B2. In a top hat, there are a hundred tickets numbered from 1 to 100. You want to have three tickets with the property that each of the three numbers is smaller than the sum of the other two numbers. For example, the three tickets with numbers 10, 15, and 20 would be suitable (since $10<15+20, 15<10+20$, and $20<10+15$), but the tickets with numbers 3, 4, and 7 would not (since 7 is not less than $3+4$). You may (without looking at the numbers) take a number of tickets from the hat.

How many tickets must you take at a minimum to be sure that there are three tickets among them that meet your requirement?
2. 11

Suppose we have drawn 11 lottery tickets with numbers $a_{1}<a_{2}<\cdots<a_{11}$ on them such that no three tickets meet our wish. We know first that $a_{1} \geqslant 1$ and $a_{2} \geqslant 2$. Since the triplet $a_{1}, a_{2}$, and $a_{3}$ does not meet our wish, it must be that $a_{3} \geqslant a_{1}+a_{2} \geqslant 1+2=3$. We then proceed with the triplet $a_{2}, a_{3}$, and $a_{4}$ and find $a_{4} \geqslant a_{2}+a_{3} \geqslant 2+3=5$. If we continue this way, we find

$$
\begin{gathered}
a_{5} \geqslant 3+5=8, \quad a_{6} \geqslant 5+8=13, \quad a_{7} \geqslant 8+13=21, \quad a_{8} \geqslant 13+21=34, \\
a_{9} \geqslant 21+34=55, \quad a_{10} \geqslant 34+55=89 \quad \text { and } \quad a_{11} \geqslant 55+89=144 .
\end{gathered}
$$

The largest number on a ticket, however, is 100. So this is impossible, and we see that there must always be three tickets that meet our wish if we draw 11 tickets.

Moreover, 10 tickets are not enough. If we draw the tickets $1,2,3,5,8,13,21,34,55$ and 89, then we have no triplet of tickets that meets our wish.