$AX_i=4(1-\frac{1}{4^i}), BX_i=\frac{1}{4^{i-1}}$ $CY_i= 6(1-\frac{1}{2^i}),BY_i=\frac{3}{2^{i-1}}$ $AZ_i=6(1-\frac{1}{2^i}),CZ_i=\frac{3}{2^{i-1}}$ $AY_a,BZ_b,CX_c$ are concurrent if $\frac{BY_a}{CY_a} \frac{CZ_b}{AZ_b} \frac{AX_c}{BX_c}=1$ or $\frac{3}{2^{a-1}} \frac{3}{2^{b-1} }4(1-\frac{1}{4^c})=6(1-\frac{1}{2^a})6(1-\frac{1}{2^b})\frac{1}{4^{c-1}}$ or $ 4^c-1= (2^a-1)(2^b-1)$ $(2^a-1)(2^b-1) =2^{a+b}-2^a-2^b+1 <2^{a+b}-1$ so $2c < a+b$ So $2c \leq a+b-1$ and $2^{a+b}-2^a-2^b+1 \leq 2^{a+b-1}-1 \to 2^{a+b-1}-2^a-2^b+2\leq 0 \to (2^{a-1}-1)(2^{b-1}-1) \leq 0 \to a=1$ or $b=1$ If $a=1$ then we have $4^c-1=2^b-1 \to b=2c$ $b=1 \to a=2c$ So $(a,b,c)=(1,2c,c),(2c,1,c)$

Alternatively: Let WLOG $b \leq a$. Take modulo $2^b$ to see that $(-1)^2 \equiv -1 (mod 2^b)$ (because $b\leq 2c$), so $b=1$, $a=2c$.

The most overkill solution goes to me! Get to the equation $4^{c}-1=(2^{a}-1)(2^{b}-1)$ as shown above. Now we use the following well-known $\textbf{Lemma:}$ $2^{k}-1\mid 2^{\ell}-1\Longrightarrow k\mid \ell$. $\textbf{Proof:}$ Assume the contrary. Then there exist positive integers $p,q$, such that $\ell=pk+q$, $0<q<k$. Notice that: $$2^{k}-1\mid 2^{pk}-1\Longrightarrow 2^{pk}\equiv 1 (\text{mod }2^{k}-1)$$$$\Longrightarrow 1\equiv 2^{\ell}=2^{pk}\times 2^{q}\equiv 2^{q}(\text{mod }2^{k}-1)\Longrightarrow 2^{k}-1\mid 2^{q}-1$$However, $0<q<k$, so $0<2^{q}-1<2^{k}-1$, contradiction. Now since $(2^{a}-1)(2^{b}-1)=4^{c}-1=2^{2c}-1$, the Lemma implies that $a\mid 2c$ and $b\mid 2c$. If $a=2c$ or $b=2c$ we get the solutions $(a,b,c)=(1,2c,c)$ and $(a,b,c)=(2c,1,c)$. If $a\neq 2c\neq b$, then we reach a contradiction by: \[2^{2c}-1=(2^{a}-1)(2^{b}-1)\leq (2^{\frac{2c}{2}}-1)(2^{\frac{2c}{2}}-1)=(2^{c}-1)(2^{c}-1)<(2^{c}-1)(2^{c}+1)=2^{2c}-1\]

@above the most overkill is perhaps with Zsigmondy lol