Let $ABC$ be an equilateral triangle with side length 16. Three circles of the same radius $r$ are tangent to each other in pairs, and each of these circles is tangent to two sides of the triangle. The radius $r$ can be written as $r=\sqrt{a}-b$ where $a$ and $b$ are integers. Determine $a+b$.

Soit $A B C$ un triangle équilatéral de côté 16. Trois cercles de même rayon $r$ sont tangents entre eux deux à deux, et chacun de ces cercles est tangent à deux côtés du triangle. Le rayon $r$ s'écrit $r=\sqrt{a}-b$ où $a$ et $b$ sont des entiers. Déterminer $a+b$.

Let $ABC$ be an equilateral triangle with side length 16. Three circles of the same radius $r$ are tangent to each other in pairs, and each of these circles is tangent to two sides of the triangle. The radius $r$ can be written as $r=\sqrt{a}-b$ where $a$ and $b$ are integers. Determine $a+b$.
Answer: 52. Let $A^{\prime}$ be the center of the circle closest to $A$, etc. Let $H$ be the projection of $B^{\prime}$ on $(B C)$. We have $\tan (30)=B^{\prime} H / B H=r / B H$ so $B H=r \sqrt{3}$. Similarly, if we denote $K$ as the projection of $C^{\prime}$ on $(B C)$, then $C K=r \sqrt{3}$, so $16=B C=2 r+2 r \sqrt{3}$, which gives $r=\frac{8}{\text { Response1 }+\sqrt{3}}=\frac{4(\sqrt{3}-1)}{\text { Response }(\sqrt{3}-1)(\sqrt{3}+1)}=4(\sqrt{3}-1)=\sqrt{48}-4$, therefore $a+b=48+4=52$.