$a)$ Easy to see $(I)$ is the $A-$mixtilinear circle (outside) $\implies$ $\triangle{DA_bP},\triangle{A_cA_bD},\triangle{A_cDN}$ are similar ($D$ midpoint $PN$ so $D$ is the excenter) From here using ratios and $AM-GM$ for $2$ variable we receive $A_bP+A_cN \ge NP$. Simiarly ,sum them up we get the result. $Q.E.D$ $b)$ Denote the triangle formed by 3 lines is $\delta$. Simple angle chasing gives $BA_bA_cC,AB_aB_cC,AC_aC_bB$ are cyclic $\implies$ $\delta$ is homothetic to the orthic triangle $\psi$. If $A_bA_c \cap B_aB_c$ at $L$ $\implies$ $\triangle{LA_bB_a}$ isosceles. Our mission now is to prove $PA_b=PB_a$ (Important part in the problem) $Z$ midpoint $PM$ By similar triangles $A_bP=PD\frac{A_bD}{DA_c}$ $B_aP=PZ\frac{B_aZ}{B_cZ}$ Construct the extriangle $\triangle{UVT}$ Easy to see $\triangle{UCB} \sim \triangle{DA_bA_c}$ and similar to other triangles. We need to show $\frac{PD}{PZ}\frac{A_bD}{DA_c}=\frac{B_aZ}{B_cZ}<=>\frac{PN}{PM}\frac{CU}{BU}=\frac{VC}{VA}$ $\triangle{MNP} \sim \triangle{UBC} \sim \triangle{AVC} \sim \triangle{ABT}$. Plug in the ratios $\implies$ $P$ midpoint $A_bB_a$. Similarly so $I$ is also the incenter of $\delta$. Observe $H$ is the incenter of the orthic circle $\psi$. $K,O_a$ are circumcenters of $\delta,\psi$($O_a$ is the nine-point circle center and it is well known that $\overline{O,O_a,H}$). By homothety, we are done. $Q.E.D$

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Comment: A bit better than P4, but still tricky to find the solution, especially in the examination room. Also, someone please add these problem to the problem collection soon.

Extra problem on this configuration (Old) (Teacher Nguyen Van Linh) $J$ is the radical center of $\Omega_A,\Omega_B,\Omega_C$. $G_e$ is the Gergonne point $\implies$ $\overline{G_e,I,J}$.

Paramizo_Dicrominique wrote: $a)$ Prove $B_cC_b+C_aA_c+A_bB_a \ge NP+PM+MN$. It appears that $B_cC_b=B_cM+C_bM={\left[r\cot\frac{B}2-{\left(r\csc\frac{B}2-r\right)}\right]}+{\left[r\cot\frac{C}2-{\left(r\csc\frac{C}2-r\right)}\right]}$ and $NP=2r\cos\frac{A}2$; isn't it \(B_cC_b+C_aA_c+A_bB_a \le NP+PM+MN\)?

b) Redefine $A_b, A_c$ as $AA_b \cdot AB=AA_c \cdot AC=AA' \cdot AM$. Then by inversion with radius $AP=AN$ we get $AA_bA'A_c$ is tangent to $(I)$. Thus $BA_bA_cC$ is cyclic which implies orthic triangle $DEF$ and $\Delta$ are homothetic (parallel sides) where $\Delta$ is the triangle formed by the lines $A_bA_c,B_cB_a,C_aC_b$. $OH$ is the $"OI"$ line of $DEF$ and $IK$ is the $"OI"$ line of $\Delta$ which gives the result.