Internal Angle Bisectors

For a triangle with sides $a_1$ , $a_2$ , and $a_3$ , an internal angle bisector $t_1$ has the equation $t_1^2 = a_2a_3(1 - \frac{a_1^2}{(a_2 + a_3)^2})$ . Solving the equations

$(7595)^2 = a_2a_3(1 - \frac{a_1^2}{(a_2 + a_3)^2})$

$(4704\sqrt{2})^2 = a_1a_3(1 - \frac{a_2^2}{(a_1 + a_3)^2})$

$(14880\sqrt{2})^2 = a_1a_2(1 - \frac{a_3^2}{(a_1 + a_2)^2})$

gives $a_1 = 21700$ , $a_2 = 20832$ , and $a_3 = 6076$ , for a perimeter of $21700 + 20832 + 6076 = \boxed{48608}$ .