Prove that only triangles, quadrilaterals, and hexagons will Tesselate the plane
So, I have almost completed the proof, I will write it down all the way to the part I'm stranded in.
Theorem: only triangles, quadrilaterals, and hexagons will Tesselate the plane
So, the angles of a regular polygon with $n$ sides have a measure of $(\frac{180(n-2)}{n})°$ when several "$n$-gons" meet at one point, $a$ angles meet, so the measure of all this a angles totals $(\frac{180(n-2)}{n}(a))°$; if $(\frac{180(n-2)}{n}(a))°>360°$ the polygons will overlap, whilst $(\frac{180(n-2)}{n}(a))°<360°$ will result in empty spaces. So $$(\frac{180(n-2)}{n}(a))°=360°$$ which has to be simplified.
$\frac{180(n-2)}{n}(a)=360$
$180(n-2)a=360n$
$(n-2)a=2n$
$na-2a=2n$
$na-2a-2n=0$
$(a-2)n-2a=0$
So, this is this the immediately aforementioned linear system :
$\begin{bmatrix} (a-2) & -2 \end{bmatrix}\begin{bmatrix} n \\ a \end{bmatrix}=\begin{bmatrix} 0 \end{bmatrix}$
Which I solve
$\begin{bmatrix} (a-2) & -2 & | & 0 \end{bmatrix}$
$\begin{bmatrix} 1 & \frac{-2}{a-2} & | & 0 \end{bmatrix}$
So turns out
$\begin{bmatrix} n \\ a \end{bmatrix}=\begin{bmatrix} \frac{2t}{a-t} \\ t \end{bmatrix}$
But I get stuck in here; I know that $n$ is the number of sides of the $n$-gons we are tiling the plane with, and $a$ is the number of angles that meet at a point, so I know that I have to make sure that $a∈ℤ∧n∈ℤ∧n≥3$; I already know that this only happens when
$\begin{bmatrix} n \\ a \end{bmatrix}=\begin{bmatrix} 3 \\ 6 \end{bmatrix}$
$\begin{bmatrix} n \\ a \end{bmatrix}=\begin{bmatrix} 6 \\ 3 \end{bmatrix}$
and
$\begin{bmatrix} n \\ a \end{bmatrix}=\begin{bmatrix} 4 \\ 4 \end{bmatrix}$
But, I only do, because I know that beforehand; the only thing I'm missing to complete the proof, is that those are the only combinations that won't result in a number of sides or angles that make no sense.
How would you do this?
Thanks in advance.
(I'm assuming the scope of this problem only includes regular polygons.)
Most of your later steps are unnecessary. Once you have
$$a(n-2)=2n$$
or $(n-2)|2n$, we have
$$(n-2)|(2(n-2)+4).$$
$$(n-2)|4.$$