First note that $A_1,B_1$ lie on $BC,AC$ and $AI=A_1I, BI=B_1I$. calculating the medians in two triangles $AIB, A_1IB_1$, we see that $$4IN^2=2AI^2+2BI^2-A_1B_1^2, 4IM^2=2AI^2+2BI^2-AB^2$$so its enough to prove $AB>A_1B_1$, Set the middles of arcs $ACB, AC$ on the circumcircle as $T,R$. by a spiral similarity, Triangles $TAB,TB_1A_1$ are similar and isosceles. so it is enough to prove that $TA>TA_1$. now note that $R$ must lie on arc $AT$ of the circle so $A$ and $T$ lie on different sides of line $BI$. however $BI$ is the perpendicular bisector of $AA_1$, This give us the desired $AT>A_1T$.

trivialized by appolonius (the only problem i could do)

I felt it was a bit straightforward, even for a $P4$, but on the other hand $P5$ and $P6$ greatly made up for that. 1) $A_1$, $C$, and $B$ are in one line and $B_1$, $C$, and $A$ are in one line. Let the midpoints of $AA_1$ and $BB_1$ be $F$ and $G$ respectively, since $BI$ is the bisector of $\angle CBA$ and $BI \perp AA_1$ we'll have that $A_1$, $C$, and $B$ are in one line. Analogously for $B_1$, $C$, and $A$. 2) FNGM - Rhombus By midlines it's easy to see that $FN = \frac{1}{2}AB_1=GM$ and $NG= \frac{1}{2}A_1B=FM$ (Also parallel). But we also have that $A_1B=AB=AB_1$ so $FN=GM=NG=FM$, which means $FNGM$ is a rhombus. 3) $I$ and $M$ lie on the same side of $FG$ For $I$ and $M$ to lie on the same side of $FG$, we just need to prove that the perpendiculars from $A$ and $B$ to $BI$ and $AI$ lie on the rays $BI$ and $AI$, respectively and not their segments. Which is easy to see since $\angle AIB = 90^{\circ}+\frac{A}{2}$ which means the perpendiculars can never land on the segments. This is enough to say that $IN>IM$ since $N$ and $M$ are symmetric to each other with respect to $FG$.

Using the median formula this is equivalent to show that $A_1B_1<AB$. By the reflections we have $A_1B=AB=AB_1$ and this means that there is a spiral similarity centered at $D$, the midpoint of the arc $BAC$ on $(ABC)$ that sends $A_1B_1 \to AB$ also note that $A_1$ lies on $BC$ and $B_1$ lies on $AC$. Case 1: At least one of the $A_1,B_1$ is inside the segments $AC,BC$. In this case WLOG $B_1 \in \overline{BC}$ we clearly have that if $A_1B_1 \ge AB$ then $DB_1 \ge DB$ which is not possible. Hence herr $A_1B_1<AB$ holds always. Case 2: Both $A_1,B_1$ are outside $\triangle ABC$. By an easy angle chasing we get $BB_1CI, AA_1CI$ cyclic and $\angle B_1IA_1=\frac{3 \angle ACB}{2}-90$. Now assume that $A_1B_1 \ge AB$ hence that would mean that $\angle B_1A_1A \le \angle B_1AA_1$ and $\angle A_1B_1B \le \angle B_1BA_1$, adding both and using the cyclic quads we get that $\angle B_1IA_1 \ge \angle BB_1A_1+\angle AA_1B_1$ now by angle chase we have $\angle BB_1A_1+\angle AA_1B_1=180-\angle ACB+90+\frac{\angle ACB}{2}=270-\frac{\angle ACB}{2}$ and using the prevoius result we have that $\angle ACB \ge 180$ which is not possible hence $A_1B_1<AB$ holds always.

The easiest way to prove $A_1B_1<AB$ is with Law of Cosines in triangle $A_1B_1B$ (this is for the case $\beta \geq \gamma$, the other cases are similar). The inequality (using $BA_1=c$, $BB_1=2c.sin(\alpha/2)$) is equivalent to $sin(\alpha/2) \leq sin(\alpha/2+\gamma)=cos((\beta-\gamma)/2)=cos(\angle A_1BB_1)$, and this is obvious since the both angles are in $[0, \pi/2]$

By Stewart's theorem, it suffices to show that $A_1B_1 < AB$. Next, we clearly have $IA_1 = IA$ and $IB_1 = IB$. By angle chasing, $\angle AIB = 90^\circ + \frac{1}{2}\angle C$ and $\angle A_1IB_1 = |360^\circ - \angle A_1IA - \angle AIB - \angle BIB_1| = |\frac{3}{2}\angle C - 90^\circ |$. It follows that $\angle A_1IB_1 < \angle AIB$, and we are done by LoC in triangles $A_1IB_1$ and $AIB$.

Let $A_0$ and $B_0$ be the feet of the bisectors. By symmetry $IA_1 = IA$, $IB_1 = IB$, so by the median formula it suffices to show $AB > A_1B_1$. But then by the Cosine Law and $\angle AIB = \angle A_0IB_0$ it suffices to show $\angle A_0IB_0 > \angle A_1IB_1$, which is clear, as the rays $IA_1$ and $IB_1$ lie inside $\angle A_0IB_0 > 90^{\circ}$.

Let Q and S be the midpoints of the segmets BA1 and AB1 (sorry, i dont know how to use latex, so just pretend that "1" is in index). We get SN//AA1//MQ and NQ//BB1//MS which means that MSNQ is parallelogram. İts not hard to see that IQ=IM=IS. Let W be a circle centered at I with radius IM, we know that Q , S and M all lie on W. İt's enough to show that N lies outside of the circle W, which is equal to showing (<SNQ + <SMQ) is less than 180. Let AA1 and BB1 intersect at R. İt is not hard to prove that (<BAC + <CBA)/2 = <ARB = <SNQ = <SMQ and from here, the rest is obvious.

Notice that both $IN$ and $IM$ are medians, then we need to show that: $$\frac{1}{2}\sqrt{2IB_1^2+2IA_1^2-A_1B_1^2} > \frac{1}{2}\sqrt{2IB^2+2IA^2-AB^2}$$which implies that $AB > A_1B_1$, thus if we show this we are done. By angle chase, we have that $\angle A_1IB_1 \equiv \frac{3}{2}\gamma - 90 \pmod{\pi}$ which is in both cases less that $\angle AIB = 90 + \frac{1}{2}\gamma$. Since triangles $AIB$ and $A_1IB_1$ have two same sides and because $\angle A_1IB_1 < \angle AIB$, we have that $AB > A_1B_1$

The only reason I did not like this problem is that it can be easily solved by vectors (probably by complex numbers too, with similar solution). On the other hand, maybe it is good to have geo problems, which can be solved with vectors

Really liked this problem

$\angle ABI = \angle CBI = \angle A_1BI \implies $ $B,C,A_1$ collinear.Similarity $A,C,B_1$ collinear. $$AI= A_1I = x , BI = B_1I = y$$$$IN = \frac{1}{2} \cdot \sqrt{2(x^2+y^2)- A_1B_1^2} $$$$IM = \frac{1}{2} \cdot \sqrt{2(x^2+y^2)- AB^2} $$it suffices to prove $AB > A_1B_1$ $\angle ACB = \alpha,AB =c,BC = a,CA =b$ $$A_1B_1^2 = (c-b)^2+(a-c)^2+2(c-b)(a-c) \cdot cos\alpha,AB^2 = a^2+ b^2 - 2ab cos\alpha$$$$c^2-2cb + b^2 + a^2 -2ac + c^2 + 2(ca - c^2- ab +bc) cos \alpha < a^2+ b^2 - 2ab cos\alpha$$$$2c^2(1-cos \alpha) + 2bc(cos \alpha -1)+ 2ac(cos\alpha-1) <0$$$$(1-cos\alpha) > 0 \implies 2bc - 2c^2+ 2ac >0 \implies a+b >c$$$\blacksquare$

$IN^2 = \frac{{IA_1}^2}{2} + \frac{{IB_1}^2}{2} - \frac{{A_1B_1}^2}{4}$. $IM^2 = \frac{{IA}^2}{2} + \frac{{IB}^2}{2} - \frac{{AB}^2}{4}$. so we need to prove $A_1B_1 < AB$. Let $AI = IB_1 = x$ and $BI = IA_1 = y$. $AB^2 = x^2 + y^2 - 2xy \cos{(\angle 90 + \frac{\angle{c}}{2})}$. $A_1B_1^2 = x^2 + y^2 - 2xy \cos{(\angle 90 - \frac{\angle{3c}}{2})}$. so $A_1B_1 < AB$. we're Done.

Cute way to prove that $AB>A_1B_1$ We have that $A_1B=AB$ and $A_1'B=A_1B_1$, where $A_1'$ is reflection of $A_1$ in $AI$. Since $A_1$ lies on the opposite side of $AI$ to $B$, $A_1B>A_1'B$ (by triangle inequality) so we are done.