There is a circumference with 14 points $\{p_{1}, p_{2}, ... p_{14}\}$. These points are assigned numbers 1 to 14 randomly. It must be proven that if points are taken three-by-three, these triplets being formed by consecutive points, there will be at least one triplet which has a sum bigger than 29.
This is how I operated:
There are 14 sums $\{s_{1}, s_{2}, ... s_{14}\}$ where $$s_{1} = p_{1} + p_{2} + p_{3}$$$$s_{2} = p_{2} + p_{3} + p_{4}$$$$...$$$$s_{14} = p_{14} + p_{1} + p_{2}$$ So each number appears 3 times, this is $$s_{1}+s_{2}+...+s_{14} = 3*(1+2+3+4+5+6+7+8+9+10+11+12+13+14) = 315$$ And, $$14*29 = 406$$ This is where I get stuck and don't know how to prove the statement using the pigeonhole principle.