Baltic Way 1993 · Problem 13

Solution:

Figure 2

An example for $7$ vertices is shown in Figure 2. Now assume we have chosen $8$ vertices satisfying the conditions of the problem. Let the height of each small triangle be equal to $1$ and denote by $a_{i}, b_{i}, c_{i}$ the distance of the $i$th point from the three sides of the big triangle. For any $i=1,2, \ldots, 8$ we then have $a_{i}, b_{i}, c_{i} \geq 0$ and $a_{i}+b_{i}+c_{i}=10$. Thus, $\left(a_{1}+a_{2}+\cdots+a_{8}\right)+\left(b_{1}+b_{2}+\cdots+b_{8}\right)+\left(c_{1}+c_{2}+\cdots+c_{8}\right)=80$. On the other hand, each of the sums in the brackets is not less than $0+1+\cdots+7=28$, but $3 \cdot 28=84>80$, a contradiction.