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

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SUMMARY

The discussion centers on the inequality \(AO + BO + CO \geq 6r\), where \(r\) is the radius of the inscribed circle in triangle ABC. Participants clarify that the problem pertains to the triangle's inscribed circle, not a triangle inscribed in a circle. The initial confusion arose from interpreting "ABC trigon" as a triangle inscribed in a circle, leading to the incorrect conclusion that \(AO + BO + CO = 3r\). The correct interpretation confirms that the sum of distances from the triangle's vertices to the incenter must satisfy the stated inequality.

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solakis1
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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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