Let $\triangle ABC$ be a triangle with the fixed distances from its vertices to a given point $P$，where $PA,PB,PC$ are not all equal. Is there a maximum value for the perimeter of triangle ABC, and if so, indicate the location of point $P$.
Fixed points $B, C$, we can get when the perimeter takes maximum value,$AP$ must be the bisector of angle $BAC$. Can we say the answer of this question is the incenter of $\triangle ABC$？
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If I'm not mistaking, the point $P$ you're referring to, is the centre of the circumscribed circle. It that is correct, you can just use the unity circle, take three points on that, and determine the maximum value of the perimeter of the triangle, defined by those three points.
You can then easily take $A=(1,0)$ as one of the three points, the other two are then defined by the angle to the $X$-axis, so you get the points $B=(\cos(\alpha),\sin(\alpha))$ and $C=(\cos(\beta),\sin(\beta))$. The cosine-rule can be used in order to determine the length of the sides of the triangle and you derive your maximum from there.
If $AP$ bisects $BPC$, then $AB$ and $AC$ are the same length. This is because the points lie on a circle with centre $P$ and the angles $APB$ and $APC$ must be the same. The same argument applies to all the sides, so the triangle must be regular. It is then obvious that $P$ is the incentre.