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{"id":442,"date":"2020-11-27T17:24:17","date_gmt":"2020-11-27T17:24:17","guid":{"rendered":"https:\/\/www.foldworks.net\/?page_id=442"},"modified":"2020-11-30T12:23:54","modified_gmt":"2020-11-30T12:23:54","slug":"2-why-does-a-paper-work","status":"publish","type":"page","link":"https:\/\/www.foldworks.net\/home\/articles\/the-mathematics-behind-the-folds\/2-why-does-a-paper-work\/","title":{"rendered":"2. Why does \u201cA\u201d paper work?"},"content":{"rendered":"

The problem from the first part of this series was to prove the method of folding the diagonal of a rectangle. Figure 2-1 shows the result of the first step. The diagonal of the rectangle is the line that goes through two opposite corners, and therefor the centre. Folding together the rectangle\u2019s other opposite corners makes an isosceles triangle \u2013 the two equal sides are shown in heavy orange line. When we fold the pinch marks together (shown with filled circles), we find the centre of the triangle\u2019s base, which is the same as the centre of the rectangle. Also, we have constructed the perpendicular bisector which goes through the apex of the isosceles triangle, which coincides with two corners of the rectangle. We have therefore made a fold through the centre of the rectangle and one corner \u2013 hence folded the diagonal. Can you think of a better proof?<\/p>\n

Whilst working on this I asked myself what the obtuse angle at the top left was \u2013 how do it relate to the right\u2013angled triangle in the rectangle? Can you work out the relationship?<\/p>\n

\"\"<\/a>
Figure 2-1 The key step of folding the diagonal of a rectangle exactly<\/figcaption><\/figure>\n

We now move onto some mathematics of ISO \u201cA\u201d paper. For many folders \u201cA\u201d paper is common and convenient for folding. We use the term \u201cA\u201d paper interchangeably with \u201csilver rectangle\u201d because usually the proportions<\/i> are important, not the specific size of the rectangle.<\/p>\n

Why is \u201cA\u201d paper the proportion that it is? Take two silver rectangles. Fold one in half, rotate it 90 degrees and place it on the second rectangle, aligning the bottom left corners. You will see that the diagonals are also aligned. This means that the rectangles have the same proportions (figure 2-3).<\/p>\n

\"\"<\/a>
Figure 2-2 Taking two sheets of A4, fold one in half and align corners to show that the two rectangles have the same proportions<\/figcaption><\/figure>\n
\"\"<\/a>
Figure 2-3 Extension of the previous method: after folding in half, cut in half and repeat the process<\/figcaption><\/figure>\n

Only silver rectangles behave like this. If you cut the second sheet in half, you can carry on and make a very convincing demonstration <\/i>(figure 2-3). But what about a proof<\/i>?<\/p>\n

I shall attempt a proof that illuminates and generalises to other cases. There are four points to note:<\/p>\n

    \n
  1. \n

    To scale any rectangle into a new rectangle with the same proportions we multiply all edge lengths by the same number.<\/p>\n

      \n
    1. \n

      To double the area of a rectangle, we multiply the short sides by and also the long sides by.<\/p>\n<\/li>\n

    2. \n

      We can also double the area by multiplying by twice in the same direction<\/i>. However the rectangles will not necessarily be of the same proportion.<\/p>\n<\/li>\n<\/ol>\n<\/li>\n

    3. \n

      If we take a square and multiply<\/i> one edge by 5, the result is of the same proportions as result of dividing<\/i> one edge of the square by 5. (Of course the number 5 can be replaced by any non-zero number.)<\/p>\n<\/li>\n

    4. \n

      So let\u2019s take a square and make a bigger rectangle by multiplying horizontally by. We make a second smaller rectangle by dividing by. These two rectangles are the same proportion (point 3) and one is double the area of the other (point 2b).<\/p>\n<\/li>\n<\/ol>\n

      \u0007<\/span><\/h1>\n
      \"\"<\/a>
      Figure 2-4 Illustration of point 4 \u2013 the square (of white and pale blue) can be transformed into the white rectangle and also the whole rectangle<\/figcaption><\/figure>\n

      \b<\/span><\/h1>\n

      Figure 2\u20114 Illustration of point 4 \u2013 the square (of white and pale blue) can be transformed into the white rectangle and also the whole rectangle<\/i><\/p>\n

      Do you that think that was a lot work of to prove something you thought was \u201cobvious\u201d? Of course there are many other proofs. Here\u2019s an algebraic one:<\/p>\n

      \"\"<\/a>
      Figure 2-5 A rectangle of proportion a:2. The height is a and the length 2.<\/figcaption><\/figure>\n

      The smaller rectangle is of proportion 1:a. The larger rectangle is of proportion a:2. Proportions are equal so<\/p>\n

      Take the square roots of both sides and out pops the answerbut I would suggest that this proof leaves a feeling of mystery.<\/p>\n

      The power of algebra is that it lets us easily generalise. For example, what are the proportions of a rectangle such that when cut into three<\/i> equal rectangles, the new rectangles are of the same proportion? What about n<\/i> equal parts?<\/p>\n

      .<\/p>\n

      For n=4, we see that a=2 and therefore see that a 1:2rectangle contains four 2:4 rectangles. Did you know that already? If not, this shows the power of mathematics. If you did, were you aware of how self-similarity could be generalised?<\/p>\n

      A silver rectangle has proportions ; a bronze rectangle .So what is a silver triangle<\/i>? What is a bronze triangle<\/i>? How far can you generalise? We\u2019ll look at this in the next part.<\/p>\n","protected":false},"excerpt":{"rendered":"

      The problem from the first part of this series was to prove the method of folding the diagonal of a rectangle. Figure 2-1 shows the result of the first step. The diagonal of the rectangle is the line that goes through two opposite corners, and therefor the centre. Folding together the rectangle\u2019s other opposite corners […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":438,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/P9nCEd-78","_links":{"self":[{"href":"https:\/\/www.foldworks.net\/wp-json\/wp\/v2\/pages\/442"}],"collection":[{"href":"https:\/\/www.foldworks.net\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.foldworks.net\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.foldworks.net\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.foldworks.net\/wp-json\/wp\/v2\/comments?post=442"}],"version-history":[{"count":3,"href":"https:\/\/www.foldworks.net\/wp-json\/wp\/v2\/pages\/442\/revisions"}],"predecessor-version":[{"id":472,"href":"https:\/\/www.foldworks.net\/wp-json\/wp\/v2\/pages\/442\/revisions\/472"}],"up":[{"embeddable":true,"href":"https:\/\/www.foldworks.net\/wp-json\/wp\/v2\/pages\/438"}],"wp:attachment":[{"href":"https:\/\/www.foldworks.net\/wp-json\/wp\/v2\/media?parent=442"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}