Is it possible?

Let $ABC$ be the triangle, right angled at $B$ . Let $BD$ be the median from $B$ . Since $BD$ divides angle $B$ in the ratio $1:2$ , we take $\angle ABD = 60^\circ , \angle CBD = 30^\circ$ , WLOG.

Since $\angle B = 90^\circ$ , $D$ is the circumcentre of triangle $ABC$ , so $DB=DA=DC = 2015$ , so that hypotenuse, $h = BC = 4030$ . Also, $DB= DA$ gives $\angle BAC = \angle BAD = \angle ABD = 60^\circ$ , so our triangle is a $30^\circ-60^\circ-90^\circ$ triangle.

Therefore perimeter , $P = h + \frac {\sqrt {3} h }{2} + \frac {h}{2} = 4030 + 2015\sqrt3 + 2015 = 6045 +2015\sqrt 3$ . Therefore $a+b+c =8063$ .