Mumbai University > Computer Engineering > Sem 3 > Discrete Structures

Marks: 5 Marks

Year: Dec 2013

1. Let T be the equilateral triangle △ABC.
2. We are given that AB = BC = CA = 3 unit.
3. Let D, E, and F be the mid points of the sides AB, BC, CA respectively.
4. Join DE, EF, FD. Now, we have AD = DB = BE = EC = AF = FC = DF = DE = EF =1 1/2 unit.
5. The triangle ABC is divided into 4 equilateral triangles ADF, DBE, EFC and DEF, each possessing sides of length 1 1/2 units.
6. Now, each of the 10 points chosen must belong to one of these triangles (Pigeonholes).
7. Here we may consider the points as the pigeons.
8. Since there are only 4 triangles and any five points chosen from a triangle are not more than 1 unit apart. Then according to Pigeonhole principle, if 10 points are chosen, then two of them must be no more than 1 unit apart (because two points must belong to a single triangle).