## Friday, March 19, 2010

### Fox 266

This simple start will lead to a good one submitted by Lou.
There should be quite a few pure geometric solutions.

1. 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.

2. 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.

3. 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

4. Correction: With A as center draw a circle passing through thru. B & C

5. 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.

6. See

http://i40.tinypic.com/2wdyq28.png

MIGUE.

7. Correct Migue! There should be at least one more geometric solution, similar to yours.