Professor Jussara likes to make reduced copies. She starts by drawing the right triangle $ABC$ below:

![](https://cdn.mathpix.com/cropped/2024_05_01_463eeeabc25d438954b1g-05.jpg?height=452&width=578&top_left_y=1286&top_left_x=816)

Next, Professor Jussara draws a segment $AP_{1}$ such that $AP_{1}$ is perpendicular to $BC$, as shown in the figure below.

![](https://cdn.mathpix.com/cropped/2024_05_01_463eeeabc25d438954b1g-05.jpg?height=426&width=555&top_left_y=1910&top_left_x=822)

Professor Jussara then claims that the triangle $AP_{1}C$ is similar to the triangle $ABC$ (i.e., it has corresponding equal angles).
a) Show that Professor Jussara is correct!

b) Calculate the length of the segment $AP_{1}$.

c) Calculate the ratio between the area of triangle $ABC$ and the area of triangle $AP_{1}C$.

d) Professor Jussara repeats the process, now drawing a segment $P_{1}P_{2}$ perpendicular to the side $AC$, as shown in the figure below.

![](https://cdn.mathpix.com/cropped/2024_05_01_463eeeabc25d438954b1g-06.jpg?height=450&width=577&top_left_y=597&top_left_x=613)

What is the ratio between the areas of triangles $P_{1}P_{2}C$ and $AP_{1}C$?

e) Professor Jussara repeats the process two more times, as shown in the figure below.

![](https://cdn.mathpix.com/cropped/2024_05_01_463eeeabc25d438954b1g-06.jpg?height=452&width=566&top_left_y=1302&top_left_x=624)

What is the area of triangle $P_{5}P_{6}C$?
Solution

a) We will use here that the sum of the internal angles of a triangle is equal to $180^{\circ}$. Consider the following figure:

![](https://cdn.mathpix.com/cropped/2024_05_01_463eeeabc25d438954b1g-29.jpg?height=423&width=550&top_left_y=1502&top_left_x=632)

Using the internal angles of triangle $A B C$, we conclude that $\alpha+\beta+90^{\circ}=$ $180^{\circ}$. Therefore, we have $\alpha+\beta=90^{\circ}$. The same reasoning gives us that $\alpha+\gamma=$ $90^{\circ}$. Thus, we have $\beta=\gamma$. Therefore, triangles $B A C$ and $A P_{1} C$ have angles with the same measures, so they are similar.

b) We saw in part a) that triangles $B A C$ and $A P_{1} C$ are similar. Thus, we obtain that the ratio between the legs adjacent to the corresponding angles must be equal to the ratio between the hypotenuses. That is,

$$
\frac{\left|A P_{1}\right|}{|A B|}=\frac{|A C|}{|B C|}
$$

It follows from this that

$$
\left|A P_{1}\right|=\frac{|A C| \times|A B|}{|B C|}=\frac{3 \times 4}{5}=\frac{12}{5}
$$

Note that

$$
\frac{\left|A P_{1}\right|}{|A B|}=\frac{|A C|}{|B C|}=\frac{4}{5}
$$

that is, the similarity ratio between these two triangles is equal to 4/5.

c) Since the similarity ratio between the triangles is equal to $4 / 5$, we have that the base and height of triangle $A P_{1} C$ measure $4 / 5$ of the base and height of triangle $A B C$ respectively. Thus we have:

$$
\begin{aligned}
\text { Area }\left(A P_{1} C\right) & =\frac{\operatorname{base}\left(A P_{1} C\right) \times \text { height }\left(A P_{1} C\right)}{2}=\frac{4 / 5 \times \operatorname{base}(A B C) \times 4 / 5 \times \text { height }(A B C)}{2} \\
& =\frac{16}{25} \frac{\text { base }(A B C) \times \operatorname{height}(A B C)}{2}=\frac{16}{25} \times \operatorname{Area}(A B C)
\end{aligned}
$$

It follows from this that the ratio between the areas of triangles $A P_{1} C$ and $A B C$ is equal to 16/25.

d) Using a similar argument to part a), we conclude that triangles $A P_{1} C$ and $P_{1} P_{2} C$ are similar. From the figure below,

![](https://cdn.mathpix.com/cropped/2024_05_01_463eeeabc25d438954b1g-30.jpg?height=423&width=555&top_left_y=1351&top_left_x=822)

we conclude that

$$
\frac{\left|P_{1} P_{2}\right|}{\left|A P_{1}\right|}=\frac{\left|P_{1} C\right|}{|A C|}
$$

Since the similarity ratio between triangles $A P_{1} C$ and $A B C$ is equal to 4/5, it follows that

$$
\frac{\left|P_{1} P_{2}\right|}{\left|A P_{1}\right|}=\frac{4}{5}
$$

From this, we conclude that the similarity ratio between triangles $P_{1} P_{2} C$ and $A P_{1} C$ is also equal to $\frac{4}{5}$. Therefore, the ratio between their areas is also equal to $16 / 25$.

e) Through the figure

![](https://cdn.mathpix.com/cropped/2024_05_01_463eeeabc25d438954b1g-31.jpg?height=418&width=555&top_left_y=235&top_left_x=629)

and a similar procedure to parts a) and d), we see that triangles $P_{5} P_{6} C$, $P_{4} P_{5} C, P_{3} P_{4} C, P_{2} P_{3} C, P_{1} P_{2} C, A P_{1} C$ and $B A C$ are all similar, with the similarity ratio between two consecutive ones being equal to 4/5. Therefore, the ratio between the areas of two consecutive ones is equal to 16/25. From this,

$$
\begin{aligned}
\text { Area }\left(P_{5} P_{6} C\right) & =\frac{16}{25} \times \operatorname{Area}\left(P_{4} P_{5} C\right)=\left(\frac{16}{25}\right)^{2} \times \operatorname{Area}\left(P_{3} P_{4} C\right)=\cdots= \\
& =\left(\frac{16}{25}\right)^{4} \times \operatorname{Area}\left(P_{1} P_{2} C\right)=\left(\frac{16}{25}\right)^{5} \times \text { Area }\left(A P_{1} C\right)= \\
& =\left(\frac{16}{25}\right)^{6} \times \operatorname{Area}(A B C)=\left(\frac{16}{25}\right)^{6} \times \frac{3 \times 4}{2}
\end{aligned}
$$