In the evening, pizza was ordered nine people sat around a round table, 50 slices of pizza were served to these nine people. Prove that there were two people sitting next to each other who ate at least 12 pizza slices.
I used the pigeon hole principle to determine 50/9 = 5.5 => 6
Therefore, at least one person ate 6 slice of pizza.
I just don't know how to prove that two people ate at least 12 slices..
Help would be greatly appreciated!
On
Since on average people ate $50/9<6$ slices, there exists a person who ate at most $5$ slices. The other eight people together ate at least $45$ slices. Can you conclude?
On
Assume the contrary, i.e., every pair of adjacent persons ate no more than $11$ slices.
There are $9$ pairs of adjacent persons, where we count persons 1 and 9 as adjacent. So if we sum up the slices eaten by each pair (persons 1 and 2, persons 2 and 3, etc.), the total is at most $9 \times 11 = 99$. But in so doing, we have counted each slice exactly twice, so the total must be $2 \times 50 = 100$.
This contradiction shows our initial assumption must be false.
Let $K_1, K_2, \dots, K_9$ be number of pizzas for individual persons. Then
$$K_1+ K_2+ K_3+K_4+K_5+K_6+K_7+K_8+ K_9 = 50$$
There are $9$ pairs of people sitting one next to other (neighbors). They ate $P_1, P_2, \dots, P_9$ piccas, where
$$P_1 = K_1 + K_2\\ P_2 = K_2 + K_3\\ \vdots\\ P_9 = K_9 + K_1\\ $$
Now
\begin{array}{lllllllllll} &P_1 + &P_2 + &P_3 + &P_4 + &P_5 + &P_6 + &P_7+ &P_8 + &P_9 \\[1em] = &K_1 + &K_2 \\ + &&K_2 + &K_3 \\ + &&&K_3 + &K_4 \\ + &&&&K_4 + &K_5 \\ + &&&&&K_5 + &K_6 \\ + &&&&&&K_6 + &K_7 \\ + &&&&&&&K_7 + &K_8 \\ + &&&&&&&&K_8 + &K_9 \\ + &K_1 + &&&&&&&&K_9 \\ \hline = &2K_1 + &2K_2 + &2K_3 + &2K_4 + &2K_5 + &2K_6 + &2K_7 + &2K_8 + &2K_9 \\[1ex] = 2(&K_1 + &K_2 + &K_3 + &K_4 + &K_5 + &K_6 + &K_7 + &K_8 + &K_9 )\\ = 2\cdot 50 = 100, \end{array}
i. e.
$$P_1+ P_2+ P_3+P_4+P_5+P_6+P_7+P_8+ P_9 = 100$$
It means, that $100$ pizzas is divided into $9$ pairs, and as $100/9 > 11,\ $at least one pair must get at least 12 pizzas.