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 $(n-1)S$.

Removing the restriction $y\geq 0$ you can prove that for $n\geq 2$, then $nS = D_n$ where $D_n$ indicates the closed disc centered in the origin with radius $n$, i.e. $\{(x,y)\in\mathbb{R}^2\mid x^2 + y^2 \leq n^2\}$.
Here's the outline of the proof:


*

*The case $n=2$ is a special case:


*

*Prove that $2S \subseteq D_2$.

*Prove that for all $z_1\in D_2$ there is $x_2\in S$ so that $d(z_1,z_2) = 1$, therefore $z_1 - z_2 \in S$. (Consider a circumference centered on $z_1$ with radius $1$, then $z_2$ is any of the intersections of said circumference and $S$).


*Case $n\geq 3$:


*

*Prove that $S + D_{n-1} \subseteq D_n$.

*Prove that for all $z_1\in D_n$ there is $x_2\in D_{n-1}$ so that $d(z_1,z_2) = 1$, therefore $z_1 - z_2 \in S$. (Consider a circumference centered on $z_1$ with radius $1$, then $z_2$ is any of the intersections of said circumference and $D_{n-1}$).
Now we can consider the restriction $y\geq 0$. Just remove the points in each step that require $z_1$ or $z_2$ to have $y\geq 0$ or the $z_2\in D_{n-1}$ that were removed in the last iteration.
