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Is there a triangle with sides that are partitioned into line segments of ratios $3:2$, $3:5$, and $10:9$ by the points of tangency of its inscribed circle?

By Ceva's Theorem, if a triangle has sides of lengths 20, 16, and 19, the cevian between the vertex and the point that partitions the side of length 20 into line segments of lengths 12 and 8, the cevian between the vertex and the point that partitions the side of length 16 into line segments of lengths 6 and 10, and the cevian between the vertex and the point that partitions the side of length 19 into line segments of lengths 10 and 9 coincide. Is this point the center of the inscribed circle?

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No, it is not. If you join the center of the inscribed circle to each of your division points on the sides, you get segments, perpendicular to the sides (since the radius of the incircle is perpendicular to the tangent). The center of the incircle is the point of intersection of angle bisectors. If an angle bisector is perpendicular to the opposite side, your triangle has to be isosceles, and the point on the side would be the midpoint of the side. But that is not the case for your triangle. Those Cevians intersect at one point by Ceva's Theorem, but that point is definitely not the incenter.

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  • $\begingroup$ Your explanation reveals that the scenario as described in a problem on this national competition in Bangladesh is contradictory. $\endgroup$ – A gal named Desire Mar 23 at 17:45
  • $\begingroup$ math.stackexchange.com/questions/3104005/… $\endgroup$ – A gal named Desire Mar 23 at 17:45
  • $\begingroup$ Is that correct? $\endgroup$ – A gal named Desire Mar 23 at 17:46
  • $\begingroup$ The problem is correct. The fact that the points are the points of tangency of the inscribed circle is completely irrelevant. Neither the circle or its center are used in any way. $\endgroup$ – GReyes Mar 23 at 18:37
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    $\begingroup$ Yes. I would say so. $\endgroup$ – GReyes Mar 23 at 20:09

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