A fractal triangle

Consider the next smaller triangle.

It has mass $\dfrac{M}{3}$ and length $\dfrac{l}{2}$ . So its Moment of Inertia about its center of mass is $I_{s} = \dfrac{A}{B}\dfrac{M}{3}\dfrac{l^{2}}{4}$

Moment of Inertia of smaller part about the center of mass of the bigger part is :- $I_{sCM} = \dfrac{A}{B}\dfrac{Ml^{2}}{12} + \dfrac{M}{3}\dfrac{l^{2}}{12}$ $I_{sCM} = \dfrac{A}{B}\dfrac{Ml^{2}}{12}+\dfrac{Ml^{2}}{36}$ The total Moment of Inertia about the center of mass is:- $I_{CM} = 3I_{sCM}$ $\implies \dfrac{A}{B}Ml^{2} = 3I_{sCM}$ Solving this, we get:- $\boxed{\dfrac{A}{B}=\dfrac{1}{9}}$

By the parallel axis theorm we can know that k equals 1/9 then you add 1 and 9 to get A+B equals to 10. If a Sierpinski triangle of mass m and side length l is rotated about an axis passing normally through the centre, then the the moment of inertia k =m l^2.Then if you solve it it would be k=1/9ml^2

i didn't solve it my self , i saw it here https://www.physics.harvard.edu/uploads/files/undergrad/probweek/sol9.pdf