Given $2n+1$ fixed points: $A_1, A_2,....,A_{2n+1}$ and point $P$. Let $B_1$ is the reflection of $P$ in $A_1$, $B_2$ is the reflection of $B_1$ in $A_2$,...., $B_{2n+1}$ is the reflection of $B_{2n}$ in $A_{2n+1}$.
My question: Could You show that midpoints of $PB_{2n+1}$ is fixed point when $P$ is moved.
The reflection of a point $p$ in a point $u$ is given (if we identify points with vectors) by $2u-p$. So the composition of two such reflections, say about points $u$ and $v$, is given by the translation $p\mapsto2v-(2u-p)=p+2(v-u)$. By induction, the composition of reflections in $u_1,v_1,u_2,v_2,\dots u_n,v_n$ is the translation $p\mapsto p+2(V-U)$ where $V$ and $U$ are the sums of the $v_i$'s and $u_i$'s, respectively. Composing with one last reflection in a point $w$, we get $p\mapsto 2w-p-2(V-U)$. The midpoint between the initial point $p$ and the result of all these reflections is therefore $w-(V-U)=w+U-V$, which is independent of $p$.
We describe the $B_i$'s vectorially. For simplicity I write $AB$ instead of $\vec{AB}$
$$ OB_1=OP+2PA_1= OP+2(OA_1-OP)=2OA_1-OP$$
$$OB_2= OB_1+2(OA_2-OB_1)= 2OA_2-OB_1=2OA_2 -2OA_1+OP $$
$$OB_3=OB_2+ 2(OA_3-OB_2)= 2OA_3-OB_2=2OA_3-2OA_2+2OA_1-OP.$$
The midpoint $M$ of $PB_3$ is given by
$$OM=\frac{1}{2}(OP+OB_3)=OA_3-OA_2+OA_1. $$
Same computation works for any odd number of $A$'s.
