, 2017

Solution:

There are three cases: $\ell$ intersects $AB, AC$, $\ell$ intersects $AB, BC$, and $\ell$ intersects $AC, BC$. These cases are essentially identical, so let $\ell$ intersect segment $AB$ at $M$ and segment $AC$ at $N$.

Then the condition is equivalent to
$$\begin{gathered} AM + MN + AN = MB + BC + CN + MN \\ AM + AN = MB + CN + 15 \end{gathered}$$
but $AN + CN = 14$ and $AM + BM = 13$, so that
$$BM + CN = 27 - AM - AN = AM - AN - 15$$
implying that $AM + AN = 21$.

Now let $\angle BAC = \theta$ for convenience, so that
$$[AMN] = \frac{1}{2} AM \cdot AN \cdot \sin \theta$$
which is maximized when $AM = AN = \frac{21}{2}$. Further we can easily calculate $\sin \theta = \frac{12}{13}$ (e.g. by LOC); note that this is why the area is maximized in this case (we want to maximize $\sin \theta$, which is equivalent to maximizing $\theta$, so $\theta$ should be opposite the largest side). Our answer is thus
$$\frac{1}{2} \cdot \frac{21}{2} \cdot \frac{21}{2} \cdot \frac{12}{13} = \frac{1323}{26}$$

Alternatively we could also calculate
$$\begin{aligned} {[AMN]} & = [ABC] \cdot \frac{AM}{AB} \cdot \frac{AN}{AC} \\ & = 84 \cdot \frac{21}{23} \cdot \frac{\frac{21}{2}}{14} \end{aligned}$$
which gives the same answer.