Given two segments: $a$ and $b$. Construct the segment

$$
c=\sqrt[4]{a^{4}+b^{4}}
$$
I. Solution: By arbitrarily selecting the unit segment, we construct the segment $a^2$ from the segments 1 and $a$, using the theorem of the altitude in a right triangle (Figure 1), which states that $a$ is the geometric mean between 1 and $a^2$.

![](https://cdn.mathpix.com/cropped/2024_05_02_8d6f6da21fa593565648g-1.jpg?height=547&width=895&top_left_y=320&top_left_x=626)

Figure 1

Similarly, we construct the segment $b^2$. By constructing a right triangle with the obtained segments $a^2$ and $b^2$ as legs, the hypotenuse is $\sqrt{(a^2)^2 + (b^2)^2} = \sqrt{a^4 + b^4} = c^2$ (Figure 1). The desired segment $c$ is obtained as the geometric mean between 1 and $c^2$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_8d6f6da21fa593565648g-1.jpg?height=385&width=596&top_left_y=1217&top_left_x=751)

Figure 2

Note: The constructed segments $a^2$, $b^2$, and $c^2$ depend on the choice of the unit segment, but the final result $c$ is independent of this choice.

Bálint Pödör (Bp. II., Rákóczi g. II. o. t.)

II. Solution: By transforming the expression for $c$, we can solve the problem without choosing the unit segment. Let the hypotenuse of the right triangle with legs $a$ and $b$ be $t$, and the altitude to the hypotenuse be $m$ (Figure 3), then $a^2 + b^2 = t^2$ and $a \cdot b = t \cdot m$,

thus

$$
c^4 = a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2 = t^4 - 2t^2m^2 = t^2(t^2 - 2m^2) = t^2q^2
$$

where

$$
q = \sqrt{t^2 - (m\sqrt{2})^2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_8d6f6da21fa593565648g-2.jpg?height=613&width=667&top_left_y=122&top_left_x=726)

Figure 3

The construction process: The segment $m\sqrt{2}$ is obtained as the hypotenuse of an isosceles right triangle with leg $m$. According to (2), by constructing a right triangle with the leg $m\sqrt{2}$ and the hypotenuse $t$, we obtain the other leg: $q$.

From (1),

$$
c = \sqrt[4]{t^2q^2} = \sqrt{tq}
$$

Thus, $c$ is obtained as the geometric mean between $t$ and $q$ (Figure 3).

Katalin Holik (Balassagyarmat, Balassa g. II. o. t.)