{"id":5652,"date":"2013-12-31T08:16:23","date_gmt":"2013-12-31T13:16:23","guid":{"rendered":"http:\/\/spinor.info\/weblog\/?p=5652"},"modified":"2014-01-08T14:57:01","modified_gmt":"2014-01-08T19:57:01","slug":"more-pointless-mathematics","status":"publish","type":"post","link":"https:\/\/spinor.info\/weblog\/?p=5652","title":{"rendered":"More pointless mathematics"},"content":{"rendered":"

So the other day, I solved this curious mathematics puzzle using repeated applications of Pythagoras’s theorem and a little bit of algebra.<\/p>\n

Now I realize that there is a much simpler form of the proof.<\/p>\n

The exercise was to prove that, given two semicircles drawn into a bigger circle as shown below, the sum of the areas of the semicircles is exactly half that of the larger circle.<\/p>\n

\"\"<\/p>\n

Again, I’m inserting a few blank lines before presenting my proof.<\/p>\n

 <\/p>\n

 <\/p>\n

 <\/p>\n

 <\/p>\n

 <\/p>\n

 <\/p>\n

 <\/p>\n

 <\/p>\n

 <\/p>\n

 <\/p>\n

 <\/p>\n

 <\/p>\n

 <\/p>\n

 <\/p>\n

\"\"<\/p>\n

Once again I am labeling some vertices in the diagram for easy reference.<\/p>\n

Our goal is to prove that the area of a circle with radius AO<\/em><\/span> is twice the sum of the areas of two semicircles, with radii AC<\/em><\/span> and BD<\/em><\/span>. But that is the same as proving that the area of a circle with radius AO<\/em><\/span> is equal to the sum of the areas of two circles, with radii AC<\/em><\/span> and BD<\/em><\/span>.<\/p>\n

The ACO<\/em><<\/span> angle is a right angle. Therefore, the area of a circle with radius AO<\/em><\/span> is the sum of the areas of circles with radii AC<\/em><\/span> and CO<\/em><\/span>. (To see this, just multiply the theorem of Pythagoras by \u03c0<\/span>.) So if only we could prove that CO<\/em> = BD<\/em><\/span>, our proof would be complete.<\/p>\n

Since AO<\/em> = BO<\/em><\/span>, they are the sides of the isosceles triangle ABO<\/em><\/span>. Now if we were to pick a point O<\/em>‘<\/span> on the line CD<\/em><\/span> such that CO<\/em>‘ = BD<\/em><\/span>, the ACO<\/em>‘<\/span> and O<\/em>‘DB<\/em><\/span> triangles will be identical (CD<\/em><\/span> being the sum of AC<\/em><\/span> and BD<\/em><\/span> by construction). Therefore, AO<\/em>‘ = BO<\/em>‘<\/span>, and the ABO<\/em>‘<\/span> triangle would be another isosceles triangle with its third vertex on the CD<\/em><\/span> line. Clearly that is not possible, so O<\/em> = O<\/em>‘<\/span>, and therefore, CO<\/em> = BD<\/em><\/span>. This concludes the proof.<\/p>\n<\/fb:like>","protected":false},"excerpt":{"rendered":"

So the other day, I solved this curious mathematics puzzle using repeated applications of Pythagoras’s theorem and a little bit of algebra. Now I realize that there is a much simpler form of the proof. The exercise was to prove that, given two semicircles drawn into a bigger circle as shown below, the sum of […]<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[30],"tags":[],"class_list":["post-5652","post","type-post","status-publish","format-standard","hentry","category-mathematics","category-30-id","post-seq-1","post-parity-odd","meta-position-corners","fix"],"_links":{"self":[{"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=\/wp\/v2\/posts\/5652","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=5652"}],"version-history":[{"count":7,"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=\/wp\/v2\/posts\/5652\/revisions"}],"predecessor-version":[{"id":5698,"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=\/wp\/v2\/posts\/5652\/revisions\/5698"}],"wp:attachment":[{"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=5652"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=5652"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=5652"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}