, 2017

Solution:

By the degenerate case of Von Aubel's Theorem we have that $YZ = AX = 6$ and $ZX = BY = 7$ and $XY = CZ = 8$ so it suffices to find the area of a $6$-$7$-$8$ triangle which is given by $\frac{21 \sqrt{15}}{4}$.

To prove that $AX = YZ$, note that by LoC we get
$$YX^2 = \frac{b^2}{2} + \frac{c^2}{2} + bc \sin \angle A$$
and
$$\begin{aligned} AX^2 & = b^2 + \frac{a^2}{2} - ab(\cos \angle C - \sin \angle C) \\ & = c^2 + \frac{a^2}{2} - ac(\cos \angle B - \sin \angle B) \\ & = \frac{b^2 + c^2 + a(b \sin \angle C + c \sin \angle B)}{2} \\ & = \frac{b^2}{2} + \frac{c^2}{2} + a h \end{aligned}$$
where $h$ is the length of the $A$-altitude of triangle $ABC$. In these calculations we used the well-known fact that $b \cos \angle C + c \cos \angle B = a$ which can be easily seen by drawing in the $A$-altitude. Then since $bc \sin \angle A$ and $a h$ both equal twice the area of triangle $ABC$, we are done.