Geometrical proof by induction 
Given a segment $AB$ of length $1$, define the set $M$ of points in the following way: it contains the two points $A,B$ and also all points obtained from $A,B$ iterating the following rule: for every pair of points $X,Y$ in $M$, the set $M$ also contains the point $Z$ of the segment $XY$ for which $YZ = 3XZ$. Prove by induction that the set $M$ consists of points $X$ from the segment $AB$ for which the distance from the point $A$ is either $$AX = \dfrac{3k}{4^n} \hspace{3 mm} \text{or} \hspace{3 mm}AX = \dfrac{3k-2}{4^n}$$ where $n,k$ are nonnegative integers.

I am confused by this question since I am not used to doing induction geometrically. Seeing as how we have to iterate each time a new point, I find it hard to formalize an inductive argument to prove this.
 A: Every point on the segment $AB$ has some distance from $A$.
So let's say you have two such points, $X$ and $Y$.
Let $x$ denote the distance of $X$ from $A$,
$y$ the distance of $Y$ from $A$.
Let $D$ be the set of distances from $A$ to each point in the
set $M$ that is constructed in the problem.
Recall the rule in the problem that says that if $X$ and $Y$ are members of
the set $M$, the point $Z$ of the segment $XY$ for which $YZ = 3XZ$ is too.
An equivalent rule is, if $x$ and $y$ are numbers in the set $D$,
then the number $z$ between $x$ and $y$ such that 
$|z - y| = 3|z - x|$ also is in the set $D$.
The theorem you are to prove is that $D$ consists entirely of
numbers of the form $\dfrac{3k}{4^n}$ or of the form 
$\dfrac{3k-2}{4^n}$.
There, now it's not a geometric induction any more. Just arithmetic.
There are still some complications, for example you can't just assume
$x < y$, you can't just assume that one of $x$ and $y$ has the form
$\dfrac{3k}{4^n}$ and the other has the form $\dfrac{3k-2}{4^n}$
(they might both have the same form),
and you can't just use one value of $k$ to write both $x$ and $y$.
You might have to consider several possible cases separately 
for the induction step.
A: Using the same starting point as David K to transform this into something more arithmetic, his expression
$|z - y| = 3|z - x|$ 
is equivalent to $z=\dfrac {3x}4 + \dfrac{y}{4} \quad\forall  x ,y \in D$.
If $y>z>x$,then $|z - y| = 3|z - x|$ is equivalent to
$y-z=3(z-x)$.
If, however, $x>z>y$, then $|z - y| = 3|z - x|$ is equivalent to
$z-y=3(x-z)$. Both are equivalent statements. Solving for $z$,
$z-y=3(x-z)$
$z-y=3x-3z$
$4z=3x+y$
$z=\dfrac{3x}4+\dfrac y4$
$z$ will always take the form of $y$ (either $\dfrac{3k}{4^n}$ or $\dfrac{3k-2}{4^n}$). Modular arithmetic may ease explanation. The numerator is either 0 (first case) or 1 (second case) mod 3. $3x=0\mod 3$ so we need only consider how $y$ affects the form of $z$.
If $x$ and $y$ have the same denominator, then the numerator of $z$ is simply $3x+y$ and $(3x+y)\mod\ 3=y\mod\ 3$.
If the denominator of $y$ is greater than that of $x$, the numerator of $z$ is $(3x)(4^n)+y$ and $(3x)(4^n)+y\mod\ 3=y\mod\ 3$.
If the denominator of $x$ is greater than that of $y$, the numerator of $z$ is $3x+y(4^n)$. In modular arithmetic $(ab)\mod c=[(a\mod c)(b\mod c)]\mod c$. $4^n=1 \mod c \quad\forall n \in \mathbb{Z}$.
Therefore, $3x+y(4^n) \mod\ 3=y\mod\ 3$.
