A plane intersects a cube at three points $A_{1}, A_{2}, A_{3}$, which lie on the edges extending from one vertex at distances $a_{1}, a_{2}, a_{3}$ from the vertex. What is the area of the triangle $A_{1} A_{2} A_{3}$?
Each side of the triangle is the hypotenuse of a right-angled triangle whose legs are two of the distances $a_{1}, a_{2}$, $a_{3}$; therefore, using the Pythagorean theorem, the sides $a, b, c$ of the triangle are:

$$
a=\sqrt{a_{1}^{2}+a_{2}^{2}}, \quad b=\sqrt{a_{2}^{2}+a_{3}^{2}}, \quad c=\sqrt{a_{3}^{2}+a_{1}^{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_9968d62b3096575841d8g-1.jpg?height=455&width=512&top_left_y=366&top_left_x=796)

The area of the triangle can be calculated using the sides. According to Heron's formula, $T=\sqrt{s(s-a)(s-b)(s-c)}$, where $s$ is half the perimeter. Substituting this into the equation, we get

$$
T=\frac{1}{4} \sqrt{[(a+b)+c][(a+b)-c][(a-b)+c][c-(a-b)]}
$$

Using the identity $(x+y)(x-y)=x^{2}-y^{2}$ multiple times, we obtain

$$
T=\frac{1}{4} \sqrt{\left(a^{2}+b^{2}-c^{2}+2 a b\right)\left(c^{2}-a^{2}-b^{2}+2 a b\right)}
$$

Substituting the expressions for $a, b, c$ from (1) and simplifying, we get

$$
T=\frac{1}{4} \sqrt{\left[2 \sqrt{\left(a_{1}^{2}+a_{2}^{2}\right)\left(a_{2}^{2}+a_{3}^{2}\right)}+2 a_{2}^{2}\right]\left[2 \sqrt{\left(a_{1}^{2}+a_{2}^{2}\right)\left(a_{2}^{2}+a_{3}^{2}\right)}-2 a_{2}^{2}\right]}=\frac{1}{4} \sqrt{4\left(a_{1}^{2}+a_{2}^{2}\right)\left(a_{2}^{2}+a_{3}^{2}\right)-4 a_{2}^{4}}=\frac{1}{2} \sqrt{a_{1}^{2} a_{2}^{2}+a_{2}^{2} a_{3}^{2}+a_{3}^{2}}
$$

where in the further transformations, we repeatedly used the previous identity.

Tamás Terpai (Fazekas M. Főv. Gyak. Ált. Isk., 8th grade)