Czech-Slovak-Polish Match

Let us inscribe an equilateral triangle $PQR$ into the boundary unit circle. If we prove that any point $X$ of our unit disc satisfies
$$|PX| + |QX| + |RX| \le 4, \quad (1)$$
then summing (1), with the given points $X = X_k$, $1 \le k \le 60$, we get
$$\sum_{k=1}^{60} |PX_k| + \sum_{k=1}^{60} |QX_k| + \sum_{k=1}^{60} |RX_k| \le 4 \cdot 60 = 240.$$
Consequently, one of the sums in the left-hand side does not exceed $240 : 3 = 80$, hence some of the points $P$, $Q$, $R$ has always the required property.

In view of symmetry, it suffices to prove (1) if $X$ lies in the sector $PSQ$, where $S$ denotes the centre of the disc. We are going to show that, in this case,
$$|PX| + |QX| \le 2, \quad (2)$$
which together with $|RX| \le 2$ will lead to (1).

Let us denote $S'$ the midpoint of the corresponding arc $PQ$ (opposite to the arc $PRQ$, Fig. 1), then clearly $PS'QS$ is a rhombus, hence it is sufficient to prove (2) only for points $X$ in the region $PS'Q$ of the given disc (bounded by segment $PQ$ and arc $PS'Q$).
Denoting $\alpha = |\angle XPQ|$, $\beta = |\angle XQP|$ we have $\alpha + \beta \le 60^\circ$ and using the law of sines in the triangle $PQX$ we get
$$\begin{aligned} |PX| + |QX| &= \frac{|PQ|(\sin \alpha + \sin \beta)}{\sin(\alpha + \beta)} = \frac{\sqrt{3} \cdot 2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}}{2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha+\beta}{2}} \\ &= \frac{\sqrt{3} \cos \frac{\alpha-\beta}{2}}{\cos \frac{\alpha+\beta}{2}} \le \frac{\sqrt{3} \cdot 1}{\sqrt{3}} = 2, \quad \text{as } \frac{\alpha+\beta}{2} \le 30^\circ. \end{aligned}$$
Thus (2) is proven.

Fig. 1

Fig. 2