Problem 1165

Suppose we extend the sides of a regular pentagon from one of its vertices to the line of the opposite side, we get a triangle as described in the problem. If we connect the vertices of the pentagon to the center of the pentagon, the original triangle is divided into seven acute triangles. We will show that the original triangle cannot be divided into fewer than seven acute triangles, and furthermore, we will show that there is no right or obtuse triangle that can be divided into fewer than seven acute triangles. We will prove this by assuming the opposite and arriving at a contradiction.
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So, let's assume that there is - or are - right or obtuse triangles that can be divided into fewer than seven acute triangles. Let's take one of these that can be divided into the fewest possible parts, and consider its minimal division. In this minimal division, an edge starts from the right or obtuse angle vertex, but it cannot reach the opposite side. Otherwise, it would cut the original triangle into two triangles, at least one of which would not be acute, and it could be divided into at least one fewer acute triangles than the original. Therefore, the original division would not be minimal.

The segment starting from the right or obtuse angle vertex can only end at an internal point. At least five triangles must connect to this point in the division, otherwise, they could not all be acute. Apart from these, only one more triangle can result from the division. Let's denote the vertices of the triangle's acute angles by $A$ and $B$, the third vertex by $C$, and the already mentioned internal point by $D$. Since the angles $A D B$, $B D C$, and $C D A$ are not all acute, the edges $A D$ and $B D$ cannot both be in the division (otherwise, we would again find that the original division was not minimal). Suppose the edge $B D$ is not in the division, then the acute triangle in the division containing $B$ cannot have $D$ as a vertex. This triangle clearly cannot have $A$ as a vertex either, so the triangle connected to $A$ must have $D$ as a vertex.

The endpoints of the additional edges starting from $D$ can only be points not mentioned so far, so at least 3 more vertices are created in the division. Among these, there cannot be another internal point, otherwise, at least five triangles would connect to it, of which at most two could connect to $D$, so we would have at least $2 \cdot 5 - 2 = 8$ triangles.

Therefore, the endpoints of the additional edges starting from $D$ are on the sides of the triangle $A B C$. If there is one on the segment $A C$, the edge leading to it would cut the triangle $A D C$ into parts, one of which could be chosen instead of $A B C$, so $A B C$ would not be minimal. If there is no point among them on $A C$, then there are two among them on $A B$ or $B C$: let's denote the one closer to $B$ by $P$ and the other by $Q$. Then the triangle $A D P$ (or if $P Q$ is on $B C$, the triangle $C D P$) is cut into two parts by $D Q$, one of which contradicts the minimality of $A B C$. This proves our statement.