## Task A-4.6.

In a triangle $A B C$ with an area of 1, a rectangle $P Q R S$ is inscribed such that points $P$ and $Q$ lie on side $\overline{A B}$, point $R$ on side $\overline{B C}$, and point $S$ on side $\overline{A C}$. Determine the maximum possible value of the area of rectangle $P Q R S$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_f93036d069fffd92d333g-35.jpg?height=611&width=1036&top_left_y=551&top_left_x=430)

Let's introduce the notations $c:=|A B|, x:=|P Q|=|R S|, y:=|Q R|=|P S|$. Additionally, let $h$ be the length of the altitude from vertex $C$ in triangle $A B C$.

From $A B \| R S$, we conclude that triangles $A B C$ and $S R C$ are similar.

Furthermore, since $A B \| R S$ and $|Q R|=y$, the lines on which the altitudes from $C$ in triangles $A B C$ and $S R C$ lie coincide, so the length of the altitude from $C$ in triangle $S R C$ is $h-y$.

From the similarity of triangles $A B C$ and $S R C$, the ratio of corresponding side lengths is equal to the ratio of the altitudes from $C$. Therefore, we have

$$
\frac{|R S|}{|A B|}=\frac{h-y}{h}
$$

Using the notations from the beginning of the solution, the above equality becomes

$$
x=\frac{c}{h}(h-y)
$$

Let $P$ be the area of rectangle $P Q R S$. Then

$$
P=x y=\frac{c}{h} \cdot y(h-y)
$$

The above expression is a quadratic function in the variable $y$. It achieves its maximum at the vertex, for the value $y=\frac{h}{2}$.

In this case, we have

$$
P=\frac{c}{h} \cdot \frac{h^{2}}{4}=\frac{c h}{4}=\frac{1}{2}
$$

since the area of triangle $A B C$ is 1.

Therefore, the upper bound on the area of rectangle $P Q R S$ is $\frac{1}{2}$.

This bound can be achieved when $y=\frac{h}{2}$, which corresponds to the case where the line $R S$ bisects the altitude from $C$, and thus also the sides $\overline{A C}$ and $\overline{B C}$. This is the case when points $R$ and $S$ are the midpoints of sides $\overline{B C}$ and $\overline{A C}$, respectively, and $P$ and $Q$ are their orthogonal projections onto the line $A B$.

Thus, the maximum possible area of rectangle $P Q R S$ is $\frac{1}{2}$.

Note: The scoring scheme is structured as follows:

- The answer that the desired maximum is $\frac{1}{2}$, which earns the last 1 point in the scoring scheme;
- The proof that the area of the rectangle cannot be greater than $\frac{1}{2}$, which earns the first 8 points in the scoring scheme;
- The description of an example of choosing points $P, Q, R$, and $S$ for which the area of rectangle $P Q R S$ is $\frac{1}{2}$, which earns the second-to-last 1 point in the scoring scheme.

After the area of rectangle $P Q R S$ is expressed in terms of one variable, the proof that its area is less than $\frac{1}{2}$ can also be done using the A-G inequality.