16) Antonio and Barbara have their birthdays on the same day. Antonio turns 7 years old and, by arranging the candles appropriately, he can divide his birthday cake with three straight cuts so that each piece contains exactly one candle (see the figure next to it). Barbara manages to do the same with her cake using 4 straight cuts, but she knows that next year 4 cuts will no longer be enough, regardless of how the candles are arranged. How old is Barbara?
$\begin{array}{ll}\text { (A) } 9 & \text { (B) } 10\end{array}$
(C) 11
(D) $14 \quad$ (E) 16

![](https://cdn.mathpix.com/cropped/2024_04_17_b5d519ff6c1a8005abb1g-2.jpg?height=317&width=317&top_left_y=1005&top_left_x=1117)
16) The answer is (C).

Suppose we have already made $k-1$ cuts. The $k$-th cut (which we can assume is different from the previous ones) will intersect the previous cuts at a certain number $h$ of distinct points inside the cake. Then the $k$-th cut will cross $h+1$ distinct parts of the cake, dividing each of them into two: the effect will be to increase the number of parts of the cake by an amount equal to $h+1$. On the other hand, since two distinct lines can intersect at most at one point, we have that $h \leq k-1$, and therefore with the $k$-th cut, the number of parts of the cake will increase by at most $k$. Starting from the initial situation (1 single part), the number of parts that can be created with 4 cuts is at most

$$
1+1+2+3+4=11 \text {. }
$$

On the other hand, the figure shows how 11 parts can be obtained with 4 cuts.

![](https://cdn.mathpix.com/cropped/2024_04_17_0a7649c17dc24dfa1af2g-3.jpg?height=414&width=419&top_left_y=867&top_left_x=1521)