Heptagon is divided into pentagons and hexagons. Prove that there are at least $27$ pentagons in this division.

Question

A heptagon is divided into convex pentagons and convex hexagons in such a way that each vertex of the heptagon is the vertex of at least three polygons of the division.

Prove that there are at least $27$ pentagons in this division.

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My solution:

We know that both pentagons and hexagons are convex, so we can use Eulers equation: $v + f = e + 2$

• $2e \geq 3v$, because we have $2$ vertices for every edge and we have at least 3 edges meeting in every vertice - we know that it's true for the vertices of heptagon, but it must be also true for the inner vertices since otherwise those would be just straight lines
• $f = f_5 + f_6 + 1$, where: $f_5$ is the number of pentagons and $f_6$ is the number of hexagons, as it was said in the comments, we need to add one more face which is the heptagon.
• $5f_5 + 6f_6 = 2e$, as @Lexponent shown in his answer.

So we get:

$$v + f = e + 2 \iff 2v + 2f = 2e + 4$$

$$2v + 2f_5 + 2f_6 + 2 = 5f_5 + 6f_6 + 4$$

$$2v = 3f_5 + 4f_6 + 2 \iff 3v = \frac{9}{2}f_5 + 6f_6 + 3$$

$$2e \geq \frac{9}{2}f_5 + 6f_6 + 3$$

$$5f_5 + 6f_6 \geq \frac{9}{2}f_5 + 6f_6 + 3 \iff 10f_5 + 12f_6 \geq 9f_5 + 12f_6 + 6 \iff f_5 \geq 6$$

And I got an inequality on $f_5$, but it's not good enought.

Answer 1

Here is a proof.

Note first that $$5f_5+6f_6+7=2e\text{,}\tag{1}$$ which we obtain by counting every side of each polygon twice. (Notice that $5f_f+6f_6$ will count most edges twice, but the seven outer edges only get counted once, so we must add $7$ to get that every edge is counted twice.)

Next, from the Euler characteristic equation we have $$f_5+f_6+1 +v = e+ 2\text{,}$$ so we multiply by $6$ to get $$6f_5+6f_6+6 +6v = 6e+ 12\text{,}\tag{2}$$ and so we subtract (1) from (2) to obtain $$f_5-1 +6v = 4e+ 12\text{.}\tag{3}$$ On the other hand, the degree of each outer vertex of the Heptagon is at least $4$, and the degrees of all the other vertices are at least $3$ (by convexity), so we have (if $V$ is the vertex set) $$2e=\sum_{p\in V}\text{degree}(p) \geq 4\cdot 7 + 3(v-7) = 3v+7\text{,}$$ and doubling both sides of this inequality gives us $$4e\geq 6v+14\text{.}\tag{4}$$

But now combining (3) and (4) we have $$f_5-1 +6v = 4e+ 12 \geq 6v+14+12 = 6v+26 \text{,}$$ whereby adding $1-6v$ to each side yields $$f_5\geq 27\text{,}$$ as desired.

Answer 2

Your approach, which involves Euler's formula, is a solid start. However, there seem to be some inconsistencies in your assumptions. Let's go step by step:

1. Vertices: First, as @RobertShore mentioned in the comments, the assumption that $v = 7$ is incorrect. The hexagon itself has 6 vertices. Based on the problem statement, only the vertices of the hexagon are guaranteed to meet at least 3 polygons. Hence, the vertices inside the hexagon can have different degrees, which might affect the total count of $v$.
2. Faces: As for the faces, we have the inner ones ($f_5$ and $f_6$), but there is also the outside face. Therefore, $f = f_5 + f_6 + 1$.
3. Edges: Let's reexamine the edge calculation:
   • The edges contributed by pentagons would be $\frac{5f_5}{2}$ since each pentagon has 5 edges but shares its edges with other polygons.
   • Similarly, the edges contributed by hexagons would be $\frac{6f_6}{2}$.

Therefore, $e = \frac{5f_5 + 6f_6}{2}$.

Using Euler's formula, $v - e + f = 2$, we can substitute the relationships from above.

4. Convexity: As @aschepler pointed out in the comments, the problem involves convexity. This is crucial since it ensures that no two pentagons share more than one edge. This constraint can be utilized when determining the minimum number of pentagons.
5. Example counterpoint: @GerryMyesson's example shows a case where there's a hexagon divided into 12 pentagons and a hexagon. This seems to contradict the question's assumption. However, without a visual representation, it's hard to discern if it adheres to all the constraints.

Suggested Approach:

1. Recount Vertices: Begin by reassessing the number of vertices, especially the internal ones. This can be deduced from the relationship between the edges and faces.
2. Edge Constraints: Revisit the relationship you developed for the number of edges. Each edge inside the hexagon will belong to two polygons. Each edge on the boundary of the hexagon will belong to one internal polygon and the external face.
3. Face Constraints: Utilize the constraint that each vertex of the hexagon is a vertex of at least 3 internal polygons. This constraint could be key in establishing a minimum count for pentagons.
4. Use Convexity: Since no two pentagons can share more than one edge, this might affect the minimum number of pentagons needed in the division. Visualize or list down possible configurations and their edge-sharing properties.

I hope this provides a clearer path forward. Remember, these problems often require some iteration and exploration. Sometimes a fresh perspective or drawing things out can help clarify the relationships and constraints. Good luck!