tag:blogger.com,1999:blog-6500033298667240354.post7863850547586420412..comments2024-02-19T00:34:12.578-08:00Comments on Always Creative Geometry Problems plus Occasionally Annoying Philosophy: Fox 3238foxeshttp://www.blogger.com/profile/09567328431908997738noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6500033298667240354.post-80567316951484849812011-01-04T23:00:44.892-08:002011-01-04T23:00:44.892-08:00That can be a conjecture I am planning to ask. Th...That can be a conjecture I am planning to ask. That is: <br />"No regular polygon -other than square- can be drawn on discrete points of the Cartesian plane" <br /><br />which could be much harder to prove.8foxeshttps://www.blogger.com/profile/09567328431908997738noreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-47058487155479537742011-01-02T17:58:29.682-08:002011-01-02T17:58:29.682-08:00I suppose you could generalize this to any simple ...I suppose you could generalize this to any simple polygon with an irrational area.sixnoreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-71083682107869846492011-01-01T22:38:38.116-08:002011-01-01T22:38:38.116-08:00Please also see six' WolframAlpha formulation ...Please also see six' WolframAlpha formulation for this problem below:<br /><br /><br />a^2 + b^2 = (b + c)^2 + d^2 = (a + d)^2 + c^2<br /><br />Wolframalpha seems to confirm that the only integer solutions to this equation are when each variable is 0.<br /><br />http://www.wolframalpha.com/input/?i=a%5E2+%2B+b%5E2+%3D+c%5E2+%2B+d%5E2+%2B+2bc+%2B+b%5E2+%3D+c%5E2+%2B+d%5E2+%2B+2ad+%2B+a%5E28foxeshttps://www.blogger.com/profile/09567328431908997738noreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-43287738913623463372010-12-16T22:37:09.500-08:002010-12-16T22:37:09.500-08:00I can buy this great proof. It makes sense.
Ration...I can buy this great proof. It makes sense.<br />Rational + Irrational = Irrational.<br /><br />Your WolframAlpha formulation also indicates the same fact.<br /><br />This fact seems to deceive perception, doesn't it?<br /><br />There are infinitely-many points and infinitely-many different equilaterals, YET, none fits on three. <br /><br />Amazing!8foxeshttps://www.blogger.com/profile/09567328431908997738noreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-23441553368610414802010-12-16T22:23:53.657-08:002010-12-16T22:23:53.657-08:00I see a simpler proof. Area of equilateral triang...I see a simpler proof. Area of equilateral triangle is s^2*sqrt(3)/4 where s is the length of any side. Since each corner of the triangle is at integer points, the area is irrational.<br /><br />On the other hand, consider the rectangle drawn around the triangle. This area minus the corner areas around the triangle must be the area of the triangle. However, the rectangle area is rational, and the corner areas are rational. This means that the triangle area is rational.<br /><br />Contradiction.sixnoreply@blogger.com