Tukukkk

Alice and Bob started to play a game with a piece of A4 paper. Alice is going to cut that paper as she wants and will give it to Bob, then Bob will fold the paper from anywhere he wants. In the game, Alice wants to maximize the visible area so cut it accordingly, whereas Bob want to minimize the area after folding.

Game 1

Let's call the area of new piece of paper is 1 unit. Bob optimally folds this piece of paper once from anywhere he wants to make the paper having the minimum possible area (included all visible area) after this procedure.

What is the maximum area possible if Bob and Alice play the game optimally?

Game 2

It is the same game but this time we start to consider our original paper dimension. That means let's call our A4 area is 1 unit and with the same rules above; (Alice cuts this A4 paper into something, Bob folds it once)

What is the maximum area possible if Bob and Alice play the game optimally?

Game 3

let's call our cut paper area is 1 unit again. and as you guessed it is the same game but this time Alice can cut our A4 paper from anywhere she wants but she can only cut once and this specific cut has to be a straight cut. (you may take any piece you want, bigger or smaller piece of paper)

What is the maximum area possible if Bob and Alice play the game optimally?

Game 2

I think Bob can always get to at least 0.5 by imagining the original A4 and folding so that the original paper would go exactly in half. No matter how Alice cuts the paper, the remaining area can be no larger than 0.5.




Note that if neither of the halving folds is available to Bob, the area after the cut can be no more than 0.25 which is way worse for us (as Alice) than 0.5.


Alice can get arbitrarily close to 0.5 by nipping an extremely thin slice off one side.

Game 1

Alice could make it arbitrarily near 1:


The line segments should be made very thin and very long. Each line segment halves the width and doubles the length of the previous one, and is rotated by a fixed small angle.

If the a-th line segment intersects (or overlaps) with the b-th where a<b, the folding point must be in line segment b or b-1. That means there could be only 2 line segments intersecting with line segments with a lower or equal number. Each line could have at most 2 intersection points with a convex shape. And we could count both vertices if they are at the endpoints. We also count the case where the folding line crosses at most 2 endpoints and 4 line segments. That's only 32 intersection points. (Actually not this many, but we don't have to prove a smaller number)

In the worst case, it's technically not a intersection point, but two line segments overlaps exactly, or a line segment overlaps with itself. But we could say they could never have an area bigger than the area of one line segment. That's still negligible if there are a large number of line segments.

A bit too late, but constructed differently.

Game 1:

Answer: arbitrarily close to 1
First, Alice should draw a point on the paper sheet. Then another. Then another. After that Alice keeps adding different random points to the sheet, somehow watching out that no 4 points become ends of a symmetrical quadrangle (rectangle, rhombus, equilateral trapezoid or this "kinda-rhombus" with a single symmetry) and that no 3 points get on the same straight line. Seeing how she has infinite options for a point, she can always find such.
Having stopped at N points, now we have such a position, that by folding the sheet we can't get more than 3 points into points (our best options are: 1) have 2 points max on fold, yet connect no more points 2) Have two points connected with fold and one on fold). That's the intention.
Next, Alice draws circles with arbitrarily small radius, centered in these points. The radius should be small enough that we can't even intersect more than three circles while folding the paper, just as with points.
Next step is obvious - we connect the circles with a web infinitely thin lines, so that the sheet is connected, yet it's area is still concentrated at N circles. Than we cut it out carefully.
Since Bob can intersect 3 circles max, the area of resulting sheet will be decreased by around 3/N with all his efforts. Taking arbitrarily large N we can create arbitrarily bad shape for Bob.