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Geometry Level 2

How many circles can pass through all the given three non-collinear points?

2 3 0 Infinitely many 1

2 solutions

Nihar Mahajan
Dec 19, 2015

Every triangle (made out of 3 non-collinear points) has one and only one circumcircle.

But I was thinking if the question had a joke, he didn't sad "at the same time", só I put the last alternative.

Fernando Sckaff - 5 years, 5 months ago

Proof of the uniqueness of circumcircle is not at all trivial.

Venkata Karthik Bandaru - 5 years, 5 months ago

Yes , it is not trivial.

Nihar Mahajan - 5 years, 5 months ago

Venkata Karthik Bandaru
Dec 19, 2015

We need to find a point $P$ such that $\overline{PA} = \overline{PB} = \overline{PC}$ , and then $P$ will be the centre of the circle we need. Since the locus of all points equidistant from $2$ given points is the perpendicular bisector of the line segment joining the $2$ points, point $P$ is the intersection of perpendicular bisectors of line segments $\overline{AB}$ and $\overline{BC}$ . Since there is only one intersection point, there is only one such $P$ we can find, and hence only one circle centred at $P$ and passing through points $A, B$ and $C$ .

Moderator note:

How do we know that there is at least one such intersection point?

I agree that it is clear 3 distinct lines can be concurrent in at most 1 point, so that will give us the uniqueness that we seek.

Response to Challenge Master Note : Since $A, B$ and $C$ aren't collinear, perpendicular bisectors mentioned do meet at a specific point.

Venkata Karthik Bandaru - 5 years, 5 months ago

Is there enough information?

2 solutions

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