Three circles with radii $2$, $3$, and $10$ are pairwise externally tangent. A fourth circle, lying in the gap among them, is externally tangent to all three. What is the radius of the fourth circle?

1. Let the circles of radii $2,3,10$ have centers $O_2,O_3,O_{10}$. The distances between centers are $O_2O_3=5$, $O_2O_{10}=12$, $O_3O_{10}=13$, so $\angle O_3O_2O_{10}=90^\circ$. Place $O_2=(0,0)$, $O_3=(5,0)$, $O_{10}=(0,12)$, and let the fourth circle have center $P=(x,y)$ and radius $r$.
2. $PO_2=2+r$, $PO_3=3+r$, $PO_{10}=10+r$. Subtracting $x^2+y^2=(2+r)^2$ from $(x-5)^2+y^2=(3+r)^2$ gives $-10x+25=5+2r$, so $x=2-\frac r5$.
3. Subtracting $x^2+y^2=(2+r)^2$ from $x^2+(y-12)^2=(10+r)^2$ gives $-24y+144=96+16r$, so $y=2-\frac{2r}3$.
4. Substituting into $x^2+y^2=(2+r)^2$ and clearing denominators gives $29r^2+420r-225=0$, i.e. $(29r-15)(r+15)=0$. So $r=\boxed{\dfrac{15}{29}}$.