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