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In triangle ABC, D is a variable point on BC. AD is then joined. Excluding ABC itself, two triangles (namely ABD and ACD), under the normal circumstances, are formed. Associated with them, two circum-circles are also formed.

Now, if D is moved to B (i. e. they coincide), then there is only one triangle (namely ACD) because ABD is a de-generated one having no three vertices and no area.

The question is how many circum-circles are there now? [2 or 1]

‘2’ because one is ABC and the other is the one using AB as diameter.

‘1’ because only ABC qualifies but the other is not for the reasons of (1) It, if exists, is a de-generated one that has no legitimate triangle to be circum-scribed and (2) It, if exists, circum-scribes a de-generated; zero-area triangle.

The idea comes from problem #1086289. According to that problem, I agree that there are two such circles, but how about considering the case without those given conditions in mind?

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  • $\begingroup$ you are still working on the problem, mick? $\endgroup$ – abel Jan 2 '15 at 18:58
  • $\begingroup$ @abel Yes. As a matter of interest. $\endgroup$ – Mick Jan 3 '15 at 4:34
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The way I see it, your main objection seems to come from the fact that through two points there can pass an infinite number of circles. But this is like objecting to the definition of the derivative based in the fact that $\dfrac00$ is undefined, having an infinite number of solutions. Or to the definition of the integral as the limit of a Riemann sum, based on the fact that $\dfrac\infty\infty$ is also undefined, having an infinite number of solutions. Notice that the key notion in all three afore-mentioned concepts is the idea of a limit. Only here you are asked to find the “limit” of the position of the circumcenter of the triangle $\Delta ABD$ where $D\to B$. Just like in calculus.

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