Three small circles

First, let the first radius be $r_a$ , the second one $r_b$ and the third one $r_c$ . Also, let $r$ be the inradius of $\triangle ABC$ . Now, with angle chasing and the general properties of the angle bisector, we get:

$r_a=r\left(\dfrac{1-\sin(\frac{A}{2})}{1+\sin(\frac{A}{2})}\right)$ , $r_b=r\left(\dfrac{1-\sin(\frac{B}{2})}{1+\sin(\frac{B}{2})}\right)$ , $r_c=r\left(\dfrac{1-\sin(\frac{C}{2})}{1+\sin(\frac{C}{2})}\right)$

Now, we will derive a relationship with $r$ , $r_a$ , $r_b$ and $r_c$ .

From $\dfrac{1-\sin(\frac{\theta}{2})}{1+\sin(\frac{\theta}{2})}$ we get $\tan^2 \left(\dfrac{\pi-\theta}{4}\right)$ with few trigonometric identities.

So, $r_a=r\left(\tan^2 \left(\dfrac{\pi - A}{4}\right)\right)$ .

Take square root to both sides and get: $\sqrt{r_a}=\sqrt{r}\tan\left(\dfrac{\pi - A}{4}\right)$

Similarly, $\sqrt{r_b}=\sqrt{r}\tan\left(\dfrac{\pi - B}{4}\right)$ and $\sqrt{r_c}=\sqrt{r}\tan\left(\dfrac{\pi - C}{4}\right)$

From the identity $\tan\left(\dfrac{A}{2}\right)\tan\left(\dfrac{B}{2}\right)+\tan\left(\dfrac{A}{2}\right)\tan\left(\dfrac{C}{2}\right)+\tan\left(\dfrac{B}{2}\right)\tan\left(\dfrac{C}{2}\right)=1$ with $A+B+C=\pi$ , we obtain multiplying the three previous equations, two at time:

$r=\sqrt{r_a*r_b}+\sqrt{r_a*r_c}+\sqrt{r_b*r_c}$

Substituting the known values of $r_a$ , $r_b$ and $r_c$ :

$r=\sqrt{\left(\dfrac{73-\sqrt{145}}{36}\right)\left(\dfrac{66-8\sqrt{29}}{25}\right)}+\sqrt{\left(\dfrac{73-\sqrt{145}}{36}\right)\left(18-8\sqrt{5}\right)}+\sqrt{\left(\dfrac{66-8\sqrt{29}}{25}\right)(18-8\sqrt{5})}$

And simplifying a lot, we obtain that $r=2$ .

Next, from the three first equation that we obtained, we're going to solve for the three sines of $\triangle ABC$ :

$\dfrac{73-\sqrt{145}}{36}=2\left(\dfrac{1-\sin(\frac{A}{2})}{1+\sin(\frac{A}{2})}\right) \Rightarrow \sin(\frac{A}{2})=\dfrac{\sqrt{145}}{145}$

$\dfrac{66-8\sqrt{29}}{25}=2\left(\dfrac{1-\sin(\frac{B}{2})}{1+\sin(\frac{B}{2})}\right) \Rightarrow \sin(\frac{B}{2})=\dfrac{2\sqrt{29}}{29}$

$18-8\sqrt{5}=2\left(\dfrac{1-\sin(\frac{C}{2})}{1+\sin(\frac{C}{2})}\right) \Rightarrow \sin(\frac{C}{2})=\dfrac{2\sqrt{5}}{5}$

Obtain the sines with the double angle formula and the Pythagoeran identity:

$\sin A=\dfrac{24}{145}$ , $\sin B=\dfrac{20}{29}$ and $\sin C=\dfrac{4}{5}$

With the inradius and the sines of the three angles of $\triangle ABC$ , it's easy to obtain every side.

We know that $[ABC]=\dfrac{ab\sin C}{2}=\dfrac{ac\sin B}{2}=\dfrac{bc\sin A}{2}$ and $r=\dfrac{2[ABC]}{a+b+c}$ , so form a equation system:

$2=\dfrac{ab *\frac{4}{5}}{a+b+c} \Rightarrow 2(a+b+c)=\dfrac{4ab}{5}$

$2=\dfrac{ac *\frac{20}{29}}{a+b+c} \Rightarrow 2(a+b+c)=\dfrac{20ac}{29}$

$2=\dfrac{bc *\frac{24}{145}}{a+b+c} \Rightarrow 2(a+b+c)=\dfrac{24bc}{145}$

From the first and the second equation, we get $b=\dfrac{25c}{29}$ ; and from the first and the third equation, we get $a=\dfrac{6c}{29}$ . Substitute them in any equation, let's choose the first one:

$2\left(\dfrac{6c}{29}+\dfrac{25c}{29}+c \right)=\dfrac{4(\frac{6c}{29})(\frac{25c}{29})}{5} \Rightarrow c=29$

With $c$ , obtain $a$ and $b$ :

$a=\dfrac{6*29}{29} \Rightarrow a=6$

$b=\dfrac{25*29}{29} \Rightarrow b=25$

Finally, $10000a+100b+c=\boxed{62529}$ .

Firstly I would have also written :

${ r }_{ a }=r(\frac { 1-sin(\frac { A }{ 2 } ) }{ 1+sin(\frac { A }{ 2 } ) } )$

${ r }_{ b }=r(\frac { 1-sin(\frac { B }{ 2 } ) }{ 1+sin(\frac { B }{ 2 } ) } )$

${ r }_{ c }=r(\frac { 1-sin(\frac { C }{ 2 } ) }{ 1+sin(\frac { C }{ 2 } ) } )$

Then I delibrately manipulated your given values like that

${ r }_{ a }=2(\frac { { (\sqrt { 145 } -1) }^{ 2 } }{ { \sqrt { 145 } }^{ 2 }-{ 1 }^{ 2 } } )=2\frac { (1-\frac { 1 }{ \sqrt { 145 } } ) }{ (1+\frac { 1 }{ \sqrt { 145 } } ) }$

${ r }_{ b }=2\frac { { (\sqrt { 29 } -2) }^{ 2 } }{ { \sqrt { 29 } }^{ 2 }-{ 2 }^{ 2 } } =2(\frac { 1-\frac { 2 }{ \sqrt { 29 } } }{ 1+\frac { 2 }{ \sqrt { 29 } } } )$

${ r }_{ c }=2(\frac { { (\sqrt { 5 } -2) }^{ 2 } }{ { \sqrt { 5 } }^{ 2 }-{ 2 }^{ 2 } } )=2(\frac { 1-\frac { 2 }{ \sqrt { 5 } } }{ 1+\frac { 2 }{ \sqrt { 5 } } } )$

Then it is easy to see that :

$r=2 \quad ,\quad sin(\frac { A }{ 2 } )=\frac { 1 }{ \sqrt { 145 } } \quad ,\quad sin(\frac { B }{ 2 } )=\frac { 2 }{ \sqrt { 29 } } \quad ,\quad sin(\frac { C }{ 2 } )=\frac { 2 }{ \sqrt { 5 } }$

To find the sides use this simple identity

$a=r(cot\frac { B }{ 2 } +cot\frac { C }{ 2 } )$

$b=r(cot\frac { A }{ 2 } +cot\frac { C }{ 2 } )$

$c=r(cot\frac { A }{ 2 } +cot\frac { B }{ 2 } )$

Now $cot\frac { A }{ 2 } =12$ , $cot\frac { B }{ 2 } =\frac { 5 }{ 2 }$ , $cot\frac { C }{ 2 } =\frac { 1 }{ 2 }$

Put the values to get :

$a=6,b=29,c=25$

Hence we have $\boxed { 10000a+100b+c=62529 }$