Gervinho has a square cake with a side length of $20 \mathrm{~cm}$, as seen from above in the following figure. He is going to divide the cake into 5 equal-area pieces to share with his 4 friends. He can only make vertical cuts, as the cake is made of different layers, so everyone will receive the same amount of each layer.

![](https://cdn.mathpix.com/cropped/2024_05_01_20684006194a352111ccg-30.jpg?height=639&width=694&top_left_y=2005&top_left_x=590)
(a) Suppose Gervinho has already made the first cut from the center to a point on the side, 7 cm away from the upper left corner. How should he make the other cuts, knowing that all must start from the center?

(b) Setting aside the previous situation, suppose Gervinho made the first cut from point $P$, which is 8 cm and 9 cm away from its nearest sides, and ended at point $Q$, which is 7 cm away from the upper left corner.

![](https://cdn.mathpix.com/cropped/2024_05_01_20684006194a352111ccg-31.jpg?height=628&width=706&top_left_y=968&top_left_x=775)

How should he make the other cuts, knowing that all must start from point $P$ to the side of the cake?

(c) On the edge of the topping, there are candies, so in each of the cases, Gervinho should receive the piece of cake that has the largest perimeter of the cake's edge. If there is more than one with this largest perimeter, then he should take any of them. In each of the previous cases, indicate the piece that Gervinho should receive.
Solution

(a) Since the area of the square is $20 \cdot 20=400$, each of the 5 friends should receive $\frac{400}{5}=80$ of the cake area. If we consider each piece of cake as a triangle with a base on the sides of the cake or a combination of triangles with a base on the sides, then the height will always be 10. Therefore, it is sufficient to make the base or the sum of the bases on the sides 16. With the part that has 7, we need to add a part with a base of $16-7=9$. This leaves a part with a base of $20-9=11$. For this part, we need to add another part with a base of $16-11=5$. Following this, we will have the following cuts:

![](https://cdn.mathpix.com/cropped/2024_05_01_20684006194a352111ccg-32.jpg?height=489&width=506&top_left_y=949&top_left_x=678)

(b) Using the same strategy as in the previous item, it is sufficient to calculate the areas of each triangle formed by cuts parallel to the sides and directed towards the vertices from the center. We thus obtain the following distribution of the cake:

![](https://cdn.mathpix.com/cropped/2024_05_01_20684006194a352111ccg-32.jpg?height=520&width=520&top_left_y=1830&top_left_x=674)

By observing the areas, we can create a procedure to assemble the pieces, one at a time. To use the two pieces of sizes 28 and 36, we need to remove $80-28-36=16$ from the triangle with an area of 54, leaving $54-16=38$. To complete the piece with an area of 80 using the piece of size 38, we need pieces that sum up to an area of $80-38=42$. This can be obtained from the second triangle with an area of 54.

The process continues until we reach the following configuration:

![](https://cdn.mathpix.com/cropped/2024_05_01_20684006194a352111ccg-33.jpg?height=577&width=531&top_left_y=331&top_left_x=865)

Below is another figure without the subdivisions of each piece:

![](https://cdn.mathpix.com/cropped/2024_05_01_20684006194a352111ccg-33.jpg?height=591&width=540&top_left_y=1029&top_left_x=861)

(c) In the case of the first item, all pieces have the same perimeter of the cake edge, so Gervinho can receive any of them. In the case of the second item, we can compare the perimeters of the edge of each piece, which we will call: $P_{A}, P_{B}, P_{C}$, $P_{D}$, and $P_{E}$.

$$
\begin{gathered}
P_{A}=7+\frac{104}{9}=18+\frac{5}{9} \\
P_{B}=\frac{76}{9}+7=15+\frac{4}{9} \\
P_{C}=13+\frac{4}{11} \\
P_{D}=\frac{160}{11}=14+\frac{6}{11} \\
P_{E}=\frac{56}{11}+13=18+\frac{1}{11}
\end{gathered}
$$

Since $\frac{5}{9}>\frac{1}{11}$, we conclude that Gervinho should get piece $A$, as it has the largest perimeter of the cake edge.