Steiner’s December Triangle

Label the equilateral triangle $ABC$ and the center of the central circle with radius $r_2$ be $O$ . Due to symmetry, center $O$ is also the centroid of $\triangle ABC$ . Let the side length of $\triangle ABC$ be $1$ . Then $AO=BO=CO=\frac 1{\sqrt 3}$ and $ON = \frac 1{2\sqrt 3}$ , where $BN$ is a median of $\triangle ABC$ . Then we have:

$\begin{cases} 3r_3 + r_2 = \dfrac 1{\sqrt 3} & ...(1) \\ 2r_1 + r_2 = \dfrac 1{2\sqrt 3} & ...(2) \end{cases} \implies (1) - (2) : \quad 3r_3 - 2r_1 = \frac 1{2\sqrt 3}$

Let the center of the left-bottom circle with radius $r_3$ be $P$ and $PM$ be perpendicular to $AC$ . Then we have

$\begin{aligned} AM + MN & = AN \\ \sqrt 3 r_3 + \sqrt{(r_3+r_1)^2 - (r_3-r_1)^2} & = \frac 12 \\ \sqrt 3 r_3 + 2 \sqrt{r_1r_3} & = \frac 12 \\ 2 \sqrt{r_1r_3} & = \frac 12 - \sqrt 3 r_3 & \small \blue{\text{Squaring both sides}} \\ 4r_1r_3 & = 3r_3^2 - \sqrt 3 r_3 + \frac 14 & \small \blue{\text{Multiply both sides by }4} \\ \blue{16r_1}r_3 & = 12r_3^2 - 4\sqrt 3 r_3 + 1 & \small \blue{\text{Since }3r_3 - 2r_1 = \frac 1{2\sqrt 3}} \\ \blue{8\left(3r_3-\frac 1{2\sqrt 3}\right)}r_3 & = 12r_3^2 - 4\sqrt 3 r_3 + 1 \\ 24r_3^2 - \frac 4{\sqrt 3} r_3 & = 12r_3^2 - 4\sqrt 3 r_3 + 1 \\ 12\sqrt 3 r_3^2 + 8r_3 - \sqrt 3 & = 0 \\ \implies r_3 & = \frac {\sqrt{64+144}-8}{24\sqrt 3} = \frac {\sqrt{13}-2}{6\sqrt 3} \\ r_2 & = \frac 1{\sqrt 3} - 3r_3 = \frac {4-\sqrt{13}}{2\sqrt 3} \\ r_1 & = \frac 1{4\sqrt 3} - \frac {r_2}2 = \frac {\sqrt{13}-3}{4\sqrt 3} \\ \implies \frac {r_2+r_3}{r_1} & = \frac {2(1+\sqrt{13})}3 \end{aligned}$

Therefore $a+b+c+d+e = 2+3+1+1+13 = \boxed{20}$ .

Since we're interested in ratios, we can take the side length of the triangle to be $1$ . Then we can write the following three relations involving $r_1 , r_2$ and $r_3$

$2 r_1 + r_2 = \dfrac{1}{2 \sqrt{3} } \hspace{24 pt} (1)$

$r_2 + 3 r_3 = \dfrac{1}{\sqrt{3}} \hspace{24 pt} (2)$

$(r_1 + r_3)^2 = (r_1 - r_3)^2 + ( \sqrt{3} r_3 - \frac{1}{2} )^2 \hspace{24 pt} (3)$

Solving the first two equations, results in:

$(r_1 , r_2, r_3 ) = ( -\dfrac{1}{4 \sqrt{3} } , \dfrac{1}{\sqrt{3} } , 0 ) + t ( \dfrac{3}{2}, -3 , 1 )$

Pluggin this into equation (3), results in a quadratic equation in $t$ , which solves to

$t = \dfrac{1}{6 \sqrt{3} } ( \sqrt{13} - 2 )$

The required ratio is $\dfrac{r_2 + r_3}{r_1} = \displaystyle \dfrac{ \frac{1}{\sqrt{3}}-2 t} {-\frac{1}{4 \sqrt{3} } +\frac{3}{2} t}$

Pluggin in the value of $t$ we obtained into the above quotient, and simplifying, results in:

$\dfrac{r_2 + r_3}{r_1} = \frac{2}{3} (1 + \sqrt{13} )$

Hence, $a = 2, b = 3, c = 1 , d = 1, e = 13$ , and this makes the answer $\boxed{20 }$

Let $r_1 = 1$ , $r_2 = x$ , $r_3 = y$ , and label the diagram as follows:

From right $\triangle ABF$ we know that $AB = \frac{BF}{\cos 60°} = 2y$ .

From right $\triangle ACE$ we know that $AC = \frac{CE}{\cos 60°} = 2x + 4$ , and since $AC = 3y + x$ , $2x + 4 = 3y + x$ or $x = 3y - 4$ .

By the law of cosines on $\triangle BCD$ , $BD^2 = BC^2 + CD^2 - 2 \cdot BC \cdot CD \cos 60°$ , or $(y + 1)^2 = (x + y)^2 + (x + 1)^2 - (x + y)(x + 1)$ .

These two equations solve to $x = \frac{1}{2}(\sqrt{13} - 1)$ and $y = \frac{1}{6}(7 + \sqrt{13})$ .

Therefore, $\cfrac{r_2 + r_3}{r_1} = \cfrac{x + y}{1} = \cfrac{2(1 + \sqrt{13})}{3}$ , so that $a = 2$ , $b = 3$ , $c = 1$ , $d = 1$ , $e = 13$ , and $a + b + c + d + e = \boxed{20}$ .