Locus of vector endpoint

Question

Let u and v be two vectors, with their starting point at the center O of the square ABCD, and their endpoints moving along the sides of the square. Determine the locus of the endpoint of u + v!

This is what I got so far:

$u =\begin{bmatrix} u_1 \\ u_2 \end{bmatrix}$ and $v = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}$, then $u + v = \begin{bmatrix} u_1 + v_1 \\ u_2 + v_2 \end{bmatrix}$.

Now let's take

$A = (-2a, 2a), B = (-2a, 0), C = (0, 0), D = (0, 2a)$.

Then, the coordinates of the O center is $O=(-a,a)$

My intuition tells me that the locus of the $u+v$ vector endpoint is going to be a circle.

How can I prove this?

Answer 1

No, this set is a (filled) square (blue line in the figure below) with sidelength doubled with respect to the initial square (in black). It is called the Minkowski addition of the two shapes. It is known that the Minkowski sum of two polygonal shapes is a polygonal shape.

More precisely, the Minkowski sum of two shapes $A \oplus B$ can be thought as the shape resulting of the sweeping (see here) of the center of shape $A$ along shape $B$, which indeed gives, in the present case, a filled square with a doubled side.

Here is a graphical representation with 5000 instances of the vector's sum (red points) covering the (big) square.

Here is the Matlab program that has generated the figure (please note how function $f$ encodes - with complex numbers - the initial square)

clear all;close all;axis equal;
f=@(t)(exp(i*t)./(abs(cos(t))+abs(sin(t)))); % square
for k=1:5000;
   t1=2*pi*rand;t2=2*pi*rand;
   T(k)=f(t1)+f(t2); % vector sum
end;
plot(T,'.r');
t=0:0.01:2*pi;
plot(f(t),'k'); % black square
plot(2*f(t),'b'); % blue square

Answer 2

Jean Marie has pointed out the answer. What follows is a proof using elementary concepts.

For simplicity, let the coordinates of the original square $ABCD$ be $A(1,-1), B(1,1), C(-1,1)$ and $D(-1,-1)$.

Then our vectors have the form: $(1, \beta), (-1, \beta), (\alpha, 1),$ and $(\alpha, -1)$, where $-1 \leq \alpha, \beta \leq 1.$

Let $S$ be our set of vectors with endpoints on square $ABCD.$

Let $T$ be the set of vectors with end point on or inside square $EFGH$ as shown in the figure.

Then $$T= \{ (x,y): -2 \leq x, y \leq 2. \}$$

We are going to prove that the sum of any $2$ vectors of $S$ is a vector in $T$, and conversely any vector in $T$ is the sum of $2$ vectors in $S$.

Proof: $ \Rightarrow$ direction

Vectors of $S$ have the form: $(1, \beta), (-1, \beta), (\alpha, 1),$ and $(\alpha, -1)$, where $-1 \leq \alpha, \beta \leq 1.$

$$\therefore (x_1,y_1), (x_2,y_2) \in S \implies -1 \leq x_1, y_1, x_2, y_2 \leq 1.$$ This implies that$$-2 \leq x_1+y_1, x_2+y_2 \leq 2.$$ Hence $$(x_1,y_1)+(x_2,y_2)=(x_1+y_1, x_2+y_2) \in T.$$

Thus the sum of any $2$ vectors in $S$ belongs to $T$.

$$$$ $\Leftarrow$ direction

Conversely, for any vectors $(x,y) \in T$, we consider $4$ different cases, corresonding to the $4$ quadrants in which $(x, y)$ lies.

Case $(1)$ $(x,y)$ lies in quadrant $1$.

Then $$0 \leq x, y \leq 2 \implies -1 \leq x-1, y-1 \leq 1,$$ $$(1,y-1), (x-1,1) \in S$$ and $$(1,y-1 )+(x-1,1)=(x,y).$$ Thus $(x,y)$ is the sum of $2$ vectors in $S$. $$$$ Case $(2)$ $(x,y)$ lies in quadrant $2$.

Then $-2 \leq x \leq 0$ and $0 \leq y \leq 2$

This implies that $-1 \leq 1+x \leq 1$ and $-1 \leq y-1 \leq 1$ and hence $(1+x,1), (-1, y-1) \in S$.

$$\because (1+x,1)+(-1, y-1)=(x,y),$$ $\therefore (x,y)$ is the sum of 2 vectors in $S$.

Cases in other quadrants are similar.