My good friend Padyta asked me if it was possible to obtain a rectangular angle from an arbitrary piece of paper. If we remember that a square is possible to be constructed from that angle, her question have a great importance.
To build a perpendicular line to a given one in two-dimension geometry demands the use of ruler and compass and it is not an easy thing. One way is tracing a circumference of an arbitrary radius, centered in an starting point A, then a second circumference centered on the intersection of the line and the first circumference (point B), the intersection of both circumferences gives us a point C, equidistant to both corners A and B, a new circumference of the same radius centered on C and its intersection with the first circumference gives us a point D and finally the intersection of a circumference centered on D with the one centered on C gives us the point E, perpendicular to the line AB.
However, the "origeometry" allows us to use the paper itself and gives us a third dimension and increas our capabilities. So, if we have a given straight line, folding the paper aligning the line on itself we can get a perpendicular line.
Now that we have our rectangular angle, is it possible to construct a perfect square from it? The answer, luckily, it's a yes. Here it comes a way.
First you need to fold a 45° line between both lines, that will allow us to find the main diagonal passing through the next corner of the square. To do it, we fold aligning both lines on their intersecton point.
Then we align the small diagonal on itself folding a diagonal that passes over the corner of the square. A good choice of that corner will allow us to optimize the use of the paper to get the biggest size possible.
We can cut the main angle and the sides of the square. As we saw on the previous post we can obtain the opposite angle and complete the square by folding the main diagonal aligning it on itself, passing through the original corner.
It's pendant to btain a method to optimize the size of the obtained square, maximizing its area on the given paper. Many regards.
To build a perpendicular line to a given one in two-dimension geometry demands the use of ruler and compass and it is not an easy thing. One way is tracing a circumference of an arbitrary radius, centered in an starting point A, then a second circumference centered on the intersection of the line and the first circumference (point B), the intersection of both circumferences gives us a point C, equidistant to both corners A and B, a new circumference of the same radius centered on C and its intersection with the first circumference gives us a point D and finally the intersection of a circumference centered on D with the one centered on C gives us the point E, perpendicular to the line AB.
However, the "origeometry" allows us to use the paper itself and gives us a third dimension and increas our capabilities. So, if we have a given straight line, folding the paper aligning the line on itself we can get a perpendicular line.
Now that we have our rectangular angle, is it possible to construct a perfect square from it? The answer, luckily, it's a yes. Here it comes a way.
First you need to fold a 45° line between both lines, that will allow us to find the main diagonal passing through the next corner of the square. To do it, we fold aligning both lines on their intersecton point.
Then we align the small diagonal on itself folding a diagonal that passes over the corner of the square. A good choice of that corner will allow us to optimize the use of the paper to get the biggest size possible.
We can cut the main angle and the sides of the square. As we saw on the previous post we can obtain the opposite angle and complete the square by folding the main diagonal aligning it on itself, passing through the original corner.
It's pendant to btain a method to optimize the size of the obtained square, maximizing its area on the given paper. Many regards.
