, 2019

Solution:
It's not difficult to see that the first triangle must connect three non-adjacent vertices (e.g. $P O I$), which covers area $\frac{1}{2}$, and leaves three $30$-$30$-$120$ triangles of area $\frac{1}{6}$ each. Then, the next three triangles cover $\frac{1}{3}$ of the respective small triangle they are in, and leave six $30$-$30$-$120$ triangles of area $\frac{1}{18}$ each.

This process continues, doubling the number of $30$-$30$-$120$ triangles each round and the area of each triangle is divided by $3$ each round. After $1+3+6+12+24=46$ triangles, the remaining area is $\frac{3 \cdot 2^{4}}{6 \cdot 3^{4}}=\frac{48}{486}=\frac{8}{81}<0.1$, and the last triangle removed triangle has area $\frac{1}{486}$, so this is the minimum number necessary.