Numbers on arcs of a circle Consider a circle where 50 people are, all of different ages. We can create a minimal pair out of two people X and Y. X and Y are not sitting next to each other, and they form two arcs on the circle of the people between them. They satisfy the conditions for a minimal pair if in at least one of the arcs, everybody in the arc is older than both of them. What is the least number of minimal pairs possible?
I've tried solving this problem, and came to the conclusion that the answer should be 47, but I have no way to actually prove it. I used the chain of numbers: 1, 0, 2, 3, 4, ... 49. This created 47 minimal pairs. I don't know how else I can prove this rigorously.
 A: Claim: No matter the arrangement of the people at the table, for $n$ people ($n>3$) there will be $n-3$ arcs as defined above.
Base Case: $n=4$. We can see there are three possible orderings around the table: $1-2-3-4, 1-2-4-3, 1-3-2-4$. In each case there is $4-3=1$ arc, respectively $1-3, 2-3, 1-2$ for the three cases.
Induction: Consider a table with $k$ people seated and $k-3$ arcs. Person $k+1$ enters and sits between two random people who were previously seated next to one another. As both of those people have a number lower that $k+1$, a new arc is formed between the two people who are now separated by $k+1$.
Now consider a random arc already present before $k+1$ sits. That arc had two sides: on one side, all numbers are greater than the pair; on the other side, they are not. If $k+1$ sits on the side with all greater numbers, the arc does not change, as $k+1$ is greater than any of the numbers on that side. But if $k+1$ sits on the other side of the arc, it doesn't stop that side containing lower numbers, nor stop the first side having only higher numbers than the pair forming the arc.
Since this is true for any arc, it is true for every arc. Hence when person $k+1$ sits down, one new arc is created, giving $k-2=(k+1)-3$ arcs as desired. QED
A: Consider the different ages as $1,2,3,...,50.$
We use induction to calculate the answer. If we had $4$ people ($1,2,3,4$), the answer would be $1$.
We claim if we have $2n$ people, then the answer is $2n-3$, where $n\ge2$.
Assume we have $2n+2$ people, and name them as $1,2, ..., 2n+1, 2n+2$.
If the people of $1, 2, ..,, 2n$ are sitting around the circle, by induction, we have at least $2n-3$ arcs, and notice that the two people of $2n+1$ and $2n+2$ don't ruin any of the accepted arcs when they are added as well. Now, if $2n+1$ and $2n+2$ are added, in any way, to the circle, we, at least, get two new accepted arcs simply because $2n+1$ and $2n+2$ are greater than any of the other numbers.
For example, if $2k+1$ and $2k+2$ are adjacent, consider the other numbers next to those two and $2k+1$ and the other number next to $2k+2$ as new arcs. In case $2k+1$ and $2k+2$ are not adjacent, just consider the numbers next to either.
Hence, the claim is verified because $2n-3+2=2(n+1)-3$.
