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Paper Folding

Problem Statement
Take a rectangular sheet of paper. Fold it in half to make a crease down the center of the sheet from top to bottom. Then, select a point on the sheet and make a crease from the upper right corner to the point; now make a crease from the upper left corner to the point. How would the point be selected so that the triangle formed by the top of the sheet and the two slant creases has the same area as each of the lateral trapezoids?

Problem setup

The goal of this investigation is to select the correct point on the vertical line which is formed by folding the paper in half so that it divides the paper into three equal sections-two trapezoids and one triangle.

Plans to Solve/Investigate the Problem

First I will divide the paper into three equal sections and then use my knowledge of equal fractional parts to begin exploring the idea of equal sections.

Investigation/Exploration of the Problem

I begin by folding the paper into two equal sections. In other words, I formed a perpendicular bisector of the side of the rectangle.



I then bisected each of the rectangles’ sides to divide the side of the paper into four equal sections.  At this point I know that I have to fold the paper in order to connect the bisectors to the opposite corners. In this way, the point where these folds meet is two-thirds down the original fold.



The height of the triangle formed is now 2/3 and the length is 1. Using the formula A=bh/2, the area of the triangle is now 1/3 the area of the paper.


Now the problem turns to determining if the areas of the two trapezoids are equal to two thirds of the paper. You can determine this by looking at the fact that you have divided the paper into two equal sections. Since the right triangle formed by each side of the equilateral triangle is equal to 1/6 of the total area, the rest of the rectangular section must add up to 1/2 the total area. 1/6 +x = 1/2, so x=1/3.


There are some other, really interesting relationships amongst the sections that are created by the folds.  By folding the paper again (by the same process of bisecting the vertical sides and then folding the paper to these points) you can fold the paper into a total of six sections. By doing this, now you can look at both the triangles and the quadrilaterals that are formed and their relationship to the total area of the paper.

Here are some relationships that I found:



Now that I have cut the figure into six equal sections, the square sections are now 1/6 of the total area. The equilateral triangle in the center has a side length that is 1/2 of the length of the larger equilateral triangle. As a result, the area of the smaller triangle is 1/3 the area of the smaller triangle, or 1/9 the total area of the rectangle.

Another relationship is between the trapezoid and the right triangle that is formed. Since the smaller equilateral triangle is 1/9 the total area, the right triangle that is formed by the bisection of the figure yields 1/18 of the total area or 1/6 of each of the three sections.

Extensions of the Problem

The extensions above can be taken even further by dividing the three sections again into 12 and then 24 sections and look at the figures that are created by these divisions.