tag:blogger.com,1999:blog-6527727667243463517.post7514514838765609783..comments2024-01-15T00:38:25.082-08:00Comments on Fun with num3ers: n circles with radius r ...BenVitalehttp://www.blogger.com/profile/08252218231924085041noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6527727667243463517.post-80244531602864870212011-05-23T04:22:45.431-07:002011-05-23T04:22:45.431-07:00n=6 is an easy one. The radius of small circle is...n=6 is an easy one. The radius of small circle is 1/3. the area of the small hex formed is 2/Srqt[3]. The section of the circles removed is 2/9 * Pi leaving the shaded area as<br />0.4565688<br /><br />Paul.Paulhttps://www.blogger.com/profile/14265011659682148188noreply@blogger.comtag:blogger.com,1999:blog-6527727667243463517.post-75144527090416553822011-05-23T02:44:53.914-07:002011-05-23T02:44:53.914-07:00There are some quite long expressions when n=5. We...There are some quite long expressions when n=5. We use sin 36 degs as Sqrt[10 - 2 Sqrt[5]]/4 to find r which equates to, -5 + 2 Sqrt[5] + Sqrt[50 - 22 Sqrt[5]] or 0.370192.<br /><br />We then have a Pentagon whos side length is 2r and area of, 5 (30 - 14 Sqrt[5] + Sqrt[1885 - 842 Sqrt[5]]) or 0.943111.<br /><br />We have a combines circle area to remove of 3/2 * Pi r^2. (3/2 because the area of 1 circle section is (108/360) and we have 5 of them). This area is quite nasty giving<br />(285 \[Pi])/2 - 63 Sqrt[5] \[Pi] - 15 Sqrt[50 - 22 Sqrt[5]] \[Pi] + <br /> 6 Sqrt[5 (50 - 22 Sqrt[5])] \[Pi]. or 0.64579<br /><br />and finaly the remaining area is<br /><br />0.297316<br /><br />The expresion is <br />5 (30 - 14 Sqrt[5] + Sqrt[1885 - 842 Sqrt[5]]) - <br /> 3/2 (95 - 42 Sqrt[5] + 4 Sqrt[250 - 110 Sqrt[5]] - <br /> 10 Sqrt[50 - 22 Sqrt[5]]) \[Pi].<br /><br />Working on n=6 to follow.<br /><br />PaulPaulhttps://www.blogger.com/profile/14265011659682148188noreply@blogger.comtag:blogger.com,1999:blog-6527727667243463517.post-77628060995441860002011-05-22T15:57:26.963-07:002011-05-22T15:57:26.963-07:00Nice Paul. Now, let's contemplate with n > ...Nice Paul. Now, let's contemplate with n > 4BenVitalehttps://www.blogger.com/profile/08252218231924085041noreply@blogger.comtag:blogger.com,1999:blog-6527727667243463517.post-79213780933419546172011-05-21T15:34:06.979-07:002011-05-21T15:34:06.979-07:00for n=3 the radius of the circles are 2 Sqrt[3]-3...for n=3 the radius of the circles are 2 Sqrt[3]-3 and area of shaded section would be <br />21 sqrt[3] - 36 - 21/2 Pi + 6 sqrt[3] Pi which is approx 0.0347327.<br /><br />For n=4 the radius would be sqrt[2] - 1 and the shaded secion works out at 12 - 8 sqrt[2] + 2 sqrt[3]* Pi - 3 * Pi which is approx 0.147279<br /><br />PaulPaulhttps://www.blogger.com/profile/14265011659682148188noreply@blogger.com