Hey I am from germany and I participated at the BWM. Let $A_1,A_2,\dots,A_{12}$ be the vertices of the 12-gon. Now look at the center $X$ and the circumcenters $B_1,B_2,\dots,B_6$ of the triangles $\triangle A_1A_2X,\triangle A_3A_4X,\dots,\triangle A_{11}A_{12}X$ and $M_1,M_2,\dots,M_6$ the centers of $A_1A_2,A_3A_4,\dots,A_{11}A_{12}$. Obviously, $\boxed{\overline{A_kB_k}=\overline{A_{k+1}B_k}=\overline{B_kX}}$ for $k=1,2,\dots,6$ (overline indicates length). Because $\angle A_kB_kA_{k+1}=2\cdot\angle A_kXA_{k+1} = 2\cdot\frac{360^\circ}{12}=60^\circ$ due to the inscribed angle theorem, the isoceles triangle $\triangle A_kA_{k+1}X$ is even quadliteral, thus $\boxed{\overline{A_kB_k}=\overline{A_{k+1}B_k}=\overline{A_{k}A_{k+1}}}$. As $M_kX$ or $B_kX$ halves $\angle A_kXA_{k+1}$, $\angle B_kXB_{k+1}=\angle B_kXA_{k+1}+\angle A_{k+1}XA_{k+2}+\angle A_{k+2}XB_{k+1}=\frac{\angle A_kXA_{k+1}}{2}+\angle A_{k+1}XA_{k+2}+\frac{\angle A_{k+2}XA_{k+3}}{2} = \frac{\frac{360^\circ}{12}}{2}+\frac{360^\circ}{12}+\frac{\frac{360^\circ}{12}}{2}=60^\circ$. Hence, the isosceles triangle $\triangle B_kXB_{k+1}$ is eve quadliteral, thus $\boxed{\overline{B_kX}=\overline{B_{k+1}X}=\overline{B_kB_{k+1}}}$. Altogether, we have $\boxed{\overline{A_kB_k}=\overline{A_{k+1}B_k}=\overline{B_kX}=\overline{A_kA_{k+1}}=\text{\textbf{constant}}}$. So we can put 7 coins with the centers $X,B_1,B_2,\dots,B_6$ inside the ring of the twelve coins, and they are just touching because their distance of their centers is $2r$. Now we have proved that it is possible by giving an example and showing it works. I apologize for my bad english, have a nice day!