HMMT November

Solution:

Answer: $\frac{80}{3}$

Note that by equal tangents, $B'A = B'C$, $C'A = C'B$, and $A'B = A'C$. Moreover, since the line segments $A'B'$, $B'C'$, and $C'A'$ are tangent to the circumcircle of $ABC$ at $C$, $A$, and $B$ respectively, we have that $\angle A'BC = \angle A'CB = \angle A$, $\angle B'AC = \angle B'CA = \angle B$, and $\angle C'BA = \angle C'AB = \angle C$.

By drawing the altitudes of the isosceles triangles $BC'A$ and $AC'B$, we therefore have that $C'A = 2 / \cos C$ and $B'A = 3 / \cos B$.

Now, by the Law of Cosines, we have that
$$\begin{aligned} & \cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{25 + 16 - 36}{2 \times 5 \times 4} = \frac{1}{8} \\ & \cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{25 + 36 - 16}{2 \times 5 \times 6} = \frac{3}{4}. \end{aligned}$$
Therefore,
$$B'C' = C'A + B'A = 2 \left(\frac{4}{3}\right) + 3 \times 8 = \frac{80}{3}$$