Given points $\mathrm{P}_{1}, \mathrm{P}_{2}, \ldots, \mathrm{P}_{\mathrm{n}}$ on a line we construct a circle on diameter $\mathrm{P}_{\mathrm{i}} \mathrm{P}_{\mathrm{j}}$ for each pair $\mathrm{i}, \mathrm{j}$ and we color the circle with one of $\mathrm{k}$ colors. For each $\mathrm{k}$, find all $\mathrm{n}$ for which we can always find two circles of the same color with a common external tangent.
Answer: $\mathrm{n}>\mathrm{k}+1$.

There are $\mathrm{n}-1$ circles with diameter $\mathrm{P}_{\mathrm{i}} \mathrm{P}_{\mathrm{i}+1}$. Obviously, each pair has a common tangent. If $\mathrm{n}-1$ $>\mathrm{k}$, then two of them must have the same color.

If $\mathrm{n}-1 \leq \mathrm{k}$, then color all circles with diameter $\mathrm{P}_{\mathrm{i}} \mathrm{P}_{\mathrm{j}}$ and $\mathrm{i}<\mathrm{j}$ with color $\mathrm{i}$. Then if two circles have the same color, then both have a tangent at one of the points. Hence one lies inside the other and they do not have a common external tangent.