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There are two solutions. The first one is easier and shorter. The second one has more theory in it.

Solution 1:

Here, let $FD=FE=r$ and $AB=AE=R$

$AE=EF+FA=r+\sqrt{FD^2+DA^2} = r+\sqrt{r^2+(\frac{R}{2})^2}$

$\implies R=r+\sqrt{r^2+(\frac{R}{2})^2}$

$\implies (R-r)^2 = r^2+\frac{R^2}{4}$

$\implies R^2+r^2-2Rr = r^2+\frac{R^2}{4} $

$\implies 3R^2 = 8Rr \implies \frac{r}{R}=\frac{3}{8}$

From the centers of the circles, $BC=AB=AC=R$

So, $\angle BAC = 60$.

So, the arc length of $BC$ is one sixth of the circumference of $\Gamma$.

So, the circumference of $\Gamma$ is $6\cdot 12=72$.

If the circumference of the circle $DEG$ is $t$ and the circumference of the circle with center $A$ through $B$ is $T$, we know that

$\frac{t}{T} = \frac{3}{8}$ because circumference is proportionate to radius.

Given that $T=72$ from the question, $t=\frac{72\cdot 3}{8}=27$

So, the circumference of the circle $DEG$ is 27.

Solution 2:

Lemma:

Let $AB$ be a chord in circle $\gamma$. A circle $\omega$ touches $\gamma$ at $X$ and $AB$ at $Y$. If $Z$ is the intersection of $XY$ and $\gamma$, then

$Z$ is the midpoint of arc $AB$.

$ZA^2=ZX\cdot ZY$

This lemma is commonly known as Shooting Lemma.

Proof:

For the first part, 

a homothety centered at $X$ of ratio $\frac{ZX}{YX}$ takes $\omega$ to $\gamma$. This also takes $Y$ to $Z$.

Now, homothety takes a line to another line parallel to the previous. So, if $AB$ is taken to $CD$, then $AB||CD$ and $CD$ is tangent to $\gamma$ at $Z$.

    $$\angle ZAB = \angle CZA =\angle ZBA $$

From this, we have $ZA=ZB \implies$ $Z$ is the arc midpoint of arc $AB$.

For the second part,

\[\angle AXZ = \angle ABZ =\angle BAZ =\angle YAZ\]

So, $ZA$ is tangent to the circumcircle of $AZB\implies ZA^2=ZX\cdot ZY$.

Solution:

Let the circle through $B$ with center $A$ be $\Gamma$ and the circle through $A$ with center $B$ be $\Gamma'$.

Let the radius of $\Gamma$ be $a$.

From the centers of the circles, $BC=AB=AC=a$

So, $\angle BAC = 60$.

So, the arc length of $BC$ is one sixth of the circumference of $\Gamma$.

So, the circumference of $\Gamma$ is $6\cdot 12=72$.

Let the circle tangent to $AB$ and the arcs $BC,AC$ be $\Omega$

Let the point where $\Omega$ touches $BC$ and $AB$ be $F$ and $E$ respectively.

Let $AB$ intersect $\Gamma$ at $H$ and $FD$ at $E$.

Using the first part of the shooting lemma, $D$ is the midpoint of arc $BH$. 

Also, $A$ is the midpoint of segment $BH \implies \angle DAB=90$

$\Gamma$ and $\Gamma'$ are symmetric which implies $E$ is the midpoint of $AB$.

So,

$BD=\sqrt{BA^2+DA^2}=\sqrt{2}a$

$ED=\sqrt{EA^2+DA^2}=\frac{\sqrt{5}}{2}a$

Using the 2nd part of the shooting lemma,

\[DE\cdot DF = DB^2\implies DF=\frac{DB^2}{DE}=\frac{2a^2}{\frac{\sqrt{5}}{2}a} = \frac{4}{\sqrt{5}}a\]

Now, $EF=DF-DE=\frac{4}{\sqrt{5}}a-\frac{\sqrt{5}}{2}a=(\frac{8-5}{2\sqrt{5}})a=\frac{3}{2\sqrt{5}}a$.

If the circumference of $\Omega$ is $K$, using thr homothety centered at $F$ with ratio $\frac{FE}{FD}$, we have

\[\frac{K}{72}=\frac{EF}{DF}=\frac{EF}{DF}=\frac{\frac{3}{2\sqrt{5}}a}{\frac{4}{\sqrt{5}}a}=\frac{3}{8}=\frac{K}{72}\implies K=27\]

So, the circumference of $\Omega$ is $27$.