Minkowski sum of two disks An open disk with radius $r$ centered at $\mathbf{p}$ is $D(\mathbf{p}, r)=\{\mathbf{q} \mid d(\mathbf p, \mathbf q) < r\}$, and the Minkowski sum of two sets $A$ and $B$ is $A \oplus B=\{\mathbf p + \mathbf q \mid \mathbf p \in A, \mathbf q \in B \}$.
How can you show that $D(\mathbf{a}, r_a) \oplus D(\mathbf{b}, r_b) = D(\mathbf{a} + \mathbf{b}, r_a + r_b)$?
Attempt:
\begin{align}
D(\mathbf{a}, r_a) \oplus D(\mathbf{b}, r_b) &=
\{\mathbf p + \mathbf q \mid \mathbf p \in D(\mathbf{a}, r_a), \mathbf q \in D(\mathbf{b}, r_b) \} \\&=
\{\mathbf p + \mathbf q \mid \mathbf p \in \{\mathbf{x} \mid d(\mathbf a, \mathbf x) < r_a\}, \mathbf q \in \{\mathbf{y} \mid d(\mathbf b, \mathbf y) < r_b\} \\&=
\{\mathbf p + \mathbf q \mid d(\mathbf a, \mathbf p) < r_a, d(\mathbf b, \mathbf q) < r_b \}
\end{align}
And here I got stuck. As best as I can tell, now I would need to prove that $$d(\mathbf a, \mathbf p) < r_a, d(\mathbf b, \mathbf q) < r_b \iff d(\mathbf a + \mathbf b, \mathbf p + \mathbf q) < r_a + r_b$$ but this seems false to me. I tried adding the two inequalities together, but that doesn't seem to give me that condition unless $\mathbf a - \mathbf p$ and $\mathbf b - \mathbf q$ are parallel.
 A: First, notice that $D(p,r)= \{ p+ u \mid u \in D(0,r) \}$, hence
$$D(a,r_a)+D(b,r_b)= \{ a+b+u+v \mid u \in D(0,r_a), v \in D(0,r_b)\}.$$
Therefore, it is sufficient to show that $D(0,r_a)+D(0,r_b)=D(0,r_a+r_b)$. But $$\|u+v\| \leq \|u \| + \|v\| <r_a+r_b, \ \text{if} \ u \in D(0,r_a) \ \text{and} \ v \in D(0,r_b),$$
so $D(0,r_a)+D(0,r_b) \subset D(0,r_a+r_b)$. Then,
$$D(0,r_a+r_b)= \{ (r_a+r_b)u \mid u \in D(0,1) \}= \{\underset{\in D(0,r_a)}{\underbrace{r_au}} + \underset{\in D(0,r_b)}{\underbrace{r_bu}} \mid u \in D(0,1)\},$$
so $D(0,r_a+r_b) \subset D(0,r_a)+D(0,r_b)$.
A: To show $A = B$ it is usually the easiest to split into two parts $A \subseteq B$ and $A \supseteq B$.
First, if $p \in D(a,r_a)$ and $q \in D(b,r_b)$ then $p + q \in D(a+b, r_a+r_b)$. This is simple using the triangle inequality: $$|(p+q)-(a+b)| < |p-a| + |q-b|.$$
Secondly, if $u \in D(a+b, r_a + r_b)$, then we need to find a point $v$ such that $v \in D(a,r_a)$ and $u-v \in D(b,r_b)$. A good candidate is to split the distance between $u$ and $a+b$ in ratio $r_a : r_b$, that is (the $-b$ is to adjust for the center of the disk)
$$v = \frac{r_a\cdot u + r_b \cdot (a+b)}{r_a + r_b}-b.$$
Now, 
$$|v-a| = \left|\frac{r_a\cdot u + r_b \cdot (a+b)}{r_a + r_b}-b-a\right|
=  r_a\frac{|u-(a+b)|}{r_a+r_b}\leq r_a$$
and
$$|(u-v)-b| = \left|\frac{r_b\cdot u -r_b\cdot(a+b)}{r_a + r_b}+b-b\right|
=  r_b\frac{|u-(a+b)|}{r_a+r_b}\leq r_b.$$
I hope this helps ;-)
A: I would take the following route. Hopefully it is more intuitive. 
It is clear that $$D(b, r_b)\oplus D(0, r_0)=D(b, r_b+r_0)$$ 
and that
$$a+D(b, r)=D(a+b, r).$$
(Here $a+ \text{"some set"}$ denotes translation). So we write 
$$D(a, r_a)=a+D(0, r_a).$$
Therefore 
\begin{equation}\begin{split}
D(a, r_a)\oplus D(b, r_b) &= [a+D(0, r_a)]\oplus D(b, r_b) \\
&=a+[D(0, r_a)\oplus D(b, r_b)]\\
&=a+ D(b, r_a+r_b) \\
&= D(a+b, r_a+r_b).
\end{split}
\end{equation}
A: How about using triangular inequality ?
$$d(u,w) \leq d(u,v) + d(v, w)$$ 
Hint: use $a+(b-q)$.
A: $D(\mathbf{a},r_a)\oplus D(\mathbf{b},r_b)=\{\mathbf{p}+\mathbf{q}|d(\mathbf{a},\mathbf{p})<r_a,d(\mathbf{b},\mathbf{q})<r_b\}$
$D(\mathbf{a}+\mathbf{b},r_a+r_b)=\{\mathbf{s}|d(\mathbf{a}+\mathbf{b},\mathbf{s})<r_a+r_b\}$
$\forall \mathbf{p} \in D(\mathbf{a},r_a) \forall \mathbf{q}\in D(\mathbf{b},r_b) \exists \mathbf{s}=\mathbf{p}+\mathbf{q} \in D(\mathbf{a}+\mathbf{b},r_a+r_b)$
$d(\mathbf{a},\mathbf{p})<r_a, d(\mathbf{b},\mathbf{q})<r_b \Rightarrow d(\mathbf{a+b},\mathbf{p+q})=d(\mathbf{a+b-q},\mathbf{p}) \leq d(\mathbf{a},\mathbf{p})+d(\mathbf{a},\mathbf{a+b-q})=d(\mathbf{a},\mathbf{p})+d(\mathbf{q},\mathbf{b})<r_a+r_b$
$\forall \mathbf{s} \in D(\mathbf{a}+\mathbf{b},r_a+r_b) \exists \mathbf{p}\in D(\mathbf{a},r_a), \mathbf{q}\in D(\mathbf{b},r_b), \mathbf{s=p+q}$
$\mathbf{p}=\mathbf{a}+(\mathbf{s-(a+b)}) r_a/(r_a+r_b)$,
$\mathbf{q}=\mathbf{b}+(\mathbf{s-(a+b)}) r_b/(r_a+r_b)$
