Japan Junior Mathematical Olympiad

Let us consider the situation, as indicated in the figure below, where $2$ circles and $2$ straight lines which are outer tangent to both of the circles and $1$ inner common tangent line are involved.



Let $T$ and $U$ be the points of intersection of the common inner tangent line and $2$ outer common tangent lines to the circles. Denote by $M,N$ the centers of the $2$ circles and let $m,n$ be the corresponding radius, respectively. Let $I$ and $J$ be the foot of the perpendicular line to the line $TU$ drawn from $M$ and $N$, respectively, and put $TI=mt$, $UI=mu$. Since the line $TM$ bisects one of the angles formed by the outer common tangent and inner common tangent to circles through $T$, and the line $TN$ does the same with respect to the other angle formed by these tangents, we see that $\angle MTN=90^{\circ }$ holds. This fact, together with $\angle TIM=\angle NJT=90^{\circ }$

implies that we have $\angle MTI=90^{\circ }-\angle NTJ=\angle TNJ$ and therefore, we conclude that the triangles $MTI$ and $TNJ$ are similar. Consequently, we obtain $TJ=NJ\times \frac{MI}{TI}=\frac{n}{t}$. Similarly, we get $UJ=\frac{n}{u}$. It then follows that we have $TU=m(t+u)=n\left(\frac{1}{t}+\frac{1}{u}\right)$, and therefore, $n=mtu$.

$$x=ars=bsp=cpq=dgr,$$ from which we obtain $$x^2=ars\times cpq=acpqrs,$$ and $$x^2=bsp\times dqr=bdpqrs.$$ Hence, we conclude that $ac=bd$ holds and since $a=2$, $b=1$, $c=4$, we see that $d=8$ is the desired answer.