Japan 2013 Initial Round

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, $$r^2-x^2-y^2=40(r+x),r^2-x^2-y^2=12(r-y),r^2-x^2-y^2=2d(r+y).$$ Consequently, if we let $t=r^2-x^2-y^2$, then we have $$r-x=\frac{t}{10},r+x=\frac{t}{40},r-y=\frac{t}{12},r+y=\frac{t}{2d}.$$ 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 $r\ne 0$, $\frac{1}{10}+\frac{1}{40}=\frac{1}{12}+\frac{1}{2d}$. Thus we obtain $d=12$ for the desired answer for the problem.