Dona Bete draws a triangle with sides 3, 4, and 5, as shown in the figure below:

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-09.jpg?height=450&width=600&top_left_y=1208&top_left_x=800)

Then, Dona Bete marks the midpoints of each side and draws a new triangle as shown in the figure below.

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-09.jpg?height=417&width=613&top_left_y=1864&top_left_x=801)

a) Calculate the area of triangle \(A_{1} B_{1} C_{1}\).
b) Seu Maurício notices an interesting pattern and repeats the process, as shown in the figure, always marking and connecting the midpoints of each new triangle.

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-10.jpg?height=409&width=608&top_left_y=393&top_left_x=603)

Calculate the area of triangle \(A_{2014} B_{2014} C_{2014}\).
Solution

a) The points $A_{1}, B_{1}$, and $C_{1}$ are the midpoints of the sides $B C, A C$, and $A B$, respectively, as shown in the figure below.

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-30.jpg?height=420&width=619&top_left_y=1815&top_left_x=798)

The large triangle $A B C$ is a right triangle with height 3 and base 4. Therefore, its area is $\frac{3 \times 4}{2}=6$. The four triangles $C B_{1} A_{1}, A_{1} C_{1} B_{1}, B_{1} A C_{1}$, and $C_{1} A_{1} B_{1}$ are all congruent, so they have the same area, which must be $\frac{1}{4}$ of the total area. Therefore, the area of the triangle $A_{1} B_{1} C_{1}$ is equal to $\frac{6}{4}=\frac{3}{2}$.
b) From the previous item, each time we draw a triangle by connecting the midpoints of the previous triangle, the new central triangle has an area equal to $\frac{1}{4}$ of the previous one. Therefore,

$$
\text { Area of triangle } A_{1} B_{1} C_{1}=\frac{1}{4} \cdot 6=\frac{3}{2} \text {. }
$$

Repeating the process,

$$
\text { Area of triangle } A_{2} B_{2} C_{2}=\frac{1}{4} \cdot \frac{3}{2} \text {. }
$$

Again,

$$
\text { Area of triangle } A_{3} B_{3} C_{3}=\frac{1}{4} \cdot \frac{1}{4} \cdot \frac{3}{2}=\left(\frac{1}{4}\right)^{2} \cdot \frac{3}{2} \text {. }
$$

Repeating the process one more time,

$$
\text { Area of triangle } A_{4} B_{4} C_{4}=\frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{3}{2}=\left(\frac{1}{4}\right)^{3} \cdot \frac{3}{2}
$$

and so on. Therefore, the area of the triangle $A_{2014} B_{2014} C_{2014}$ will be equal to $\left(\frac{1}{4}\right)^{2013} \cdot \frac{3}{2}$.