, 2013

Let $DX = d$. Choose xy-coordinate axis in such a way that $X = (0,0)$, $A = (5,0)$, $B = (0,6)$, $C = (-20,0)$, $D = (0,-d)$ are satisfied. Let $\Gamma$ be the circle tangent to each of the circles $O_1, O_2, O_3, O_4$, and let $P = (x, y)$ be the center of $\Gamma$ and $r$ be its radius. Since the circle $O_1$ touches the circle $\Gamma$ tangentially from inside, we see that $AP = r - 5$, which means that $(x - 5)^2 + y^2 = (r - 5)^2$ is satisfied. Simplifying this, we obtain
$$r^2 - x^2 - y^2 = 10(r - x).$$
If we use the fact that each of the circles $O_2, O_3, O_4$ also touches the circle $\Gamma$ tangentially from inside, we obtain in the same way as above,
$$r^2 - x^2 - y^2 = 40(r + x), \quad r^2 - x^2 - y^2 = 12(r - y), \quad r^2 - x^2 - y^2 = 2d(r + y).$$
Consequently, if we let $t = r^2 - x^2 - y^2$, then we have
$$r - x = \frac{t}{10}, \quad r + x = \frac{t}{40}, \quad r - y = \frac{t}{12}, \quad r + y = \frac{t}{2d}.$$
Since $2r = (r-x) + (r+x) = (r-y) + (r+y)$, we obtain
$$2r = t \left( \frac{1}{10} + \frac{1}{40} \right) = t \left( \frac{1}{12} + \frac{1}{2d} \right),$$
from which we get, as $r \neq 0$, $\frac{1}{10} + \frac{1}{40} = \frac{1}{12} + \frac{1}{2d}$. Thus we obtain $d = 12$ for the desired answer for the problem.