HMMT, 2013

Solution:

Answer: $\frac{288}{5}$.

Place $ABC$ in the coordinate plane so that $A = (0,6)$, $B = (0,0)$, $C = (8,0)$. Consider a point $P = (a, b)$ inside triangle $ABC$. Clearly, $d(P, AB) = a$, $d(P, BC) = b$.

Now, we see that the area of triangle $ABC$ is $\frac{6 \cdot 8}{2} = 24$, but may also be computed by summing the areas of triangles $PAB$, $PBC$, $PCA$. The area of triangle $PAB$ is $\frac{6 \cdot a}{2} = 3a$, and similarly the area of triangle $PBC$ is $4b$. Thus, it follows easily that $d(P, CA) = \frac{24 - 3a - 4b}{5}$.

Now, we have
$$(x, y) = \left( \frac{24}{5} - \frac{8}{5}a - \frac{4}{5}b, \frac{24}{5} - \frac{3}{5}a - \frac{9}{5}b \right)$$
The desired region is the set of $(x, y)$ obtained by those $(a, b)$ subject to the constraints $a \geq 0$, $b \geq 0$, $6a + 8b \leq 48$.

Consequently, our region is the triangle whose vertices are obtained by evaluating $(x, y)$ at the vertices $(a, b)$ of the triangle. To see this, let $f(a, b)$ output the corresponding $(x, y)$ according to the above. Then, we can write every point $P$ in $ABC$ as $P = m(0,0) + n(0,6) + p(8,0)$ for some $m + n + p = 1$. Then, $f(P) = m f(0,0) + n f(0,6) + p f(8,0) = m\left(\frac{24}{5}, \frac{24}{5}\right) + n(-8, 0) + p(0, -6)$, so $f(P)$ ranges over the triangle with those three vertices.

Therefore, we need the area of the triangle with vertices $\left(\frac{24}{5}, \frac{24}{5}\right)$, $(0, -6)$, $(-8, 0)$, which is easily computed (for example, using determinants) to be $\frac{288}{5}$.