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 $$|PX|+|QX|+|RX|\le 4,(1)$$ then summing (1), with the given points $X=X_k$, $1\le k\le 60$, we get $$\sum _{k=1}^{60}|P{X}_k|+\sum _{k=1}^{60}|Q{X}_k|+\sum _{k=1}^{60}|R{X}_k|\le 4\cdot 60=240.$$ 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, $$|PX|+|QX|\le 2,(2)$$ which together with $|RX|\le 2$ will lead to (1).

Let us denote $S^{\prime }$ the midpoint of the corresponding arc $PQ$ (opposite to the arc $PRQ$, Fig. 1), then clearly $P{S}^{\prime }QS$ is a rhombus, hence it is sufficient to prove (2) only for points $X$ in the region $P{S}^{\prime }Q$ of the given disc (bounded by segment $PQ$ and arc $P{S}^{\prime }Q$). 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 $$\begin{array}{rl}|PX|+|QX|& =\frac{|PQ|(\sin \alpha +\sin \beta )}{\sin (\alpha +\beta )}=\frac{\sqrt{3}\cdot 2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}}{2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha +\beta }{2}}\\ & =\frac{\sqrt{3}\cos \frac{\alpha -\beta }{2}}{\cos \frac{\alpha +\beta }{2}}\le \frac{\sqrt{3}\cdot 1}{\sqrt{3}}=2,\text{as }\frac{\alpha +\beta }{2}\le 30^{\circ }.\end{array}$$ Thus (2) is proven.



Fig. 1



Fig. 2