The B-Humpty point is the centroid. Since the humpty point lies on the apollonius circle, by apollonius formula, we have either a=c, or $a^2+c^2=2b^2$. If a=c, the triangle is equilateral, if not, $cos \beta = \frac{b^2}{2ac}$ By law of cos we have $\frac{b^2+2ac cos\beta}{2ac} = \frac{a^2+c^2}{2ac} \stackrel{AM-GM}{\ge} 1$, and using $cos\beta$, we have $cos\beta=\frac{b^2}{2ac}\ge\frac{1}{2}$. So, $\beta\le$60

What is minimum value?

RagvaloD wrote: What is minimum value? Let $E$ be the midpoint of $AC$, then $EA_1= EC_1=\frac{1}{2}AC= \frac{b}{2}$ and $EA_1$, $EC_1$ are tangent to $\odot (BA_1C_1H)$. Thus \[ \frac{1}{4} b^2 = EA_1^2 = EG \cdot EB = \frac{1}{3} EG^2 \]Note that $EG^2 = \frac{2a^2+2c^2 -b^2}{4}$, we must have \[ a^2+c^2 = 2b^2 . \tag{$\star$}\]On the other hand $b^2 = a^2+c^2 -2ac \cos B$. Therefore, \[ a^2 + c^2 -4ac \cos B =0 \tag{$\star \star$}\]Consider $(\star \star)$ as a quadratic equation in $a$, a solution exists iff $4 \cos^2 B \geq 1 $, i.e. $\cos B \geq \frac{1}{2}$. Now, assume that $c>a$, then the assumption of acute triangle requires $a^2+b^2 >c^2$, together with ($\star$) , we have $2a^2 > b^2 = 2ac \cos B$, hence $a > c \cos B$. On the other hand, from ($\star \star$), we have \[ a = 2c \cos B - c\sqrt{4\cos^2 B -1 } \]Basic calculus implies that $\cos^2 B \leq \frac{1}{3}$. Therefore, $\cos B \in (\frac{1}{\sqrt{3}}, \frac{1}{2}]$.