17) Davide and Golia live in a building whose floor plan is a regular pentagon with a side of 10 meters. The entrance of the building is located at one of the vertices of the pentagon, and the building is surrounded, for several kilometers, by flat ground. Golia steals David's slingshot, exits through the entrance, travels no more than 20 meters (without re-entering the building) and leaves the slingshot on the ground. What is the maximum area in which David will have to search before finding his slingshot?
(A) $320 \pi \mathrm{m}^{2}$,
(B) $160 \pi \mathrm{m}^{2}$,
(C) $100\left(4 \pi-\frac{\sqrt{5}-1}{2}\right) \mathrm{m}^{2}$,

(D) $100\left(2 \pi+\frac{\sqrt{5}-1}{2}\right) \mathrm{m}^{2}$,

(E) $280 \pi \mathrm{m}^{2}$
17. The answer is $(\mathbf{A})$.

In the figure below, $P$ indicates the entrance of the building, and the circle centered at $P$ with a radius double the side of the pentagon, and the two circles centered at $P$ and $Q$ with a radius equal to the side of the pentagon, have been drawn.

![](https://cdn.mathpix.com/cropped/2024_04_17_e415f128e14398de33bcg-5.jpg?height=506&width=525&top_left_y=1843&top_left_x=820)

Goliath could have left the slingshot in any point of the circular sector of the circle centered at $P$, delimited by the radii $P V$ and $P U$ and not containing the pentagon. The internal angles of a regular pentagon measure $108^{\circ}$, so the central angle of the circular sector we are interested in measures $252^{\circ}$, which is $7/10$ of a full angle. Therefore, the area of the circular sector is

$$
\frac{7}{10} \pi 400 \mathrm{~m}^{2}=280 \pi \mathrm{m}^{2}
$$

On the other hand, Goliath could have walked for $10 \mathrm{~m}$ along one of the sides of the building starting from $P$, for example $P Q$, and then moved away from $Q$, up to another ten meters, in any direction allowed by the presence of the building. Therefore, Goliath could have reached any point in the circular sectors centered at $Q$ and $R$, with a radius of $10 \mathrm{~m}$ and delimited by the segments $Q T, Q U$ and $R S, R V$ respectively. The central angles of these two sectors are complementary to the internal angles of the pentagon and are thus $180^{\circ}-108^{\circ}=72^{\circ}$, which is $1/5$ of a full angle. Therefore, the area of each of these two sectors is

$$
\frac{1}{5} \pi 100 \mathrm{~m}^{2}=20 \pi \mathrm{m}^{2}
$$

Overall, the surface that David must explore, at most, has an area of

$$
(280+20+20) \pi \mathrm{m}^{2}=320 \pi \mathrm{m}^{2} .
$$

[Problem proposed by D. Lombardo.]