Couldn't solve this as well

I spent a lot of time thinking it synthetically . Still i ended up bashing it. Heres a solution involving some trignometric bash + inequalities. We denote the sides opposite to the vertices $A,B,C$ by $a,b,c$ and let $p$ be the semiperimeter i.e. $p=\frac{a+b+c}{2}$ Proof: Firstly we redefine some points. Rename $A'$ as $J$ and $K$ as $L$ . Now let $IL \cap AB=G$ and $IL \cap BC=K$. Now let $AB=c,BC=a,CA=b$. Note that $G,K$ lie on the $B$-mixtillinear incircle( follows from definition). From this link property 1 we have that$$BG=BK=\frac{ac}{p}$$. Now by sine rule in $\Delta BKG$ we have$$\frac{GK}{\sin (\angle B)}=\frac{BG}{\sin (90-\frac{\angle B}{2})}$$Since $BG=BK \implies \angle BGK=\angle BKG=90-\frac{\angle B}{2}$. For convinience we'll denote $\frac{\angle B}{2}=\alpha$. So rephrasing our last condition we have $$ \frac{GK}{\sin 2\alpha}=\frac{BG}{\sin (90-\alpha)}$$. Now well known fact that $\sin (90-\alpha)=\cos \alpha$ . So rephrasing our last condition we have$$ \frac{GK}{\sin 2\alpha}=\frac{BG}{\cos \alpha }$$. Now using the identity $\sin 2\alpha=2\sin\alpha \cdot \cos \alpha $ we can rephrase our last condition as $$ \frac{GK}{2\sin\alpha \cdot \cos \alpha}=\frac{BG}{\cos \alpha } \implies \frac{GK}{2\sin\alpha}=BG \implies GK=BG \times 2\sin \alpha=\frac {2ac \sin \alpha}{p}$$. Let $D,E$ be the midpoint of $AB,AC$ respectively. Note that$$GD=BG-BD= \frac{ac}{p}-\frac{c}{2}$$. Now let $DE \cap GK=T$ note that since$$DT \parallel BC \implies \frac{GD}{GB}=\frac{GT}{GK}$$by thales theorem. Now inputting the values we got earlier we have$$\frac{\frac{ac}{p}-\frac{c}{2}}{\frac{ac}{p}}=\frac{GT}{\frac {2ac \sin \alpha}{p}} \implies GT=2 \sin \alpha (\frac{ac}{p}-\frac{c}{2}).$$Now let $I_AJ \cap GK=L $ . Clearly since $\angle LIB=\angle LJB=90^\circ \implies ILBJ$ concyclic. Now observe that by power of point theorem $KI \times KL=KJ \times KB$. Now we'll prove something trivial. We prove that $BJ=\frac{a+b-c}{2}$. Let $AB \cap A$-excircle= $M$ and $AC \cap A$-excircle =$N$ . Note that we have $AM=AN$. We may write $AM=AB+BJ$ and $AN=AC+CJ$ . Now note that on subtracting the two equation $\implies c-b=CJ-BJ$ . Now since $BJ+CJ=BC \implies CJ=BC-BJ $ . Substituting this in previous equation $\implies c-b=BC-2BJ=a-2BJ \implies BJ =\frac{a+b-c}{2}$ Done. Now this together with $BK=\frac{ac}{p}\implies KJ=BK-BJ=\frac{ac}{p}-\frac{a+b-c}{2}$ . Also note that since $\angle GBI=\angle IBK,\angle GIB=\angle BIK=90,BI=BI \implies IG=IK=\frac{ac \sin \alpha}{p}$.Now putting these values in the power of point condition we have$$KI \times KL=KJ \times KB \implies \frac{ac \sin \alpha}{p} \times KL= (\frac{ac}{p}-(\frac{a+b-c}{2}))(\frac{ac}{p})\implies $$$$KL=\frac{\frac{ac}{p}-(\frac{a+b-c}{2})}{\sin \alpha}$$and so $$ GL= GK-KL=\frac{2ac \sin \alpha}{p}-\frac{\frac{ac}{p}-(\frac{a+b-c}{2})}{\sin \alpha}$$. Now we claim that if $\angle B \geq 60$ then infact $GL \geq GT$ and equality holds only when $\angle B=60$. Now if we are able to show this then we can have that $L$ never lies on the medial line iff $\angle B \leq 60$ which is what was intended to be proved. Because this is really a really powerful claim we state it clearly in the next line. Lemma: Whenever $\angle B \geq 60^ \circ \implies GL \geq GT $. Equality holds when $\angle B =60^ \circ$. Proof: We need to prove$$GL \geq GT \iff \frac{2ac \sin \alpha}{p}-\frac{\frac{ac}{p}-(\frac{a+b-c}{2})}{\sin \alpha} \geq 2 \sin \alpha (\frac{ac}{p}-\frac{c}{2}) \iff$$$$\frac{\frac{ac}{p}-(\frac{a+b-c}{2})}{\sin \alpha} \leq c \sin \alpha \iff \frac{\frac{ac}{p}-(\frac{a+b-c}{2})}{\sin^2 \alpha} \leq c $$. Now note that since $180^\circ > \angle B \geq 60 ^ \circ \implies 90^\circ > \alpha \geq 30 \implies 1 > \sin \alpha \geq 1/2$ So we have that$$\frac{\frac{ac}{p}-(\frac{a+b-c}{2})}{\sin^2 \alpha} \leq \frac{\frac{ac}{p}-(\frac{a+b-c}{2})}{(\frac{1}{2})^2}$$. So if we are able to prove$$\frac{\frac{ac}{p}-(\frac{a+b-c}{2})}{(\frac{1}{2})^2}=\frac{4ac}{p}-(2a+2b-2c) \leq c \iff \frac{4ac}{p} \leq 2a+2b-c \iff (2a+2b-c)(a+b+c) \geq 8ac$$we would be done. Note that by the cosine law for $\Delta ABC$ we have$$b^2=a^2-2ac \cos B +c^2 \geq a^2-ac+c^2$$since $\cos \angle B \leq \cos 60=\frac{1}{2}$. This would also imply $$b \geq \sqrt{ a^2-ac+c^2}$$Now expandng our inequality we have to prove$$2b^2+b(4a+c)+2a^2-7ac-c^2\geq 0 $$. Now using the bound on $b$ we obtained on the cosine law we have to prove$$2b^2+b(4a+c)+2a^2-7ac-c^2 \geq 2(a^2-ac+c^2)+\sqrt{ a^2-ac+c^2}(4a+c)+2a^2-7ac-c^2=$$$$\sqrt{ a^2-ac+c^2}(4a+c)+4a^2-9ac+c^2 \geq 0 \iff \sqrt{ a^2-ac+c^2}(4a+c) \geq -(4a^2-9ac+c^2)$$. Note that $$(4a+c)\sqrt{a^2-ac+c^2}=\sqrt{((4a-c)^2+16ac)((c-a)^2+ac)}\geq$$$$\geq(4a-c)(c-a)+4ac= -(4a^2-9ac+c^2) \square .$$where the last part follows from the cauchy schwartz inequality . Now for the equality case. Clearly here the equality would occur at $\angle B=60^\circ$ since in the beginning when we removed $\sin \angle \alpha$ we added that equality hols iff $\angle B=60^ \circ$ and this would happen for some specific sidelengths of $a,b,c$(infact the triplet $b=\frac{7a}{5},c=\frac{8a}{5}$.) Hence we are done with the problem.$\blacksquare$.

I think we could remove one step.

Sadly could not find any synthetic solution.

Attachments:
Problem 13.pdf (38kb)

The official was synthetic and i liked it since it made the use of a clever idea. Here it is. Proof: Let $AH_A$ be the altitude of triangle, $M$ be the midpoint of this altitude, and $N=AH_A \cap BI$. Then $A',I,M$ are the projections of $K$ to $BC, BI, AH_A$ respectively which are collinear, therefore, $BKNH_A$ is a cyclic quadrilateral and $\angle BKH_A = \angle BNH_A =90^\circ - \frac{\angle B}{2}$ Since the midpoint $M_C$ of $AB$ is equidistant from $B$ and $H_A$, and $M_CK \parallel BH_A$, we obtain that $\angle BKH_A \leq \angle BM_CH_A = 180^\circ-2\angle B$, this yields the required inequality $\blacksquare$.

A brilliant synthetic sol, hats off to Mr.Zaslavsky

Drawing the altitude $AH$ of $\triangle ABC.$ Let $KI $ intersects $AB, BC$ at $T, S. $ Let $(I)$ be the incircle of $\triangle ABC, $ let $(I) $ touches $BC $ at $D, $ let $ID$ intersects $(I)$ at $L,$ $AK $ intersects $BC$ at $R, $ $A'I$ intersects $AH$ at $M\implies BI\perp ST.$ $BI$ is the interior bisector of $\widehat{B}.$ $BI$ is the perpendicular bisector of the segment $ST. $ Therefore, $BST $ is an isoceles triangle at $B\implies BT=BS. $ $(I) $ is the midpoint of two segments $ST, LD. $ We need to prove that $SLTD $ is a parallelogram. $SL\parallel BA'. $ Let $Y, Z$ be the projection of $(I) $, $(I_A)$ at $AB.$ Applying Thales' theorem $\frac {AI}{AI_A}=\frac {IY}{I_AZ}=\frac {IL}{I_AA'}. $ $\widehat {AIL}=\widehat {AI_AA'}\implies $ $\triangle AIL\sim \triangle AI_AA'\implies\widehat{IA\ell}=\widehat {I_AA\ell}=\widehat {I_AAA'}.$ $3$ points $\overline {A,\ell, A'}. $ $D\ell$ is the diameter of $\odot(I), $ $DI $ passes through $\odot(I),$ $(I) $ is the incentre of $\triangle ABC.$ Note, $I $ is the midpoint of $D\ell. $ Again applying Thales', $\frac {I\ell}{AM}=\frac {A'I}{A'M}=\frac {ID}{MH}.$ $M $ is the midpoint of $AH.$ Consider $P, Q $ be the midpoints of $AB, AC. $ $Q $ is the midpoint of $AC,$ $MQ\parallel BC.$ Note that, $PQ\parallel BC $ $\overline {M,P,Q}, KM\parallel RH.$ $K $ is the midpoint of $AR.$ $KM\parallel CB, KA'\parallel CB\implies KM\parallel A'H, $ and $KA'\parallel MH.$ So $KMHA'$ is a parallelogram $\implies MK=A'H.$ $K $ is the midpoint of $AR, $ $KA'\parallel AH\implies $ $A'$ is the midpoint of $RH. $ By Menelaus' theorem on $\triangle ARB\implies\overline {K, S, T}.$ $\frac {BS}{AS}\cdot\frac {AK}{RK}\cdot\frac {TR}{TB}=1. $ Because $AK=RK, TB=SB, AS=RT. $ Applying $\frac {A'H}{A'T}=\frac {MK}{A'T}=\frac {IM}{IA'}=\frac {\ell A}{\ell A'}=\frac {AS}{SB}, $ then $\frac {A'R}{A'T}=\frac {TR}{TB}=\frac {TR-A'R}{TB-A'T}=\frac {A'T}{A'B}\implies A'T^2=A'R\cdot A'B=A'H\cdot A'B.$ Claim- For $2$ non-negative integers $x $ and $y $ we will have an inequality $4xy\le (x+y)^2$ from $(x-y)^2\ge 0, $ $x^2-2xy+y^2\ge 0.$ So, $x^2+2xy+y^2\ge 4xy, 4xy\le (x+y)^2.$ Proof- Applying the inequality for the non-negative integers, $HA'$ and $A'B$ gives $4\cdot A'H\cdot A'B\le\left(A'H+A'B\right)^2=HB^2. $ Thus, $HB^2\ge 4\cdot A'T^2, $ hence, $HB\ge 2\cdot A'T.$ $HB $ and $A'T>0.$ Note that, $A'H=A'R $ as $A'$ is the midpoint of $RH $. $\implies 2\cdot A'T=A'H+HT+RT-A'R=HT+RT $ Thus, $HB\ge TH+RT\implies BH+RT-HT\ge 2TR, RT>0$ Also, $HB+HR\ge 2TR, RB(RH+BH)\ge 2BR\cdot TR,$ and $RB>0.$ $RB=BH-RH\implies RB(BH+RH)=(BH-RH)(BH+RH)=HB^2-R^2H.$ So, $HB^2-HR^2\ge 2BR\cdot TR.$ Applying Pythagoras's theorem $HB^2-HR^2=AB^2-AH^2-(AR^2-AH^2)=AB^2-AR^2, $ we get $AB^2-AR^2\ge 2BR\cdot TR.$ After that, $AB^2-BR\cdot 2RT\ge AR^2.$ $BT-BR+AS=BT-BR+AB-SB=AB-RB=2RT\implies AB^2-BR\cdot(AB-RB)\ge AR^2.$ Thus, $AB^2+BR^2-AB\cdot RB\ge AR^2.$ By cosine theorem on $\triangle ABR,$ $AB^2+BR^2-2AB\cdot BR\cos\widehat{B}=AR^2. $ $AB^2+BR^2-BA\cdot RB\ge AB^2+ BR^2-2AB\cdot RB\cos\widehat {B}.$ Hence, $-BA\cdot BR\ge -2AB\cdot RB\cos\widehat{B},$ and $\cos\widehat{B}\ge\tfrac 12.$ It is proved that $\widehat {B}\le 60^{\circ}. $