Volcanic Sangaku

Let the sides of the equilateral triangle be $1$ , and let $p = BD$ , $q = CE$ , $t = DE$ , and $u = CD$ .

The area of $\triangle BCD$ is $T_{\triangle BCD} = \frac{1}{2} \cdot p \cdot 1 \cdot \sin 60° = \frac{\sqrt{3}}{4}p$ , and the semiperimeter of $\triangle BCD$ is $s_{\triangle BCD} = \frac{1}{2}(p + u + 1)$ .

The area of $\triangle ADE$ is $T_{\triangle ADE} = \frac{1}{2} \cdot (1 - p) \cdot (1 - q) \cdot \sin 60° = \frac{\sqrt{3}}{4}(1 - p)(1 - q)$ , and the semiperimeter of $\triangle ADE$ is $s_{\triangle ADE} = \frac{1}{2}(2 - p - q + t)$ .

The area of $\triangle EDC$ is $T_{\triangle EDC} = T_{\triangle ABC} - T_{\triangle BCD} - T_{\triangle ADE} = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4}p - \frac{\sqrt{3}}{4}(1 - p)(1 - q) = \frac{\sqrt{3}}{4}(1 - p)q$ , and the semiperimeter of $\triangle EDC$ is $s_{\triangle EDC} = \frac{1}{2}(t + u + q)$ .

Since the inradius is $r = \frac{T}{s}$ and each incircle is equal,

$r = \cfrac{\frac{\sqrt{3}}{4}p}{\frac{1}{2}(p + s + 1)} = \cfrac{\frac{\sqrt{3}}{4}(1 - p)(1 - q)}{\frac{1}{2}(2 - p - q + t)} = \cfrac{\frac{\sqrt{3}}{4}(1 - p)q}{\frac{1}{2}(t + u + q)}$

By the law of cosines on $\triangle BCD$ , $u = \sqrt{p^2 + 1^2 - 2 \cdot p \cdot 1 \cos 60°}$ , or

$u = \sqrt{p^2 - p - 1}$

and by the law of cosines on $\triangle ADE$ , $t = \sqrt{(1 - p)^2 + (1 - q)^2 - 2 \cdot (1 - p) \cdot (1 - q) \cos 60°}$ , or

$t = \sqrt{p^2 + q^2 - pq - p - q + 1}$

Substituting $u$ and $t$ into the inradius equations $\cfrac{\frac{\sqrt{3}}{4}p}{\frac{1}{2}(p + u + 1)} = \cfrac{\frac{\sqrt{3}}{4}(1 - p)(1 - q)}{\frac{1}{2}(2 - p - q + t)}$ and solving for $q$ gives:

$q = \cfrac{-4p^3+4p^2+p-1+(4p^2-2p)\sqrt{p^2-p+1}}{7p^2-2p-1}$

and substituting $u$ and $t$ into the inradius equations $\cfrac{\frac{\sqrt{3}}{4}p}{\frac{1}{2}(p + u + 1)} = \cfrac{\frac{\sqrt{3}}{4}(1 - p)q}{\frac{1}{2}(t + u + q)}$ and solving for $q$ gives:

$q = \cfrac{8p^4-21p^3+16p^2-11p+4+(-8p^3-4p^2+12p-4)\sqrt{p^2-p+1}}{3(7p^3-9p^2+p+1)}$

Therefore, $q = \cfrac{-4p^3+4p^2+p-1+(4p^2-2p)\sqrt{p^2-p+1}}{7p^2-2p-1} = \cfrac{8p^4-21p^3+16p^2-11p+4+(-8p^3-4p^2+12p-4)\sqrt{p^2-p+1}}{3(7p^3-9p^2+p+1)}$ , which solves to $p = \frac{1}{5}(1 + 2\sqrt[3]{2} - \sqrt[3]{4})$ or $p = \frac{5}{8}$ for $0 < p < 1$ . However, $p = \frac{5}{8}$ leads to $q = 1$ , so it must be that $p = \frac{1}{5}(1 + 2\sqrt[3]{2} - \sqrt[3]{4})$ .

If $p = \frac{1}{5}(1 + 2\sqrt[3]{2} - \sqrt[3]{4})$ , then $u = \sqrt{p^2 - p - 1} = \frac{1}{5}(-1 + 3\sqrt[3]{2} + \sqrt[3]{4})$ .

The orange circle is an excircle to $\triangle BCD$ , so the ratio of the radii of the orange and cyan circles is $\cfrac{s_{\triangle BCD}}{s_{\triangle BCD} - p} = \cfrac{\frac{1}{2}(p + u + 1)}{\frac{1}{2}(p + u + 1) - p} = \sqrt[3]{2^4} - 1$

Therefore, $A = 2$ , $B = 4$ , $C = 3$ , $D = 1$ , and $ABCD = \boxed{24}$ .