14) Gigi arranges six red, round tokens, identical to each other and with a radius of 10 $\mathrm{cm}$, on a table so that they touch each other in pairs without overlapping and that their centers are arranged at the vertices of a regular hexagon. Then she notices that there is still space in the middle to place a blue token, round, in such a way that it touches all six red tokens without overlapping them. What is the radius of the blue token?
(A) $5 \sqrt{3} \mathrm{~cm}$,
(B) $10 \mathrm{~cm}$
(C) $10 \sqrt{3} \mathrm{~cm}$
(D) $15 \sqrt{3} \mathrm{~cm}$,
(E) $20 \mathrm{~cm}$.
14. The answer is $\mathbf{( B )}$.

Let $r$ be the length of the radius of the blue token and $R$ be the length of the radius of the red tokens.

By symmetry, the center of the blue token must be at the center $C$ of the hexagon. The triangle $A B C$ is equilateral because the centers of the red tokens are arranged at the vertices of a regular hexagon. In particular, $\overline{A B}=\overline{B C}=20 \mathrm{~cm}$. On the other hand, $\overline{B C}=r+R=r+10$ $\mathrm{cm}$, from which $r=10 \mathrm{~cm}$.

Alternatively, we can observe that $\widehat{A B D}=120^{\circ}$, that $\widehat{B C D}=$ $\widehat{C A B}=60^{\circ}$, and thus that $A B C D$ is a parallelogram. Therefore, $\overline{A B}=\overline{C D}$, and we can proceed as before.

![](https://cdn.mathpix.com/cropped/2024_04_17_4701f996ee50400fb9c5g-4.jpg?height=434&width=469&top_left_y=297&top_left_x=1496)