Twins in the Trigon

From the figure :

$\begin{aligned} \tan{\theta} = \dfrac{3}{4} = \dfrac{y}{x} \hspace{10mm} & \boxed{4y-3x=0} \\ (3-y-r)+2r+(4-x-r) = 5 & \hspace{2mm} \boxed{x+y=2} \end{aligned}$

$\hspace{1mm} y = \dfrac{6}{7} \hspace{1mm},\hspace{1mm} x = \dfrac{8}{7}$

$\hspace{1mm} x^2+y^2 = 4r^2 \\ \hspace{1mm} \boxed{r = \dfrac{5}{7}}$

From the figure above,

$tan$ $2Ф=$ $\frac{4}{3}$

$Ф=26.56505118$

$x=$ $\frac{r}{tan 26.56505118}$

$tan$ $2β=$ $\frac{3}{4}$

$β=18.43494882$

$y=$ $\frac{r}{tan18.43494882}$

$x + y + 2r = 5$

$\frac{r}{tan 26.56505118} +$ $\frac{r}{tan18.43494882} + 2r = 5$

$7r = 5$

$r =$ $\frac{5}{7}$

$A+B=5+7=$ $\boxed{12}$

Let's consider the circle tangent to the $x$ -axis. Its center $C_1=(x,r)$ , where $r$ is the radius and $x$ an unknown coordinate. The circle is also tangent to the hypotenuse, which equation is $y=-\frac{3}{4}x+3$ . Hence, we have

$\displaystyle \frac{|\frac{3}{4}x+r-3|}{\sqrt{\frac{9}{16}+1}}=r \hspace{5pt} \Longrightarrow \hspace{5pt} x=4-3r$

Considering $|\frac{3}{4}x+r-3|=-\frac{3}{4}x-r+3$ .

So, $C_1=(4-3r,r)$ . The same applies for the other center, which coordinates are $C_2=(r,y)$ . Hence

$\displaystyle \frac{|\frac{3}{4}r+y-3|}{\sqrt{\frac{9}{16}+1}}=r \hspace{5pt} \Longrightarrow \hspace{5pt} y=3-2r$

Considering $|\frac{3}{4}r+y-3|=-\frac{3}{4}r-y+3$ .

So, $C_2=(r,3-2r)$ . The tangent circles have the same radius, so the distance between $C_1$ and $C_2$ is exactly

$\displaystyle \sqrt{(4-3r-r)^2+(r-3+2r)^2}=2r \hspace{5pt} \Longrightarrow \hspace{5pt} r=\frac{5}{7}$

considering the smallest of the two possible solutions. Eventually

$\displaystyle r=\frac{A}{B}=\frac{5}{7} \hspace{5pt} \Longrightarrow \hspace{5pt} A+B=\boxed{12}$

From the figure equating the length of the hypotenuse we get,

$2r+\dfrac{r}{\tan\dfrac{\theta}{2}}+\dfrac{r}{\tan\dfrac{\phi}{2}}=5 \rightarrow(1)$

also we have $\tan\dfrac{x}{2}=\dfrac{1-\cos x}{\sin x}$

here $\sin\theta=\dfrac{4}{5},\cos\theta=\dfrac{3}{5},\sin\phi=\dfrac{3}{5},\cos\phi=\dfrac{4}{5}$

thus we get, $\tan\dfrac{\theta}{2}=\dfrac{1-\dfrac{3}{5}}{\dfrac{4}{5}}=\dfrac{1}{2}$

and $\tan\dfrac{\phi}{2}=\dfrac{1-\dfrac{4}{5}}{\dfrac{3}{5}}=\dfrac{1}{3}$

Substituting in $(1)$ we get,

$2r+2r+3r=5$ or $r=\dfrac{5}{7}$

thus the required answer is $5+7=\boxed{12}$