Answer:
√3
Step-by-step explanation:
You want the radius of the circle through A, O, and C in the given diagram, where angle B is 60°, segment AC has length 3, and O is the incenter of ∆ABC.
Special case
If we assume that the radius of circle AOC is constant, then we can choose to do our analysis assuming ∆ABC is equilateral. In that case, its altitude will be BX = (3/2)√3, and AO will be √3.
Center
The center Z of circle AOC will lie at the intersection of the perpendicular bisectors of AC and AO. We already have defined BX as the perpendicular bisector of AC. Line YZ will be the perpendicular bisector of AO.
Note that angle YOX is the same (60°) in ∆AOX and ∆ZOY. We know that centroid O of ∆ABC divides the medians into segments having a 1:2 ratio. This tells us that YO is congruent to XO, and ∆AOX is congruent to ∆ZOY by the ASA congruence postulate.
Hence radius OZ is congruent to segment AO, already established to be √3.
The radius of circle AOC with center Z is √3.
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Additional comment
We used a geometry application to demonstrate point Z does not move regardless of the position of point B. Proving Z is invariant in the general case is somewhat more work.
