there are a total of $1+b$ available vertices to join to $a$. But $1+b \le 4$ while $a$ has degree $5$. So this is impossible.
When $a = 2$, then $b \le 1$. In this case the total degree is $$
18+4b
$$and we need $$
17+4b.
$$
So let us examine the two cases, $(a,b) =(2,0)$ and $(a,b) = (2,1)$.
If $b = 0$, then one of the interior vertex has degree at least 5. But there are only 4 other vertices, so this is impossible.
If $b=1$, then one of the interior vertex $A$ with degree at least 5 must join to the four other vertices (3 corners, one vertex on boundary, one interior vertex). This interior vertex $A$ cuts the triangle into four regions, and the other interior vertex is in one of the regions. But in each region there is only 3 other vertices to connect to, it is impossible for the other interior vertex to have degree 5 or more!
Alternatively, this would give a graph that is nonplanar: The two interior vertices $A,A'$ must be adjacent to all other vertices, and there are four vertices along the boundary and corner: $B_{1},B_{2},B_{3},B_{4}$.
So we have edges $B_{i}B_{2}$, $B_{2}B_{3}$, $B_{3},B_{4}$, $B_{4},B_{1}$, $AB_{i}$, $A'B_{i}$, and $AA'$. Note this graph contains an isomorphic copy of $K_{3,3}$, so it is not planar!
Note $A$ is adjacent to $A', B_{1},B_{3}$;
$B_{2}$ is adjacent to $A', B_{1}, B_{3}$;
and $B_{4}$ is adjacent to $A',B_{1},B_{3}$. This gives $K_{3,3}$ subgraph! (Kuratowski)
Hence, it is impossible to dissect an obtuse triangle into 6 or less many acute triangles!
Although this seemingly is a very geometric problem, I like we can through enough topology and combinatorics at it to resolve it.