Prove that in a triangle, the Nagel point and incenter concur if and only if the triangle is equilateral.

\( \textbf{Note:} \) Call the tangent point of the A-excircle to side \(BC\) be \(D\). Connect \(AD\). Do so similarly for the other two sides. The intersection of these three segments is the Nagel point.

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## Comments

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TopNewestThe question is easy. Firstly it is obvious that if triangle is equilateral then then the Nagel point and the incentre coincide. So we only need to prove the other side that if the Nagel point and the incentre coincide then the triangle is equilateral.

So my proof is as follows. Firstly denote the point where AB and AC touch the excircle as P and Q respectively. Observe that AP=AQ. Also observe that BD=BP,CD=CQ. Hence we obtain that AB+BD=AC+CD. Also by angle bisector theorem we have AB/BD=AC/CD. Solving we obtain that BD=DC. Substituting we get AB=AC. Same applies for some other excircle and so AB=AC=BC. Hence proved :)

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Here is another solution:

In barycentric coordinates, the Nagel point is \((s-a : s-b: s-c)\) and the incenter is \((a:b:c)\). Thus, they concur if and only if \[\frac{s-a}{a} =\frac{s-b}{b} = \frac{s-c}{c} \implies a = b = c\]

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Hmm ..but i don't know actually what barycentric coordinates are . But i would like if u would post some more geometry problems.

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Can you define what the Nagel point is?

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I have defined it.

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