For some weird reason I fake solved it horribly and I proved that $IK>JK$ Sorry for such a fakesolve

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SharyginP24.pdf (17kb)

Let $\ell$ be the circumcentre of $\triangle BCD.$ A line is passing through $\ell$ and $\perp\odot(BCD)$ cuts the circumsphere of $ABCD $ at $K'.$ Note, $K'B=K'C=K'D.$ Let $I'$ be the intersection of the sphere of centre $K,$ radius $KB$ and $AK. $ Distances from all the points on $\odot(BCD)$ to $K$ are equal. $AK$ lies on the bisector of $\odot(ABC), \odot(ABD).$ Bisector of $\odot(ABC)$ and $\odot(ABD);$ $\odot (ABD), \odot(ACD);$ $\odot(ACD), \odot(ABC)$ intersects the points from the line $AK'$ respectively, because $I $ and $J$ lies on the plane formed by interesection points($\textbf {mentioned as above}$). Note, $\overline {A,K',I,J}.$ So, $K\equiv K'.$ Let $\mathcal {X}=AK\cap\odot (BCD),$ $B\mathcal{X}$ intersects the circumsphere of $ABCD$ at $B'.$ Similarly, we can define $D'$ and $C'.$ $A,B,I',B',\mathcal {X},K$ are on the same plane, $B,B',K',A $ all lie on a same circle. Note, $KI'=KB\implies \widehat{BI'K}=\widehat {KBI'}.$ So, $\widehat{ABI'}+\widehat {BAK}=\widehat{B'BK}+\widehat{I'BB'}\implies\widehat {BAK'}=\widehat{B'BK},\widehat{I'BB'}=\widehat{ABI'}.$ $BI'$ is the angle bisector of $\widehat{ABB'}\implies I'$ is the incentre of $\triangle ABB'.$ Similarly, we can show that $I'$ is the incentre of $ACC'$ and $ADD'.$ Hence, $I'$ is also the incentre of $ABCD, now $ $I'\equiv I\implies KI=KB.$ We can show that $KB=KJ=KI$ similarly.$2$ segments $KJ $ and $KI$ are equal.

But for regular tetrahedron, K and J coincide right?. So won't your claim be false in that case?

Yes , that's why I changed my mind to prove $IK>JK$ , instead of $IK=JK$ .

Official solution. Proof: Consider the plane passing through line $AIJ$ and perpendicular to the plane $BCD$. It intersects both spheres by their great circles. Let the tangents from $A$ to these circles meet $BCD$ at points $X$ and $Y$ . Then $I$ and $J$ are the incenter and the excenter of triangle $AXY$ , thus the midpoint of $IJ$ lies on the arc $XY$ of the circumcircle of this triangle. But $X$ and $Y$ lie inside the circumsphere of the tetrahedron, therefore arc $XY$ also lies inside it and $IK > IJ/2 > JK$. $\blacksquare$

Wow. That is simply amazing. Did you give the same proof too?

No @above. I missed the official proof. And yeah it is infact splendid.

@above could you please send a link of official solutions?