tag:blogger.com,1999:blog-6500033298667240354.post4040804942247898477..comments2024-02-19T00:34:12.578-08:00Comments on Always Creative Geometry Problems plus Occasionally Annoying Philosophy: Fox 2078foxeshttp://www.blogger.com/profile/09567328431908997738noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6500033298667240354.post-72182113850614420592009-12-17T12:17:25.370-08:002009-12-17T12:17:25.370-08:00draw lines from point P, parallel to three sides. ...draw lines from point P, parallel to three sides. additional equilaterals will be formed. simple equalities proves the relationship. a nice problems! thanks.<br /><br />-binaryAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-8436782422123414192009-12-15T08:13:14.396-08:002009-12-15T08:13:14.396-08:00I am tryingI am tryingAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-87813726681457846112009-12-15T07:50:30.317-08:002009-12-15T07:50:30.317-08:00Very good! But what about "just drawing line...Very good! But what about "just drawing lines in the sand" ? There is a geometric solution without any equation - or should say with simple additions...8foxeshttps://www.blogger.com/profile/09567328431908997738noreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-52090783229576504622009-12-15T00:10:13.362-08:002009-12-15T00:10:13.362-08:00By Phytagoras theorem it can be prove;
a^2+b^2+c^2...By Phytagoras theorem it can be prove;<br />a^2+b^2+c^2=q^2+r^2+s^2<br />(a^2-q^2)+(b^2-r^2)+(c^2-s^2)=0<br />Let<br />a+q = b+r = s+c = x<br />(a-q)x+(b-r)x+(c-s)x=0<br />a+b+c=q+r+sAnonymousnoreply@blogger.com