Is it always possible to cut out a piece of the square with $\frac{1}{5}$ the area? Let there be a square that has $n+1$  notches on each edge (corners included)  to divide each edge into $n$ equal parts. We can make cuts on the square from notch to notch. Is it always possible to cut out a connected piece with area $\frac{1}{5}$ the area of the original square if $n≥2?$ It is possible for multiples of $2, 3,$ or $5,$ but I don't know any other numbers for which this is possible. If it is possible for $n,$ it is possible for any multiple of $n.$
$2$: Make four cuts with slope $2$ and $-1/2$ and take the piece in the middle.
$3$: Make four cuts from a corner with slope $3$ and $-1/3$ and use a corner-to-corner cut to cut the center piece into two pieces, each with area $1/5.$
$5$: It's obvious.
Here's a link to the sequel.
 A: Yes, it is always possible.
Let $k$ be the smallest positive integer such that $20k^2 > n^2$. We will start with a $2k\times 2k$ axis-aligned square, which is a little larger than we need. From there, we'll need to remove a total area of $A/5$, where $A$ is a nonnegative integer equal to at most $5(4k^2-(2k-1)^2) = 20k-5$.
Our plan for removing this area is to make a single cut of slope $s/5$ for $0\le s < 15$ connecting the north and west sides of the square (as in Penguino's diagram for $n=11$), and translate our starting square along the north edge of the square until we have a $1\times s/5$ triangle. The remaining three cuts, with slope $\pm 1$, can be made at whatever integer depth we want, and will safely align with the notches no matter where they go.
Suppose the depth of the other three cuts are $a, b,$ and $c$ respectively, each at most $k$ (so that the cuts don't intersect each other). (We also need $k\ge 3$ to make sure they don't intersect the $1\times s/5$ triangle.) Then we want to obtain
$$\frac A5 = \frac{s}{10}+\frac{a^2+b^2+c^2}{2}$$
$$2A = s+5(a^2+b^2+c^2)$$
by choosing $s\in\{0,1,2,3,\ldots,14\}$, $a,b,c\in\{0,1,2,\ldots,k\}$.
First, let's see that all positive integers are expressible in the form $s+5(a^2+b^2+c^2)$ if $a,b,c$ are unrestricted. Given a target $t$, we subtract a value of $s$ with the right congruence class and express the integer $\frac{t-s}{5}$ as the sum of three squares. The only way this can fail is if either $t-s$ is negative, in which case $t<s$ and we could have simply chosen $s=t, a=b=c=0$, or if $\frac{t-s}{5}$ is not expressible as a sum of three squares, ie is of the form $4^a(8b+7)$ by Legendre's three-square theorem. But since no three consecutive integers are of this form, one of our three possible candidates for $0\le s<15$ will succeed.
So, if $2A\le 5k^2$, we'll have no way to solve this equation by choosing any of $a,b,c$ to be larger than $k$, which means that the solution we know exists must do so in our desired ranges for $a,b,c$. Since we know $A\le 20k-5$, it will suffice for
$$20k-5\le 5k^2\iff k^2-4k+1\ge 0 \iff (k-2)^2 \ge 3 \iff k-2 \ge 2 \iff k\ge 4$$
so we need $n$ large enough that $\left\lceil \frac n{\sqrt{20}}\right\rceil \ge 4\iff \frac{n}{\sqrt{20}}>3 \iff n>3\sqrt{20}\approx 13.4$. Since other users have already solved the cases for $n\le 13$, this completes the proof.
A: It is also possible with $n=7$ using three cuts.

and also possible for $n=11$.

My hypothesis is that it is possible for any $n.$
A: I solved $13.$
The shape is cut with the lines connecting $(0,5)$ and $(2,0),$
$(0,4)$ and $(13,4),$
and $(6,0)$ and $(13,4).$
Edit: I solved $17.$
The shape is the cut with the lines connecting $(0,5)$ and $(2,0),$
$(0,4)$ and $(17,4),$
and $(14,0)$ and $(17,4).$
Edit 2: I just hit gold. Here's $19.$
The shape is the cut with the lines connecting $(0,10)$ and $(4,0),$
$(0,13)$ and $(19,13),$
$(7,19)$ and $(19,11),$
and $(0,9)$ and $(19,9).$
Edit 3: ...and $23.$
The shape is the cut with the lines connecting $(0,10)$ and $(4,0),$
$(0,9)$ and $(23,9),$
$(13,0)$ and $(23,10),$
$(0,22)$ and $(22,0),$
and $(14,0)$ and $(14,23).$
