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 Ajit 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.
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. Thank you.
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. Ajit
There should be quite a few pure geometric solutions.
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
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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.
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.
ReplyDeleteThank you.
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.
ReplyDeleteAjit
Correction: With A as center draw a circle passing through thru. B & C
ReplyDeleteYep, 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.
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ReplyDeletehttp://i40.tinypic.com/2wdyq28.png
MIGUE.
Correct Migue! There should be at least one more geometric solution, similar to yours.
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