# a special problem about numbers assigned on polygons

So the problem is stated as follow:

We have $a$ numbers of regular $b$-sided polygons. We place them in a fashion such that the sides of polygons are parallel and the vertex of every polygon connects with other vertexes of other polygons with a straight line. Now we place numbers of ${1,2,3,...,ab}$ onto the vertex of the polygons in a way such:

1. The sum of vertexes of one $b$-sided polygons are equal to every other polygons' sum of vertexes.

2. The straight lines that connects the vertexes of the polygon all have the same sum.

With intermediate steps, come up with a way such that we can combine/use $a$ numbers of $b$-sided polygon with a $c$ numbers of $d$ sided polygon to create $ac$ numbers of $bd$ sided polygon.

(Added picture to explain how this works)

So I am quite frustrated with this-- it does not guide me what I should do for the preparation steps. I know that the sum of the vertexes of one polygon will be $\frac{(ab+1)(b)}{2}$and sum of lines will be $\frac{(ab+1)(a)}{2}$ but I just cannot manipulate these two equations to create $ac$ numbers of $bd$ sided polygon. Some help will be greatly appreciated!

So the key idea is to manipulate this equation: $\frac{(ab+1)(b)}{2}$. We want this equation to get to $\frac{bd(abcd+1)}{2}$. It is clear that we need to multiply $\frac{(ab+1)(b)}{2}$. by $d$-- it is just splitting the b-sided polygon into a $bd$ sided polygon. However the problem that I am now struggling with is how to manipulate $\frac{(ab+1)(bd)}{2}$ into $\frac{bd(abcd+1)}{2}$: the difference between them is $\frac{bd(abcd+1-ab-1)}{2}$ or $\frac{bd(ab(cd-1))}{2}$, which I just have no idea how to "compensate" this difference.