# Is it always possible to cut out a piece of the square with $\frac{1}{5}$ of its 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.

• It's probably worth including the solutions for $n=2,3$. Commented Nov 30, 2022 at 21:43
• 2: Make four cuts with slope 2 and -1/2 and take the piece in the middle. Commented Nov 30, 2022 at 21:50
• 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. Commented Nov 30, 2022 at 21:51
• I assume you mean a connected piece, otherwise we can probably reduce to the cases $$n=2,3$$. Commented Dec 1, 2022 at 2:03
• I mean a connected piece, but how can we reduce to those cases if disconnected pieces are allowed? Commented Dec 1, 2022 at 2:06

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 $$t and we could have simply chosen $$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.

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 $$(14,23).$$

• Have you found any kind of pattern to these? Commented Dec 13, 2022 at 19:28
• I brute-forced it by using lines with slope $2/5,$ just like Penguino did. Commented Dec 13, 2022 at 19:31

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 $$n.$$