## Task B-3.7.

In a puzzle, a hexagon needs to be assembled from 5 pieces into the shape of a rhombus, as shown in the figure. Pieces $T, U, V$ and $Z$ are congruent, and each has an area of $\sqrt{2017}$. The piece marked $\mathrm{s} X$ is lost, and Ana wants to make a new one. For precision and easier manufacturing, let the area of this piece be an integer $P$. How many different values can the number $P$ take? Among all such pieces $X$, determine the one with the largest area, where the lengths of the diagonals are rational numbers.

![](https://cdn.mathpix.com/cropped/2024_05_30_7713364b1a15726862ddg-15.jpg?height=334&width=571&top_left_y=2209&top_left_x=685)
## First solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_7713364b1a15726862ddg-16.jpg?height=363&width=574&top_left_y=298&top_left_x=684)

The area of the rhombus $X$ is

$$
P=4 \cdot \frac{d \cdot h}{2}=2 d h=2 h \sqrt{a^{2}-h^{2}}=\sqrt{4 h^{2} a^{2}-4 h^{4}}
$$

From the area of congruent rhombi $a \cdot h=\sqrt{2017} \Rightarrow a^{2} h^{2}=2017$, thus

$$
P=\sqrt{4 \cdot 2017-4 h^{4}}=\sqrt{8068-4 h^{4}}
$$

The number $P$ is an integer only if the expression under the square root is a perfect square.

Since $8100=90^{2}$ and $7921=89^{2}$, there are at most 89 such numbers.

The maximum value that $P$ can take is 89, followed by $88, 87, \ldots$

For $P=89$, the length of the diagonal $2 h$ is not a rational number.

If $P=88$ then

$$
4 h^{4}=8068-88^{2}=324 \Rightarrow h=3 \Rightarrow d=\frac{P}{2 h}=\frac{44}{3}
$$