8 men and 8 women in a circle There are 8 men and 8 women sitting in a circle. Prove that there are 8 adjacent people, so that 4 of them are men and 4 are women. (This is obvious, but how do I prove it?)
 A: Label each vertex with the number of men less the number of women in the eight people starting at that vertex. Confirm that this will be an even number between $-8$ and $8$ and that labels on adjacent vertices differ by either $0$ or $2$.
Pick a vertex. If its label is $0$ you are done. If the label $L$ is not $0$ confirm that the opposite vertex has label $-L$.
Now moving clockwise from the chosen vertex to the opposite vertex, the label changes sign. It can't change sign without becoming zero somewhere. This is a discrete version of the intermediate value theorem.
A: Hint: Treat the problem as a discrete version of the Intermediate Value Theorem.  Take a block of $8$ consecutive people, compute the excess or shortage of men, and then start moving the block, say clockwise, until you get to the complementary block of $8$.
A: Arrange the 16 people as specified in a circle;  make the spacing uniform.
Draw a diameter that runs from space  to space (no one on the dividing line)
Assume that There are more men in the part of the semicircle to the left.  Call this the area of interest.
Then the semi-circle to the right has more women.
Then rotate the dividing line clockwise 180 degrees.  The formerly "man-rich" area of interest has become the "woman-rich" area.  It must do this by passing through a balanced state...
Edit:  late by less than a minute!...
