MHB AO+BO+CO≥6r where r is the radius of the inscribed circle

AI Thread Summary
The discussion revolves around proving the inequality AO + BO + CO ≥ 6r, where O is the center of the inscribed circle in triangle ABC and r is the radius of that circle. Participants clarify that the problem pertains to the inscribed circle within a triangle, not a triangle inscribed in a circle. The initial confusion arises from interpreting "ABC trigon" as a triangle inscribed in a circle, leading to the incorrect conclusion that AO + BO + CO equals 3r, which is less than 6r. The correct understanding emphasizes the relationship between the triangle's vertices and the inscribed circle's center. The focus remains on establishing the validity of the inequality based on the properties of the triangle and its inscribed circle.
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From the entrance examinations to Ghana University ,from high school, i got the following problem:

If O is the center of the inscribed circle in an ABC trigon,then prove that: $$AO+BO+CO\geq 6r$$ where r is the radius of the inscribed circle.
 
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I first thought that your "ABC trigon" is what I would call a "triangle" but that would make the problem impossible. If ABC is a triangle inscribed in a circle, of radius r, then OA, OB, and OC are equal to r so that OA+ OB+ OC= 3r which is less than 6r.
 
HallsofIvy said:
I first thought that your "ABC trigon" is what I would call a "triangle" but that would make the problem impossible. If ABC is a triangle inscribed in a circle, of radius r, then OA, OB, and OC are equal to r so that OA+ OB+ OC= 3r which is less than 6r.

The question is about the circle inscribed in a triangle, not about a triangle inscribed in a circle.
 
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