This topic is linked to Problem 3.
Y by Adventure10
Notice that if we do the opposite procedure, that is, place the next break in the unit length in the smaller segment, we end up with no possible triangle. This is because the larger segment must have length , and as this segment is not broken up any further, it remains larger than the semiperimeter. Thus, as given before, the area expected is , as noted before. The same applies to the second scenario in which we place the next cut in the rightmost segment.