An equilateral triangle $A B C$ is divided into 100 congruent equilateral triangles. What is the greatest number of vertices of small triangles that can be chosen so that no two of them lie on a line that is parallel to any of the sides of the triangle $A B C$?
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, \((a_{1}+a_{2}+\cdots+a_{8})+(b_{1}+b_{2}+\cdots+b_{8})+(c_{1}+c_{2}+\cdots+c_{8})=80\). 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.