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Let's take for example $\triangle ABC$ with $\angle A = \angle B = 1^o$. How can a triangle like this have a circumcircle? My confusion is with triangles like this in general, with very long sides.

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    How many non-collinear points do you need so that a unique circle passes through them? :)2012-12-18
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    No matter how long you make the sides, I can make a circle big enough to circumscribe your triangle. We have all of infinity to play with, so size matters not (to paraphrase Yoda).2012-12-18
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    Maybe visualise it like this: imagine any triangle with a circumcircle, with one vertex at the top. Now imagine moving the other two points round the circle - getting nearer to the top. That way you can get a triangle with 2 angles of just one degree. Then just enlarge the diagram.2012-12-18
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    Oh, I think I visualize it now.. so those points would be relative close together on the circle?2012-12-18
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    Yes, they would need to be comparatively close together.2012-12-18
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    Note that the smallest circle containing an obtuse angled triangle will only go through two of the vertices and will be smaller than the circumcircle; in your example, much smaller. But there is still a circumcircle2012-12-18
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    Would this also mean that the points on circumcircle of an equilateral triangle are furthest apart?2012-12-18
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    Once you understand this, and if you know a little trigonometry, try this problem. If a triangle has sides 1, 1, and a little less than 2, so that its angles are $1^\circ$, $1^\circ$, and $88^\circ$, then what is the radius of the circumcircle?2012-12-18
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    @GEdgar $1+1+88-90 ;)$.2012-12-18
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    OK, we need 1,1,178.2012-12-18

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