## Fall 2016, problem 21

Determine all values of $n$ such that it is possible to divide a triangle into $n$ smaller triangles such that there are not three collinear vertices and such that each vertex belongs to the same number of segments.

1 year ago

edit: The possibilities are:

• $n=3$: Just insert a point $z$ in the interior of the original triangle. Connect it to the original vertices.
• $n=7$ : Morley's figure
• $n =19$: Something more complicated. See figure at the end.

[You get these figures, by taking the surface of a tetrahedron, octahedron or icosahedron, deleting one face and then flattening out the rest. As an aside, the other two platonic solids would give you a subdivision of a quadliteral into smaller quadliterals (cube) or of a pentagon into smaller pentagons (dodecahedron). Both subdivisions satisfy the rest of the problem requirements.]

Here's what I wrote before. It's wrong:

The obvious possibility is $n=3$: The possibility $n=1$ is ruled out, since it is specified, that the triangles after subdivision be smaller than the original. We will show, that $n=3$ is actually the only possibility:

Notation: After subdivision...

• ...$v$ is the number of vertices
• ...$e$ is the total number of segments
• ...$k$ is the number of segments any given vertex belongs to

Here's a formal Proof:

Euler's formula for planar graphs says: $$n + v - e = 1 \quad (Eq.1)$$

We sum over all segments the number of vertices on that segment (the latter is always 2). The result is $2e$, of course. Counting this way, we count every vertex $k$ times from the definition of $k$. Therefore:

$$2e = k v \quad (Eq.2)$$

edit: We sum over all triangles the number of vertices of that triangle (the latter is always 3). The result is $3 n$, of course. Counting this way, we count every vertex of the original triangle $(k - 1)$ times and we count all other vertices $k$ times (the difference between these two types of vertices comes from the fact, that the outside of the original triangle is not a triangle). Therefore:

$$3 n = (k - 1) \times 3 + k (v - 3) \quad (Eq.3)$$

Here's what I wrote before. It's wrong:

We sum over all triangles the number of vertices of that triangle (the latter is always 3). The result is $3 n$, of course. Counting this way, we count every vertex of the original triangle two times and we count all other vertices 3 times (the difference between these two types of vertices comes from the fact, that the outside of the original triangle is not a triangle). Therefore:

$$3 n = 2 ¸\times 3 + 3 (v - 3) \quad (Eq.3)$$

Putting together Equations 1 through 3 yields:

$$n (6 - k) = 5 k - 6$$.

The solutions of this equation by positive integers are $(k,n) \in \{(3,3), (4,7), (5,19), (2, 1)\}$. The solution (2,1) does not correspond to a valid subdivision, since in the problem it is specified, that after subdivision the triangles be smaller.

Thanks to rubs, who found a serious bug in my original answer and who brought the morley figure to my attention.

Here's an illustration of the $n=19$ case:

What about the figure in the theorem of Morley ?

rubs 1 year ago

@rubs Thank you. You are absolutely right. The Morley figure is another possibility. Gonna fix this.

Nelix 1 year ago