tag:blogger.com,1999:blog-8784538487783668007.post4412715567855218239..comments2024-02-27T05:48:58.685+00:00Comments on Web Optimisation, Maths and Puzzles: Geometry: A Circle in the corner of a circleDavid Leesehttp://www.blogger.com/profile/15375658779528098646noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-8784538487783668007.post-66055339790777064282022-01-27T18:41:11.442+00:002022-01-27T18:41:11.442+00:00I just draw two circles and the difference is abou...I just draw two circles and the difference is about 5,9 times bigger diameter on the big compared to the smaler one. 5.9 : 1Anonymoushttps://www.blogger.com/profile/15439631054589513144noreply@blogger.comtag:blogger.com,1999:blog-8784538487783668007.post-4131376235145770792022-01-26T23:07:43.150+00:002022-01-26T23:07:43.150+00:00I say to my student: Always draw the task first. M...I say to my student: Always draw the task first. Messure ecsact the little and the big cirkle.Its easy to se that about 10% is very wrong. He should start all over again when he got that answer. He should compare the answer with the reality. Anonymoushttps://www.blogger.com/profile/15439631054589513144noreply@blogger.comtag:blogger.com,1999:blog-8784538487783668007.post-73560826778200065112021-09-18T17:15:03.632+01:002021-09-18T17:15:03.632+01:00What happens in 3-D? N-D? Bear in mind that the le...What happens in 3-D? N-D? Bear in mind that the length of the longest diagonal of an 'N-cube', ie a 4-D cube, 5-D cube etc. is given by the square root of the dimensionality. (So a 4D hypercube has a longest diagonal of exactly 2 units, always a bit of a surprise!). You'll find that although the diameter of the big circle and its higher dimensional remains constant at 2 units, that ofAnonymoushttps://www.blogger.com/profile/16982274895669187483noreply@blogger.comtag:blogger.com,1999:blog-8784538487783668007.post-65848846711728772382018-03-06T12:44:52.260+00:002018-03-06T12:44:52.260+00:00Thanks, John - I will revisit my calculations and ...Thanks, John - I will revisit my calculations and check back with you!David Leesehttps://www.blogger.com/profile/15375658779528098646noreply@blogger.comtag:blogger.com,1999:blog-8784538487783668007.post-66057167886587448062018-03-05T02:17:04.859+00:002018-03-05T02:17:04.859+00:00Hi David, I get 0.343... too. Thanks for sharing t...Hi David, I get 0.343... too. Thanks for sharing the problem. it's a good one.John (Transum Mathematics)https://www.blogger.com/profile/05661667198384649519noreply@blogger.comtag:blogger.com,1999:blog-8784538487783668007.post-87439989298565434372015-12-27T11:04:08.064+00:002015-12-27T11:04:08.064+00:00Hi David - I get the result being 0.343... not 0....Hi David - I get the result being 0.343... not 0.216 - so not so small as you thought?Anonymoushttps://www.blogger.com/profile/13320832879992045155noreply@blogger.comtag:blogger.com,1999:blog-8784538487783668007.post-31418199287564192032015-08-17T15:01:08.033+01:002015-08-17T15:01:08.033+01:00I like it. Short, simple and straightforward - I ...I like it. Short, simple and straightforward - I like the way you've broken down the lengths of the diagonal instead of comparing their ratios.David Leesehttps://www.blogger.com/profile/15375658779528098646noreply@blogger.comtag:blogger.com,1999:blog-8784538487783668007.post-18841005427119616832015-08-11T15:11:48.308+01:002015-08-11T15:11:48.308+01:00Nice example, but isn't the solution overtly c...Nice example, but isn't the solution overtly complex? If we denote the small radii with r, all it takes is to notice that AC = r*sqrt(2), CD = r, DE = 1, AE = sqrt(2). Therefore, on one hand we have AD = AE - DE = sqrt(2) - 1, and on the other: AD = AC + CD = r(sqrt(2)+1). So r(sqrt(2)+1) = sqrt(2)-1 and r = (sqrt(2)-1)/(sqrt(2)+1).Jasminahttp://www.mathworks.co.uknoreply@blogger.comtag:blogger.com,1999:blog-8784538487783668007.post-65024761107999192822015-06-17T12:30:57.351+01:002015-06-17T12:30:57.351+01:00Nice post, and an an elegant solution.
I can see ...Nice post, and an an elegant solution.<br /><br />I can see another way to work this out which is quite fun. If you first imagine putting an identical circle in the upper right corner of the unit square, then adding two smaller circles in the remaining spaces in these corners, then two more, and two more, and so on, the infinite sum of the diameters of all these ever decreasing circles should be Thomas Oléron Evanshttp://www.mathistopheles.co.uknoreply@blogger.com