Euclidian Proof

Phrak

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.

Attached Files

Euclidian Geometry.doc (24.0 KB, 5 views)

tiny-tim

Hi Phrak!

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.

Phrak

And I was just bemoaning the fact that no one responded. But wait. What's a circumcentre, and what angle from whence and to where?

tiny-tim

I think most PF members don't like "real" geometry!
he he

The 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.

Phrak

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=180-beta 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?

tiny-tim

ooh, i'm sorry, Phrak …

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"

Phrak

No problem at all. Though it's been a while since high school for me. The classical studies are more common in the UK than the US, that makes the difference.

And thanks for the proof.