
#1
Jan2209, 04:37 PM

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No, this isn't a homework problem.
Given any triangle with a fixed length for one edge, and a fixed internal angle for the opposite vertex, the vertex will lay on a circle, where the other edges are allowed to vary in length. But I can't seem to construct a geometrical proof. Any takers? It's deceptively simple. 



#2
Jan2309, 04:06 AM

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The angle to the circumcentre is fixed (it's double the given angle, and you can prove it using isoceles triangles), and the circumcentre lies on a fixed line. 



#3
Jan2309, 04:13 AM

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#4
Jan2309, 05:34 AM

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Euclidian ProofThe circumcentre is the centre of the circumcircle, which is the circle which goes through all three vertices of a triangle, and the angle is the angle subtended by the original line at the circumcentre. join CAMREG … the CAMpaign for REal Geometry!




#5
Jan2309, 02:46 PM

P: 4,513

So true. Parallel lines that meet a a point. Hubris!
I can't imagine how you know all this. Way classical education? I had one Euclidean geometry class in High School, sans circumcentres. If I understand you correctly, I draw the pependicular bisectors through the remaining two sides. These will meet at an angle alpha=180beta at the circumcentre I should expect alpha to be constant as much as beta is expected to be constant. But I can prove alpha is constant using isosoles triangles, right? 



#6
Jan2309, 03:08 PM

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i hadn't realised so little classical geometry is taught nowadays … ok, it's best to introduce this idea the other away round, by starting with a circle, and seeing what happens inside it … one of the most useful circle theorems is that a fixed chord AB subtends the same angle ACB at any point C on the circumference on the smaller arc, and 180º minus that angle on the larger arc, and that the angle AOB subtended at the centre is twice the latter angle … to prove that, let O be the centre of the circle, and let the line CO (with C on the larger arc) poke a little beyond O to Q … then OAC and OBC are isoceles triangles ('cos it's a circle! ) and so you can prove that AOQ = 2ACO and QOB = 2OCB … and then for your problem, simply do the same proof "backwards" 



#7
Jan2309, 06:04 PM

P: 4,513

And thanks for the proof. 


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