$1$
and vertices numbered clockwise from $1$
to $6.$
She draws segments connecting each pair of vertices that differ by $2$
or $4,$
then cuts the paper along these segments. She does the same with the newly found hexagon and keeps going. If the smallest area she can cut is $\frac{1}{1000000},$
how many hexagons will she cut in total?

Have you tried drawing a diagram?

What is the area of the initial hexagon?

Try finding the side lengths of the new hexagon created.

… by deducing the length of the cuts Alice makes.

How does the ratio of side length between successive hexagons help you find the ratio of successive areas?

How could you express “the smallest area she can cut is
$\frac{1}{1000000}$
” as a mathematical inequality? 
… in terms of
$n,$
where$n$
is the number of cuts Alice makes.

We first find the area of the initial hexagon to be
$\frac{3 \sqrt 3}{2}$
(a diagram may be useful for visualisation).We then find the length of a cut that joins two vertices that differ by
$2$
or$4$
by any suitable method (e.g. using cosine rule). This gives the length of the cuts to be$\sqrt 3,$
from which we can deduce that the side length of the next hexagon is$\frac{\sqrt 3}{3}=\frac{1}{\sqrt 3}.$
Using this side length ratio, we deduce that the ratio of areas between successive hexagons is
$\frac{1}{3}.$
Hence Alice will make$n$
cuts, where$n$
is the largest number of cuts such that$\frac{3\sqrt3}{2}\frac{1}{3^n} \geq 10^{6}.$
We can rearrange this to get$3^{\frac{3}{2}n} \geq 2 \times 10^{6}$
or$n = \lfloor \frac{3}{2} + 6 \log_3 10  \log_3 2\rfloor.$
We can approximate$6 \log_3 10 \approx 12.1$
and$\log_3 2 \approx 0.6$
to get that$n=13.$