5. Elmo is drawing with colored chalk on a sidewalk outside. He first marks a set $S$ of $n>1$ collinear points. Then, for every unordered pair of points $\{X, Y\}$ in $S$, Elmo draws the circle with diameter $X Y$ so that each pair of circles which intersect at two distinct points are drawn in different colors. Count von Count then wishes to count the number of colors Elmo used. In terms of $n$, what is the minimum number of colors Elmo could have used?
Proposed by Michael Ren.
Answer. The answer is $\lceil n / 2\rceil$ colors, except when $n=3$ here the answer is 1 .
Solution. I claim that the answer for even $n$ is $n / 2$. We can let the distance between adjacent points be 1 . Label the vertices $1,2,3, \ldots, n / 2,1,2, \ldots, n / 2$ in that order, left to right (we can assume that the line the points are on is horizontal).
Now, consider the $n / 2$ circles whose diameters have endpoints with the same label. Note that these are pairwise intersecting, so we must use at least $n / 2$ colors.
For the coloring, for circles with diameter $\leq n / 2$, color them with the label of the right endpoint of the diameter. For circles with diameter $\geq n / 2$, color them with the label of the left endpoint of the diameter. By checking cases, it is not hard to confirm that this coloring works.
Now we consider $n=2 m+1$ odd. Obviously $n=3$ gives 1 . For other $n=2 m+1$ we will show the minimum number of colors is $m+1$. We can construct this by using the above construction, and coloring each circle containing $2 m+1$ with the color $m+1$.
Now, for proving it, call the vertices $1,2, \ldots, m, m+1,1,2, \ldots, m$ as earlier; we have the $m$ different colored circles from vertices of the same color. Let $f(a, b)$ denote the color of the circle with vertex color $a$ in the first $m+1$ vertices, and vertex color $b$ in the last $m+1$ vertices. Note that $f(1, m+1)=1 . f(2,1)=1,2$, but we know it is 2 due to the previous conclusion. Similarly, we show that $f(k+1, k)=k+1$ for $1 \leq k \leq m$, so in particular, we need $f(m+1, m)=m+1$, as desired.
This problem was proposed by Michael Ren. This solution was given by Mihir Singhal and James Lin.

Remark. An alternate easier version of the problem requires that circles which are tangent to each other are also distinct colors. In this case the answer is $n$.
Label the vertices $1,2, \ldots n$, and let $f(a, b)$ be the circle with diameter at vertices $a, b$. Note that $f(1,2), f(1,3), \ldots, f(1, n), f(2, n)$ are different colors, so at least $n$ colors are needed. But then we can let $f(a, b)$ be colored by color $a+b(\bmod n)$, so we are done!