## Task B-2.5.

In the figure, a rectangular picture with a frame is shown. The shaded area represents the picture, and the frame is composed of eight congruent trapezoids. The lengths of the bases of these trapezoids are natural numbers, and the area of each individual trapezoid is a prime number. If the area of the picture is less than 200 square units, what is the largest possible area of the picture?

![](https://cdn.mathpix.com/cropped/2024_05_30_4b05efd680c81b2b8cbfg-12.jpg?height=177&width=511&top_left_y=888&top_left_x=778)
## First solution.

Let $x$ and $y$, with $x > y$, be the bases of the trapezoid, and let $h$ be its height, and $p$ its area.

Then, $p = \frac{x + y}{2} \cdot h$.

![](https://cdn.mathpix.com/cropped/2024_05_30_4b05efd680c81b2b8cbfg-12.jpg?height=240&width=671&top_left_y=1365&top_left_x=196)

The longer side of the frame is $2x + y$, and the length of its shorter side is $x$. The longer side of the picture is $x + 2y$, and the length of its shorter side is $y$. The area of the picture is $P = y(x + 2y)$.

Notice that $2x + y = x + 2y + 2h$, which means $h = \frac{x - y}{2}$.

For the area of the trapezoid, we have $p = \frac{x + y}{2} \cdot \frac{x - y}{2} = \frac{(x + y)(x - y)}{4}$.

Consider the product $(x + y)(x - y) = 4p$.

Since $x$, $y$, and $p$ are natural numbers, the product $(x + y)(x - y)$ must be an even number. Since the numbers $x + y$ and $x - y$ have the same parity, we can conclude that both must be even. Also, note that $x - y > 200$ square units.

Since the area of the picture must be less than 200 square units, we conclude that the maximum possible area is 120 square units.