Answer: 715

Solution:

For a point $P,$ let $\alpha$ be the distance from $P$ to line $BC$ divided by the length of the altitude from $A,$ and define $\beta$ (to $CA$) and $\gamma$ (to $AB$) similarly; then $\alpha + \beta + \gamma = 1$ for points inside, since $[PBC] + [PCA]$ $+ [PAB] = [ABC].$ A chord parallel to $\overline{BC}$ at level $\alpha$ cuts off a triangle at $A$ similar to $ABC$ with ratio $1 - \alpha,$ so its length is $220(1 - \alpha).$ The chord of length $55$ gives $1 - \alpha = \frac{1}{4},$ so $\ell_A$ is the line $\alpha = \frac{3}{4};$ similarly $45 = 180(1 - \beta)$ puts $\ell_B$ at $\beta = \frac{3}{4},$ and $15 = 120(1 - \gamma)$ puts $\ell_C$ at $\gamma = \frac{7}{8}.$

Along any line parallel to $\overline{BC},$ the coordinate $\beta$ varies linearly, and on the chord at level $\alpha$ inside the triangle, $\beta$ runs over an interval of length $1 - \alpha$ while the chord has length $220(1 - \alpha);$ hence a segment parallel to $\overline{BC}$ with endpoints differing by $\Delta\beta$ has length $220\,|\Delta\beta|.$ The side of the new triangle on $\ell_A$ runs from $\ell_A \cap \ell_B,$ where $\beta = \frac{3}{4},$ to $\ell_A \cap \ell_C,$ where $\beta = 1 - \frac{3}{4} - \frac{7}{8} = -\frac{5}{8}.$ Its length is $$220\left(\frac{3}{4} + \frac{5}{8}\right) = 220 \cdot \frac{11}{8}.$$

Since the three lines are parallel to the sides of $ABC,$ the triangle they bound is similar to $ABC,$ here with ratio $\frac{11}{8}.$ Its perimeter is $\frac{11}{8}(120 + 220 + 180)$ $= \frac{11}{8} \cdot 520 = 715.$