The idea is to locate a point $P$ for which, when we draw any line $l$ through $P$, there are at most $20$ more of the $60$ points on one side of $l$ than on the other side (where we don't include points on $l$ in either case). In this way, if we "radar-label" the $60$ points in clockwise order by $P_1, P_2, \cdots, P_{60}$ around $P$ (if two points are on the same ray from $P$, then just label them in whatever order we want), then the triangles $P_iP_{i+20}P_{i+40}$, for $i = 1, 2, \cdots, 20,$ all contain $P.$ Therefore, it suffices to show that such a point $P$ exists. To do this, we will use Helly's theorem. For any unit vector $v,$ we will define the set $S_v$ as follows. Let's consider the direction of $v$ as "up," and consider $S_v$ to be the set of all points in the plane which are strictly higher than the $41$st highest point among the $60$ points (so that $S_v$ doesn't contain the horizontal line through this $41$st highest point). It suffices since these sets are all convex to show that any three $S_v$'s have a common point, for all three unit vectors $v.$ This follows by straightforward casework (hard without diagram, oops). $\square$

Good problem!

Attachments:

Why do such lines exist?

One of the nice problems by Yijun Yao, it has some relationship with Tverberg's theorem or Radon's theorem in 2D. The key is Helly Theorem. Let $n=20.$ We consider the all $\binom{3n}{2n+1}$ polygons with $2n+1$ vertex. It is obvious that each three of them have a common point. So by Helly Thm there exists a point $P$ inside every polygon. Now note that for every straight line passing $P,$ at most $2n$ points are in the same half plane. Otherwise there are $2n+1$ points that doesn't include $P,$ contradiction! Now label the points clockwise $A_1,\ldots ,A_{3n},$ and take triangle $A_iA_{n+i}A_{2n+i},$ its convex hull includes $P,$ so we are done.$\Box$