tag:blogger.com,1999:blog-6500033298667240354.post1022592059957511863..comments2024-02-19T00:34:12.578-08:00Comments on Always Creative Geometry Problems plus Occasionally Annoying Philosophy: Fox 2668foxeshttp://www.blogger.com/profile/09567328431908997738noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6500033298667240354.post-35743310334168308572010-03-22T20:55:53.473-07:002010-03-22T20:55:53.473-07:00Correct Migue! There should be at least one more ...Correct Migue! There should be at least one more geometric solution, similar to yours.8foxeshttps://www.blogger.com/profile/09567328431908997738noreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-45595243392335171542010-03-21T15:35:39.569-07:002010-03-21T15:35:39.569-07:00See
http://i40.tinypic.com/2wdyq28.png
MIGUE.See<br /><br />http://i40.tinypic.com/2wdyq28.png<br /><br />MIGUE.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-57807614341306947462010-03-21T10:01:56.562-07:002010-03-21T10:01:56.562-07:00Yep, that is the geometric solution we've rece...Yep, that is the geometric solution we've received from at least two more visitors. There is one more geometric solution which was not found yet.8foxeshttps://www.blogger.com/profile/09567328431908997738noreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-35989675024883646892010-03-21T04:18:40.656-07:002010-03-21T04:18:40.656-07:00Correction: With A as center draw a circle passing...Correction: With A as center draw a circle passing through thru. B & CAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-32226275331737773242010-03-21T03:12:30.939-07:002010-03-21T03:12:30.939-07:00How about this? With A as center draw a circle pas...How about this? With A as center draw a circle passing through A & B. Take any point Q in the larger sector BC. We know that /_BAC=60 deg. and hence /_BQC =30 deg. Thus for any point O on the smaller sector BC, /_BOC will always be 150 deg. since QBOC is concyclic. And for any such point O we always have AB = AO = radius of our circle.<br />AjitAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-9460292529143664532010-03-20T12:48:56.087-07:002010-03-20T12:48:56.087-07:00Joe, your solution is nice too. Although it is ri...Joe, your solution is nice too. Although it is rigorous, it adds colors. We will post geometric solutions in the following days. Let's give a little more chance to others.<br />Thank you.8foxeshttps://www.blogger.com/profile/09567328431908997738noreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-77557960091971947852010-03-19T23:19:38.295-07:002010-03-19T23:19:38.295-07:00I tried finding a simplle plane geometry proof but...I tried finding a simplle plane geometry proof but failed to do so. While that elegant answer will come later, here's my trigonometric/analytical solution: If we assume B to be the origin and /_CBO=θ with AB=z then A:[zcos(60+θ),zsin(60+θ)] while C:[zcos(θ),zsin(θ)] and OC: y=x/√3 + z[sin(θ)-cos(θ)/√3) since slope of OC is tan(30) and it passes through C. This gives us O as [(zcos(θ)-√3zsin(θ)),0] and finally AO^2 = [zcos(θ)-√3zsin(θ)-zcos(60+θ)]^2+[zsin(θ)]^2 =z^2 on expansion and using [sin(θ)]^2 + [cos(θ)]^2 = 1 and thus AB = AO<br />Ajit <br />PS: If you take a point P on AC such that /_PBC=(θ) and then prove that ABOP is concyclic then /_OBP=/_OAP=2θ from where it follows /_AOB =60+θ which makes triangle ABO isosceles. But as of now I can't prove that ABOP is concyclic.Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com