Let $ S=\{(x,y)\in\mathbb{R}^2 \ | \ x^2+y^2=1 \text{ and } y\geq 0\}$. Determine $S+S+...+S $. Let 
$$ S=\{(x,y)\in\mathbb{R}^2 \ | \ x^2+y^2=1 \text{  and  } y\geq 0\}$$
By the usual notation for sum of sets let
$$ 2S\overset{\text{not}}{=}S+S=\{(x_1+x_2,y_1+y_2) \ | \ (x_1,y_1),
(x_2,y_2)\in S \} $$
Let 
$$ nS\overset{\text{not}}{=}S+S +...+S \text{  n times.}$$
Determine $nS$.
I think I know what $2S$ and $3S$ are, and the conclusion follows intuitively for $nS$, but I couldn't write a formal proof for $nS$. Tip: compass and ruler is of much help when dealing with this problem. Hope you like it. Here's my answer for $2S.$ Again I need some notations. Let
$$ D_{(a,b)}=\{(x,y)\in\mathbb{R}^2 | (x-a)^2+(y-b)^2\leq1 \text{ and }
y\geq b \}$$
This is the "upper half" of a disc of radius $1$ and center $(a,b)$.
denote $D_{(0,0)}\overset{\text{not}}{=}D$, and
$$2\cdot D=\{(x,y)\ |\ x^2+y^2\leq 2 \text{ and } y\geq 0 \}.$$ Then the answer is:
$$ 2S=2\cdot D\setminus\left(D_{(-1,0)}\cup D_{(1,0)}\right).$$
So $2S$ is the upper half of the raidus 2 disk, out of which we remove two radius one upper-half-disks of centers $(-1,0)$ and $(1,0)$.
Edit:
The numerical solution posted below looks nice, of course you can start with any curve instead of $S$ and you'll probably get something pretty interesting for $nS$, no idea if it's useful though. Also wonder what you get in $\mathbb{R}^3$ for a semisphere.
One conjecture: We have: $nS=(n-1)S+S$, but I think it might be true that $nS=\partial ((n-1)S)+S$ where by $\partial ((n-1)S)$ we denote the boundary of $(n-1)S$.

Yes, brute force adding $10^5n$ points in a Monte-Carlo'esk fashion tells you what the solution's gonna be.
In Mathematica, you get one such point via
With[{x = RandomReal[{-1, 1}]}, {x, Sqrt[1 - x^2]}]

Here for $n=1,2,3,4$:

The way the picture naturally becomes sparse when it comes to addition of multiple very $y-$small points raises another question: What's the density on your solution surface for points generated w.r.t. a normal distribution like that?
