Let's say that $\omega_b$ is tangent to $\omega_c$. We distinguish two cases: 1) $r_b + r_c = a$. In this case, we apply the formulae of $r_b$ and $r_c$: $(p-c) \cot{\frac{\alpha}{2}} + (p-b) \cot{\frac{\alpha}{2}} = a$ $\Leftrightarrow$ $(2p-b-c) \cot{\frac{\alpha}{2}} = a$ $\Leftrightarrow$ $a \cot{\frac{\alpha}{2}} = a$ $\Leftrightarrow$ $\boxed{\alpha = \frac{\pi}{2}}$. Now, to show that: $r_a - r_b = c$ $\Leftrightarrow$ $(p-b) \cot{\frac{\gamma}{2}} - (p-a) \cot{\frac{\gamma}{2}} = c$ $\Leftrightarrow$ $(a-b) \cot{\frac{\gamma}{2}} = c$. Now, we use the identity $\cot{\frac{x}{2}} = \frac{\sin{x}}{1-\cos{x}}$: $(a-b) \cot{\frac{\gamma}{2}} = c$ $\Leftrightarrow$ $(a-b)\frac{\frac{c}{a}}{1-\frac{b}{a}}=c$, which is obviously true. Hence, $r_a - r_b = c$, i.e. $\omega_a$ and $\omega_b$ are tangent. In a similar manner, it can also be shown that $\omega_a$ and $\omega_c$ are tangent. 2) $r_b - r_c = a$. This case is done similarily to the first, except that we reverse the steps of the proof: $r_b - r_c = a$ $\Leftrightarrow$ $p \tan{\frac{\beta}{2}} - (p-a) \cot{\frac{\beta}{2}} = a$ $\Leftrightarrow$ $p \tan^2{\frac{\beta}{2}}-(p-a)=a \tan{\frac{\beta}{2}}$ $\Leftrightarrow$ $\left( \tan{\frac{\beta}{2}} - 1 \right) $ $\left( p \tan{\frac{\beta}{2}} + p - a \right) = 0$. If the left part is zero, it means that $\tan{\frac{\beta}{2}} = 1$ $\Leftrightarrow$ $\boxed{\beta = \frac{\pi}{2}}$. If the right part canceles, we obtain a contradiction, since $\tan{\frac{\beta}{2}} \geq 0$, and $\frac{a-p}{p} < 0$. From now, the solution is straight-forward: $\beta = \frac{\pi}{2}$ $\Leftrightarrow$ $b \cot{\frac{\beta}{2}} = b$ $\Leftrightarrow$ $(2p-a-c)\cot{\frac{\beta}{2}} = b$ $\Leftrightarrow$ $(p-a)\cot{\frac{\beta}{2}} + (p-c)\cot{\frac{\beta}{2}} = b$ $\Leftrightarrow$ $r_c + r_a = b$, namely that $\omega_c$ and $\omega_a$ are tangent. I the same way, it can be shown that $\omega_a$ and $\omega_b$ are also tangent. $\square$

We find insimilicentres and exsimilicentres of $w_A, w_B, w_C$. $I_AI_BI_C$ be the excentral triangle, $A_1$ be tangency of $B$ excircle with $BC$. Note, parallel through $A$ to $CI_C$ intersects $CI_B$ at the insimilicentre of $(I_B), w_C$. This point be $P$, and again by Monge d'alembert the exsimilicentre of $w_B, w_C$ is lies on the line through $P$ and the midpoint of $I_BB$ and $BC$. Let it be $A_2$. We have $\dfrac{CP}{I_CP} = \dfrac{I_BA}{I_CA} = \dfrac{s-b}{s-c}$ so we can show, either by considering the midline of $I_BC, I_BB$ and taking perspectivity with respect to the midpoint of $I_BB$ or menelaus that $\dfrac{CA_2}{BA_2}= \dfrac{s-b}{s-c}$. If $A_3B_3C_3$ is the nagel triangle, then $(B, C; A_3, A_2) = -1 \implies B_3C_3 \cap BC = A_2$. So, the insimilicentres and exsimilicentres is the nagel triangle and its perspectrix, which we will call $A_2B_2C_2$. If $w_B \cap w_C \in BC = A_3$ then it becomes simple. Let the midpoint of $I_BI_C$ be $X$. Note, $XA_3=XC_3$ and $A_3C_3 \parallel CI_C$, so easy angle chasing means $X$ is centre of it, which by last years IMO P3 implies $ABC$ is right angled at $A$. The problem follows easily, and there must always by two internally as then the problem makes no sense