I love origami, and it recently gave me an idea for a very hard but beautiful puzzle. I'm really curious whether anyone here can solve it.
So here's the puzzle. You are given a large perfectly square piece of paper with no marks on it. With this square, you have to make a square of exactly one-fifth the area of the original square. You are given no tools such as a ruler or scissors, and all you can do is fold the paper. How do you solve this?
The way to do this is:
- Fold the paper in half along both axes. You've now marked the midpoint of all four sides.
- Fold along the knight's-move diagonals, drawn here:
This creates the red square. All five colored regions have the same area, so the red square is 1/5 the size of the square you started with.
Fold the paper horizontally exactly in the middle; fold each of the two $1\times\frac 1 2$ rectangles diagonally such that the two diagonals are parallel. Rotate the paper by a quarter turn and do exactly the same. The four diagonals you have just created enclose a square of area $\frac 1 5$.
We need to show that the distance between two parallel diagonals is $\frac 1 {\sqrt 5}$. This distance equals the height over the diagonal of one of the large triangles we have created. These triangles have area $\frac 1 4$ while the base length i.e. the length of a diagonal is $\frac {\sqrt 5} 2$. The statement follows immediately.
here is a solution I think using it similarly we can have any desired square fraction.
(imp the long grey line is 1st grey line the relatively shorter one is 2nd grey line.)
1. what we do is get the blue lines first by folding in halves multiple time in this case we get 1/8th division.
2. Take five continuous such division from right edge.
3. fold paper to meet the top right corner of the full square and the point which is the bottom end of the 5th blue line (in the image one blue line overlaps the black which is the 4th blue line ).
4. we get the grey line by joining the "end of 5th blue line" and "one corner ". 5. no we have one triangle with sides x and (5/8)*x;
6. Do a similar operation for the second grey line of triangle(with sides x and (3/8)*x) , this time use the endpoint of the 3rd blue line.
7. fold the top edge of paper to get the x/8 length green line which intersects the first grey line and the right edge of the paper.(could be done easily)
8. the region of green line between the 2 grey line is length x/20. >! 9. fold the right edge to get the red line which passes from the point of intersection of the green line and 2nd grey line.
10. now we have this x/20 length measurement on one side which we can copy 4 times by folding the paper to get x/5 length and then make a square.
Now when we have x/5 lenght we will take x/5 length on one edge lets say right edge and 2x/5 length on top edge(thus these 2 length are perpendicular to each other)
this x/sqrt(5) can be used to create a square of area 1/5 of the larger;
imgur is still slow PS: I made a big mistake earlier and got 1/5 th length the edit now give 1/sqrt(5) length
PS: We can generalize it to get any fraction of area if the fraction can be written as sum of 2 sqaures mean here 5 = 22+11 , also if you are really really hardworking you can actually get any desired fractions , but you have to do these last steps multiple time .
Extending on Deusovi's answer, you can fold a square to any fraction square of the fraction $n^2/(a^2+b^2)$, where $n <= a-b$.
To achieve $1/5$, choose $n=1$, $a=2$, $b=1$.
Split the edges in $a$ equal parts. Then fold lines "knight-moves" $(a,b)$. This will generate $(a-b)^2$ sqaures of size $1/(a^2+b^2)$. Now gather $n^2$ of these to generate the desired fraction.
