Prove that there is a red segment and a black segment having the same length. A regular 888-gon has 24 vertices colored red and 37 vertices colored black. We call a segment red if it has both red ends and black if it has both black ends. Prove that there is a red segment and a black segment having the same length.
I denote $A_1,..., A_{888}$ the vertices. I noticed that $888=24\cdot 37$. I tried to use Pigenhole Principle.
 A: A Starting Point :
There are $ (24 \cdot 23) / 2 = 276$ red segments with $ (37 \cdot 36) / 2 = 666$ black segments.
Total number is $ 276+666 = 942 $ segments , which exceeds $888/2=444$ Possible Unique Segments (in terms of lengths) in the Polygon.
Hence , there must be repetition by Pigeon Hole Principle.
A Complication :
The repetition may be within the red segments or within the black segments. This must be handled !
How can we ensure that red segments are not matching the black segments ?
How can we ensure that black segments are not matching the red segments ?
Will that be Possible ?
Can you take it from here ?
Proceeding :
Let all the red vertexes be together , consecutively.
Then the red segments are all like this :
between vertex 1 & vertex 2
between vertex 1 & vertex 3
between vertex 1 & vertex 4
between vertex 1 & vertex 22
between vertex 1 & vertex 23
between vertex 1 & vertex 24 (this is the max segment size)
We do not want such segments within the black vertexes.
Hence the black vertexes must be separated by at least 24.
Then the black segments are all like this :
vertex 25 & vertex $25+24 \cdot 1$
between vertex 25 & vertex $25+24 \cdot 2$
between vertex 25 & vertex $25+24 \cdot 3$
between vertex 25 & vertex $25+24 \cdot 34 = 841$
between vertex 25 & vertex $25+24 \cdot 35 = 865$
between vertex 25 & vertex $25+24 \cdot 36 = 889$ which exceeds the Polygon vertexes.
Hence , we can not avoid repetition : we will end up with common segments between the 2 groups.
Alternatively , let us Put all the black segments together , consecutively.
Then the black segments are all like this :
between vertex 1 & vertex 2
between vertex 1 & vertex 3
between vertex 1 & vertex 4
between vertex 1 & vertex 35
between vertex 1 & vertex 36
between vertex 1 & vertex 37 (this is the max segment size)
We do not want such segments within the red vertexes.
Hence the red vertexes must be separated by at least 37.
Then the red segments are all like this :
between vertex 38 & vertex $38+37 \cdot 1$
between vertex 38 & vertex $38+37 \cdot 2$
between vertex 38 & vertex $38+37 \cdot 3$
between vertex 38 & vertex $38+37 \cdot 21 = 815$
between vertex 38 & vertex $38+37 \cdot 22 = 852$
between vertex 38 & vertex $38+37 \cdot 23 = 889$ which exceeds the Polygon vertexes.
Hence , we can not avoid repetition : we will end up with common segments between the 2 groups.
Proof works out same in which-ever way we try.
