{"id":1308,"date":"2018-02-16T20:24:50","date_gmt":"2018-02-16T20:24:50","guid":{"rendered":"https:\/\/kellyfish.me\/?p=1308"},"modified":"2018-11-04T21:15:56","modified_gmt":"2018-11-04T21:15:56","slug":"origami-math","status":"publish","type":"post","link":"https:\/\/kellyfish.me\/index.php\/2018\/02\/16\/origami-math\/","title":{"rendered":"Origami Math"},"content":{"rendered":"

[latexpage]<\/p>\n

Fun Math for Girls<\/span>
\nBy Kelly Tan<\/span><\/p>\n

\"\"Is origami one of your hobbies? It is a beautiful craft, and you can make so many models from a single sheet of paper. You can learn how to fold very easily, and make gifts for friends.<\/span><\/p>\n

Origami also lets you learn about math. Given a sheet of square paper, do you know how to fold the paper to get $\\frac{1}{3}$ the length? Let assume the length of each side is 1.<\/span><\/p>\n

It is easy to get\u00a0$\\frac{1}{2}$ the length of a side. You simply fold the left edge onto the right edge or the bottom edge onto the top edge. But folding to get $\\frac{1}{3}$ the length requires a bit of geometry and math.<\/span><\/p>\n

\"\"<\/p>\n

First, suppose you already got the point P to be the midpoint of the top edge AB. You do this by folding the left edge onto the right edge, then $AP=PB=\\frac{1}{2}$.<\/span><\/p>\n

Next, you fold the bottom left corner onto point P as shown. There are two triangles $\u2206APS$ and $\u2206BTP$. What can you say about these two triangles?<\/span><\/p>\n

\"\"<\/p>\n

They are both right angle triangles because angles A and B are $90^o$. Also, angle $SPT$ is $90^o$ So angles $APS + BPT = 90^o$. But we also know angles $APS + ASP = 90^o$ because the sum of the angles of a triangle add up to $180^o$. So we must have angles $BPT = ASP$. The last angles of both triangles $\u2206APS$ and $\u2206BTP$ must be equal too, i.e. $APS = BTP$. Therefore, the two triangles are similar.<\/span><\/p>\n

If you look again at the original sheet, you see that AS and SP form the left edge. So $AS + SP = 1$ or $SP = 1 – AS$. Using the Pythagorean Theorem, $AP2 + AS2 = SP2 = \\frac{(1 \u2013 AS)}{2}$. Expanding and simplifying this equations gives $AS = \\frac{(1-AP^2)}{2}$. Using the property that P is the midpoint of the top edge ($AP = \\frac{1}{2}$), we have $AS = \\frac{3}{8}$.<\/span><\/p>\n

Now, using the fact that they are similar triangles, you can find the lengths of the other triangle. We have $\\frac{BT}{AP} = \\frac{BP}{AS}$, and you can substitute $AS = \\frac{3}{8}$ and $AP = BP = \\frac{1}{2}$, which gives $BT=\\frac{2}{3}$. To get $\\frac{1}{3}$ the length of an edge, we simply fold BT into half, which can be done by folding the top edge so that point B is onto point T.<\/span><\/p>\n

Remember we started with P being the midpoint of the top edge, and therefore $AP = BP = \\frac{1}{2}$. In general, if P divides the top edge of the paper into a different fraction, you will get different fraction for BT.<\/span><\/p>\n

Try this with $AP=\\frac{1}{3}$ and $BP=\\frac{2}{3}$. What is the length of BT? Did you get $BT=\\frac{1}{2}$? Now try this with $AP=\\frac{2}{3}$. What did you get for BT?<\/span><\/p>\n

This technique is generally sufficient enough that you can fold many fractions out of a square paper. It is called the Haga’s Theorem.<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

[latexpage] Fun Math for Girls By Kelly Tan Is origami one of your hobbies? It is a beautiful craft, and you can make so many models from a single sheet of paper. You can learn how to fold very easily, and make gifts for friends. Origami also lets you learn about math. Given a sheet […]<\/p>\n","protected":false},"author":1,"featured_media":1309,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[28],"tags":[],"class_list":["post-1308","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-articles"],"jetpack_featured_media_url":"https:\/\/i0.wp.com\/kellyfish.me\/wp-content\/uploads\/2018\/11\/Strawberry-1.jpg?fit=6000%2C4000&ssl=1","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/posts\/1308","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/comments?post=1308"}],"version-history":[{"count":8,"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/posts\/1308\/revisions"}],"predecessor-version":[{"id":1326,"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/posts\/1308\/revisions\/1326"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/media\/1309"}],"wp:attachment":[{"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/media?parent=1308"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/categories?post=1308"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/tags?post=1308"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}