Counting how many regular polygons combinations can form 360 degrees around a point

Question

A set of regular polygons of side length 1 is chosen such that the polygons can be made to fill the 360-degree angle around a point, as a hexagon, two squares, and a triangle can. In how many ways can the polygons be chosen? (the order of placement is irrelevant)

I have used a computer program and it has calculated that there are 17 solutions but now im curious and want to prove this but I don't know where to start.As a tip, the problem was in a section involving orthocenter, tangents, homothecy, and similar/congruent triangles. Any solution/proof or even a hint will be much appreciated.

Answer 1

You can start from the facts that a regular $n-$gon has angles of $180-\frac {360}n$ degrees and you need a total of $360$ degrees at the vertex. For six polygons, you need them all to be triangles. For five polygons you can have four triangles and a hexagon or three triangles and two squares. For four polygons, let the numbers of sides be $n1 \le n2 \le n3 \le n4$. Then we have $360=720-\frac {360}{n1}-\frac {360}{n2}-\frac {360}{n3}-\frac {360}{n4}$ or $1=\frac 1{n1}+\frac 1{n2}+\frac 1{n3}+\frac1{n4}$. Clearly one solution is $n1=n2=n3=n4=4$ with four squares. You can't have three triangles because that leaves a straight angle. If you have two triangles you need $\frac 1{n3}+\frac 1{n4}=\frac 13$, which can be two hexagons or a square and a dodecagon. With one triangle you can have two squares and a hexagon. Finally you can do the same for three. The equation is $360=540-\frac {360}{n1}-\frac {360}{n2}-\frac {360}{n3}$ or $1=\frac 2{n1}+\frac 2{n2}+\frac 2{n3}$ and search for integer solutions. You don't need any geometry once you get the angles.

I find the following triples $(3,7.42),(3,8,24),(3,9,18),(3,10,15),(3,12,12),(4,5,20),(4,6,12),(4,8,8),(5,5,10),(6,6,6)$
This makes a total of $17$, as you got.

Answer 2

There are small enough to count.

The smallest angle is $60$ and the largest angle is $< 180$ so there are between $3$ to $6$ figures.

If there are $3$ figures they average to angles $120$ so one is at most $120$ and one is at least $120$.

The angle of an $n-gon$ is $180\frac {n-2}{n}$.

So if there are $3$ figures one has at most $6$ legs and one has at least $6$ legs.

If the average six legs that's solution.

Solution 1: Three hexagons.

If one of them has fewer than $6$ legs we can try if it is a triangle, square, pentagon$

So

Case 1: One is a triangle. Then the remaining $2$ will average $150$.

Solving $180 {n-2}n = 150$ give us $30n =360$ $n= 12$.

That's a solution 2. A triangle and two dodecagons. A dodecagon's angle is $180\frac {10}{12}= 150$.

Case 1a: One is a triangle. one has an angle of $180{n-2}{n} < 150$ and the third has an angle of $300- 180{n-2}{n} = 180{m-2}m > 150$.

So $\frac 53 = \frac{n-2}{n} + \frac{m-3}m;n< 12$ and so $5mn=3m(n-2) + 3n(m-2)$

$mn -6m= 6n$

$m = \frac {6n}{n-6}$ so $6< n < 12$ and you can have

$n = 7; 8; 9, 10$ and $m= 42, 24, 18, 15$ and those are several more solutions.

.... this will actually get tedious....

Case 2: One is a square and the remaining $2$ will average $135$.

Solving $180{n-2}n = 135$ give us $n=8$ and so

Solution 7: A square and two octogons.

If one has fewer $n< 8$ legs we goe $270 -180\frac{n-1}n = 180{m-2}m > 135$

So $\frac 32 = \frac{n-2}n + \frac {m-2}m$ so $3mn = 2m(n-2)+2n(m-2)$ so $mn = 4m + 4n$ and $m = \frac{4n}{n-4}$.

So $n=5,6$ and $m=20, 12$ are two more solutions.

Case 3: If one is a pentagon. We have one solution about so any other will require the second be a pentagon or higher.

We need $\frac {n}{n-2} + \frac {m}{m-2} = 252$ or average $126$ or average $6.5$ legs. So we can have two pentagons and a third with an angle of $144$ which would mean $\frac m{m-2} =\frac {144}{180} = \frac 8{10}$,

So solution 10: two pentagons and an decagon.

Or a pentagon, a hexagon and a third with an angle of $132$ which is $\frac m{m-2} = \frac {132}{180} = \frac {11}{15}$ which has not integer solution.

We can keep doing this....

Actually it gets easier as the averages are smaller.

For $4$ figures the average is $90$ or $4$ legs so that is a solution: $4$ squares.

If one has fewer than $4$ legs it must be a triangle and the remaining have $300$ and the average of the three is $100 < 108$ so one must be a triangle or a square.

If you have two triangles then the remainder is $240$ which averages $120$. So the third can have $3$ $4$ or $5$ or $6$ legs. $3$ is out as that'd make the fourth angle $180$. So are $4$ and $5$ by similar calculations.

So solution: two triangles, two hexagons.

If you only have one triangle, then you have a triangle and a square and the remaining two angles add to $210$ and average $105< 108$ so you'd have to have a triangle, two squares and .... a hexagon.

That's it for $4$.

If you have $5$ figures they average to $72< 90$ which must require at least one triangle. The leaves $300$ for an average of $75< 90$. So there must be at least two triangles. That leaves $240$ for an average of $80 < 90$ so there must be at least $3$ triangles. That leaves $180$ for an average of $90$

So solution: three triangles and two squares.

If we have a fourth less than a square we have four triangles and a remainder of $120$. So solution: four triangle and a hexagon.

Finally for $6$ figure you have an average of $60$ which can only be $6$ triangles.

Averages.... logically cut things down.

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Another way of doing this is

$\sum \frac {k_i}{k_i-2} = 2$

Where there are $3$ to $6$ summands. It's ... the same.