I am interested in the following question:
What are all the ways to completely fill a planar vertex using convex regular polygons?
For example if you place 2 squares and 3 equilateral triangles around a vertex their interal angles add to $2\pi$ and that vertex is filled. Using the fact the internal angle of a regular polygon with $n$ sides is $\pi\frac{n-2}{n}$ we can restate this as:
How many multi-sets $A$ (of integers greater than 2) satisfy: $$ 2=\sum_{n\in A}\frac{n-2}{n} $$
(Note that because $A$ is a multi-set we are not counting different ways of arranging the same polygons separately. It's fairly trivial once you have this classification to enumerate the arrangements via necklaces.)
I took a bit of time and just came up with as many as I could before I felt like it seemed complete:
$$ 3,3,3,3,3,3\\ 4,4,3,3,3\\ 4,4,4,4\\ 6,3,3,3,3\\ 6,6,3,3\\ 6,6,6\\ 8,8,4\\ 10,5,5\\ 12,4,3,3\\ 12,6,4\\ 12,12,3\\ 15,10,3\\ 18,9,3\\ 20,5,4\\ 24,8,3\\ 42,7,3\\ $$
From here I proved two fairly trivial facts:
- Solutions must have between 3 and 6 polygons.
- Solutions cannot have two values both greater than 12.
With this the search space is finite and I ran a computer search. It found a solution I missed:
$$ 6,4,4,3 $$
I felt a little silly about missing this, but that gives the complete list. However I don't think I'm alone in finding "I had a computer check all 2992 possibilities and it came up with this complete list" very satisfying as a proof.
Is there a way to prove the correct list is complete in a size that would be feasible for a human to verify on paper?
OK, I'll take it up at the point of finding all solutions, by hand, to $$ {a_1-2\over a_1}+{a_2-2\over a_2}+\cdots+{a_r-2\over a_r}=2, $$ with $3\le a_1\le a_2\le\cdots\le a_r$, $r\ge3$.
With a little bit of algebra, we can rewrite this as $$ {1\over a_1}+{1\over a_2}+\cdots+{1\over a_r}={r-2\over2} $$
Case 1: $r=3$. Changing notation a bit, we have $$ {1\over a}+{1\over b}+{1\over c}={1\over2}, $$ with $3\le a\le b\le c$.
If $a\ge7$, then the left side is at most $3/7$, which is less than $1/2$, so $3\le a\le6$.
If $a=6$, we must have $a=b=c=6$ (here and elsewhere I will leave out detailed justifications when I think it should be obvious how to do them), so we get the solution $$ 6,6,6 $$
If $a=5$, we get $$ {1\over b}+{1\over c}={3\over10}. $$ Then $1/b\ge (1/2)(3/10)=3/20$, so $b\le20/3$, so $b=5$ or $b=6$. $b=5$ leads to $c=10$, so we get $$ 5,5,10 $$ $b=6$ does not yield an integer for $c$. So, that completes $a=5$.
If $a=4$, we have $$ {1\over b}+{1\over c}={1\over4}, $$ so $4\le b\le8$. Trying each of these values of $b$ in turn, we get $$ \matrix{4,5,20\cr4,6,12\cr4,8,8\cr} $$
If $a=3$, then $$ {1\over b}+{1\over c}={1\over6}, $$ from which $7\le b\le12$. Trying each value of $b$ in turn, we find $$ \matrix{3,7,42\cr3,8,24\cr3,9,18\cr3,10,15\cr3,12,12\cr} $$
That completes the case $r=3$.
Case 2: $r=4$. We need $$ {1\over a}+{1\over b}+{1\over c}+{1\over d}=1 $$ with $3\le a\le b\le c\le d$. Also, $a\le4$.
If $a=4$, then $a=b=c=d=4$, and we get $$ 4,4,4,4 $$
If $a=3$, then $$ {1\over b}+{1\over c}+{1\over d}={2\over3} $$ $3/5<2/3$, so $3\le b\le4$.
$b=4$ gives $$ {1\over c}+{1\over d}={5\over12} $$ where the only solution is $c=4$, $d=6$, so we get $$ 3,4,4,6 $$ $b=3$ gives $$ {1\over c}+{1\over d}={1\over3} $$ which leads to $c=4$, $d=12$ and $c=d=6$. Thus, $$ \matrix{3,3,4,12\cr3,3,6,6\cr} $$ That completes Case 2.
Case 3: $r=5$. $$ {1\over a}+\cdots+{1\over e}={3\over2} $$ $5/4<3/2$, so $a=3$. $4/4<7/6$, so $b=3$. $3/4<5/6$, so $c=3$. So we are down to $$ {1\over d}+{1\over e}={1\over2} $$ which has solutions $d=3$, $e=6$ and $d=e=4$. So we get $$ \matrix{3,3,3,3,6\cr3,3,3,4,4\cr} $$ That does Case 3.
Case 4: $r=6$, $$ {1\over a}+\cdots+{1\over f}=2 $$ We have $3\le a\le\cdots\le f$, but $a=\cdots=f=3$ is a solution, so it is the only solution: $$ 3,3,3,3,3,3 $$ and we're done.