A pentagon using less than 5 lines

Question

A pen and paper challenge: Draw a pentagon on a piece of paper

• only draw straight lines
• draw less than 5 times!
• The pentagon has to be fully drawn in ink, on the paper in unfolded state.

I am not looking for creative answers, e.g.:

• do not move the paper while drawing a line
• ink does not leak though a next layer of the paper (if it is folded)!

The ideal answer:

• Shows how it is done
• Uses the minimum amount of lines possible
• Proves that is minimal

as bonus: Also tells/explains the minimum for an octagon

If a solution with a hole gives a better answer; I'd like both answers (i.e. with and without).

Answer 1

After proving that 3 strokes are necessary, here is a solution in 2 strokes!

Answer 2

Label the corners of the paper ABCD in clockwise order. Draw a line from the midpoint of AB (label it E) to the midpoint of BC (label it F). Then AEFCD is a pentagon.

Answer 3

Method:

It seems that we have a paper of finite size & folding is allowed.
We can fold the paper like D1 where the shaded area gets hidden:
When we make a single stroke with blue in D2, Perpendicular to the fold, we get two blue lines with a gap in the hidden area shown in D3, Perpendicular to the fold lines.
By changing the angle of the Stroke, we can make the two blue lines at some angle like in D4, where the 2 angles marked in Black are Equal.
By changing the angle of the folding, we get Blue lines which are not co-linear when unfolded , like in D5.
By changing both the angle of the Stroke & the angle of the folding, we get the 2 Green lines in D8.

Pentagon with 4 lines:

We can fold the paper like D5 where the shaded area gets hidden:
When we make single stroke with blue in 2, we get two blue lines with a gap in the Shaded hidden area.
Draw 3 Orange lines arbitrarily to get the Pentagon in D6.

Pentagon with 3 lines:

With one Stroke, we get 2 Blue lines in D7.
By the other fold indicated by the other Purple lines, we get 2 Green lines.
Lastly, use one Orange stroke to finish the Pentagon with 3 strokes.

OPTIMALITY:

! We can not draw Pentagon with 2 strokes in this method.
We can only get alternate lines with a Stroke but there are only 2 Pairs of alternate lines here. The last line will require one Stroke.
! Total 3 is the Minimum.

Octagon with same technique:

With 4 strokes : We can draw the Octagon with 4 strokes where each stroke gives 2 lines : Very Easy foldings !
With 3 strokes : We can get 3 alternate lines with 1 stroke by using 2 foldings at the same time : Moderately Complex foldings !
We can then get 3 more lines with 1 stroke again using 2 foldings at the same time.
We finish with the last stroke to get the last 2 lines. This Octagon uses 3 strokes.
With 2 strokes: It is Possible to get 4 alternate lines of Octagon with 1 stroke : Most Complex foldings !
We then get 4 more interleaving alternate lines with 1 stroke.
In this case we can draw the Octagon with 2 strokes.
This technique will not allow 1 Stroke.

Answer 4

Let me do the octogon (aka octagon)

Plan the octagon

Fold the paper and draw a first line.

Unfold

Fold again and draw the second line.

Note only the 3 middle segments were drawn.

Unfold and draw the third line.

The octagon is complete.

Minimality:

What is certain is that you cannot do in one line. Because you would have to "unfold" the polygon which you cannot do without tearing it. I assume tearing or cutting the paper is not allowed.

In two lines, it would actually be possible if you have a star-shaped hole in the paper. You could draw a polygon joining the points of the star and it could be stretched to a straight line, one half at a time. But I assume a hole inside the paper is not allowed.

My proof for the pentagon relied on the fact that you cannot draw two adjacent segments in one stroke because you cannot straighten the angle between the segments without tearing the paper. You cannot separate the pentagon's sides in 2 sets without any set including two adjacent sides. Therefore 3 lines are necessary.

But it doesn't work because you can split each side of the pentagon in two and get 10 segments. You can then separate these segments in two sets of 5, with no adjacent segments in either set. So the proof doesn't hold.

But let's say you have divided your polygon into a loop of segments alternating between two colors. The colors represent the 2 strokes you will draw. Suppose somehow you manage to straighten them on a line to draw all segments of one color. The loop must somehow go from one end of the line to the other and back. It must cover the distance twice. On the other side, hte length you can draw is only once the length of the line. That means the total length you can draw in one stroke cannot exceed half of the loop length. To draw all segments of the loop in two stokes, each stroke must draw exactly half of the loop length.
In other words, the line has exactly twice the length of one stroke. In that situation, when you draw the line, the whole loop must be aligned with the line you are drawing. It has no slack to do otherwise. When the loop is straightened, any 2 segments that have an angle in the original polygon, must be straightened, except the two and the ends, but this is not possible without tearing the paper.

Conclusion: It is not possible to draw any polygon in 2 strokes.
It is possible however if you allow to cut the paper or if you allow a hole in the paper.

Answer 5

This may count as too much of a "creative answer" for OP's liking, but…

One Line (and a very blobby pen)

This method works best with something like a fibre-tip or felt-tip pen, and not a ballpoint pen or fountain pen (which rely on pressure/friction between the pen and the paper)

Here is the unfolded net of the paper:

Here is the paper, folded so that all of the edges of the pentagon lie on the same edge of the result: (Note that one of the lines is folded in half, back on itself)

Squeezed together with a peg and a paperclip, to bunch everything up as much as possible, a single thick line is drawn on the edge of the folded paper:

The paper is then unfolded: