Reflection to get within convex polygon Let $P$ be a convex polygon, and let $A_1$ be a point on the same plane as $P$. 
Prove that we can find an integer $n$, and points $A_2,A_3,\ldots,A_n$, such that $A_{i+1}$ is a reflection of $A_i$ with respect to some side of $P$, and $A_n$ lies within $P$.
[Source: Hungarian competition problem]
 A: Say that you have your convex polygon $P$ and sides $A_1,A_2,\cdots A_k$. Now, reflect $P$ across side $A_i$ to $P'$. Keep doing that; reflect $P'$ across some new side, etc ,etc. It is easy to see (but you have to do some induction -where you'll also use convexity- to make it rigorous) that in this way you can cover the whole plane. This can actually be used to answer your original question:
See, the point is that if $A$ and $B$ are sides of your original polygon $P$ and $P'=s_B\cdot P$ is the reflection across $B$, then $s_B\cdot A$ is one of the new sides of $P'$. But how can we express $s_{s_B\cdot_A}$, the reflection across this new side? Well $$s_{s_B\cdot_A}=s_B\cdot s_A\cdot s_B$$ This is not difficult to see: First $s_B\cdot s_A\cdot s_B$ is a reflection because it is the composition of an odd number of reflections. Now, all that's left is to check that the fixed points of the new reflection are exactly the line $s_B\cdot A$. Indeed $$s_B\cdot s_A\cdot s_B\big( s_B(a)\big) =s_B\cdot s_A(a)=s_B(a) $$ So $s_B(a)$ is fixed for all $a$ in the line of $A$, so $s_B\cdot A$ is the fixed line of the new reflection and we are done.
That means that reflecting across one of the new lines can be expressed as a composition of reflections across the original sides of the polygon. That is, if we are allowed to reflect across the original sides of $P$, we are really allowed to reflect across any of the new sides of the new $P'$ that the above process describes. 
Then as we can fill all space with some $P'$, we pick one that contains our point and we use this (bigger but allowed) set of reflections to send this $P'$ to our original $P$ and hence our point inside $P$ as well.
