We draw squares on the sides of a right-angled triangle. Calculate the area of the hexagon defined by the outer (not coinciding with the vertices of the triangle) vertices of the squares, if we know the hypotenuse of our right-angled triangle $(c)$, as well as the sum of the legs $(d)$.
I. Solution: The hypotenuse of the right-angled triangle is $c$, one of the legs is denoted by $x$, and the other by $(d-x)$.

![](https://cdn.mathpix.com/cropped/2024_05_02_77654b872ca0b73d062ag-1.jpg?height=501&width=404&top_left_y=194&top_left_x=839)

The area of the hexagon in question is composed of three squares and four triangles. The sum of the areas of the squares is

$$
c^{2}+x^{2}+(d-x)^{2}
$$

Of the four triangles, 1 and 2 are congruent, the area of 3 is equal to the area of the III dotted triangle, and the area of 4 is equal to the area of the IV triangle. However, triangles III and IV are congruent with triangle 1, so the areas of the four triangles marked with Arabic numerals are equal, and their sum is $4 \cdot \frac{1}{2}(d-x) x$.

Therefore, the area of the hexagon is

$$
T=c^{2}+(d-x)^{2}+x^{2}+4 \cdot \frac{1}{2}(d-x) x=c^{2}+d^{2}
$$

which is equal to the sum of the square of the hypotenuse and the square of the sum of the legs.

Solved by: Csonka P. (junior), Dievald Emilia, Főző Éva, Gergely A., Kántor S., Oláh J., Papp I., Rasztovich Mária, Sajó J., Turi I., Vigh A.

The areas of triangles 3 and 4 were expressed using trigonometric methods by: Dávid P., Kálmán L., Villányi O., Zobor E.

II. Solution: We completed the hexagon to form a rectangle as shown in the diagram (with thin lines) and then subtracted the areas of the four triangles cut off at the corners from the area of the rectangle.

![](https://cdn.mathpix.com/cropped/2024_05_02_77654b872ca0b73d062ag-1.jpg?height=529&width=523&top_left_y=1470&top_left_x=779)

The area of the hexagon is then (if $a, b$ are the legs, and $c$ is the hypotenuse)

$$
\begin{gathered}
t=(2 a+b)(2 b+a)-\frac{a \cdot 2 b}{2}-\frac{a b}{2}-\frac{2 a b}{2}-\frac{a b}{2}= \\
=2 a^{2}+2 b^{2}+2 a b=\left(a^{2}+b^{2}\right)+(a+b)^{2}=c^{2}+d^{2}
\end{gathered}
$$

Lipák János