Yes. For instance, if $P$ (the initial point) is the reflection of $B$ over $A$, we can reflect $P$ over $A$ to get $P'=B$ and then we colour the midpoint of $AP'$, which is the midpoint of $AB$, red.

Tintarn wrote: Yes. For instance, if $P$ (the initial point) is the reflection of $B$ over $A$, we can reflect $P$ over $A$ to get $P'=B$ and then we colour the midpoint of $AP'$, which is the midpoint of $AB$, red. edited , $P$ should be on segment $AB$ ( Sorry for my mistake )

Let $\overline{AB}=m , \overline{PB}=\beta , \overline{PA}=\alpha ,$ WLOG$\beta>\alpha$

One can see if we are going to colour the midpoint, then there should have been a colored point with the distance of $m$ from $A$ or $B$. We will show that never happens! Let $P_n$ be the point colored in the $n^{th}$ step. Define: $ f(P_n)=min \left( \overline{P_nA},\overline{P_nB} \right) $ Claim: $ f(P_n)\leq \alpha\frac{2^{n-1}-1}{2^{n-1}} + \beta\frac{2^n-1}{2^n} $ Proof with Induction:

. For induction step, suppose for all $i<n, f(P_i)$ holds the condition. Assume $P_n$ has been resulted by reflecting $P_k<n$ wrt $A$ or $B$. We will have the following cases for $P_{k}$:

So the claim was proved for all points coming after finite steps: $$ f(P_n)\leq \alpha\frac{2^{n-1}-1}{2^{n-1}} + \beta\frac{2^n-1}{2^n}<\alpha+\beta=m$$Hence we never color the midpoint of$AB$.

Assume for contradiction, Name the $k^{th}$point:$P_k$. Suppose we coloured the midpoint of $AB$ with the $min steps=i$. Claim: If we colour the midpoint, all previous points should be out of $AB$. it's enough to proof: if $P_{i-j}$ is colourd by reflectiong $j'=i-j\geq0, P_{i-j-1}$ wrt $B$ then: $$\overline{BP_{i-j}}\geq(i-j')\overline{AB},\overline{AP_{i-j}}\geq(i-j'+1),\overline{AB}$$(And same way for $A$) Proof by induction on $j$: $j=1$, WLOG $P_i$ is coloured by reflect $P_{i-1}$ wrt $B$. $$\frac{\overline{BP_{i-1}}}{2}=\frac{AB}{2}\to\overline{BP_{i-1}}=\overline{AB}$$. Hence $P_{i-1}$ lies on $A$ or on the right side of $B$.According to assumption lies on the right side of $B$, $\overline{AB}=\overline{BP_{i-1}},\overline{AP_{i-1}}=2\overline{AB}$ Induction step: Suppose the claim holds for $P_{i-j}$. Now we prove it for $P_{i-j-1}$. WLOG $P_{i-j}$ is coloured by reflect $P_{i-j-1}$ wrt $B$. $$\overline{BP_{i-j}}\geq(i-j')\overline{AB},\overline{BP_{i-j}}=\frac{\overline{BP_{i-j-1}}}{2}$$$$(i-j'\geq2)\to\overline{BP_{i-j-1}}\geq(i-j'+2)\overline{AB}\checkmark$$Now we will have $2$ following cases:

Hence when $0<i=j\to\overline{AB}\leq\overline{AP_0} or \overline{BP_0}$. Therefore $\overline{AP_0}$or $\overline{BP_0}=\overline{AB}$. Contradiction!

Yaghi wrote: There are two distinct Points $A$ and $B$ on a line. We color a point $P$ on segment $AB$, distinct from $A,B$ and midpoint of segment $AB$ to red. In each move , we can reflect one of the red point wrt $A$ or $B$ and color the midpoint of the resulting point and the point we reflected from ( which is one of $A$ or $B$ ) to red. For example , if we choose $P$ and the reflection of $P$ wrt to $A$ is $P'$ , then midpoint of $AP'$ would be red. Is it possible to make the midpoint of $AB$ red after a finite number of moves? The answer is no! Consider A=(0,0) , B=(1,0) Claim decimal of any colored point isn't 0 or 0.5 (Base 10) Prove By Contradiction : it's obvious that the claim is true for first colored point let's say at ith turn we colored a point with 0.5 or 0 decimal and consider k decimal of point that we got from it the new point there are 2 cases Case 1: we did reflected point over A decimal of new point is decimal of k/2 Case 2 : we reflected point over B decimal of new point is decimal of 1- k/2 or k/2-1 which means k=0 or k=1/2 which means i-1> i which is contradiction notice decimal of midpoint of AB is 0,5 thus by claim 1 it won't be colored