Once again about the Pythagorean theorem

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ONCE AGAIN ABOUT THE PYTHAGOREAN THEOREM

Let us consider a vessel that has the shape of a straight line, at the base of which lies a right triangle ABC (see Fig.). Let’s fill this vessel with gas and give it the opportunity to rotate around the vertical axis OO’ (plane ABC is horizontal). Space is homogeneous, and no matter what position this vessel occupies, it will be in the same conditions. Therefore, we can assume that he is finished.

Consequently, the gas pressure forces on its side faces will mutually balance. But the forces F₁ and F₂ tend to rotate this vessel in one direction, and the force F₃ in the other. Therefore, the sum of the torques of the forces F₁ and F₂ must be equal to the torque of the force F₃.

$$F_1 \cdot \frac{AB}{2} + F_2 \cdot \frac{BC}{2} = F_3 \cdot \frac{AC}{2}.$$ (1)

But F₁ = P(AB \cdot h), F₂ = P(BC \cdot h), F₃ = P(AC \cdot h), where P is the gas pressure and h is the height of the vessel.

$$P(AB \cdot h) \cdot \frac{AB}{2} + P(BC \cdot h) \cdot \frac{BC}{2} = P(AC \cdot h) \cdot \frac{AC}{2},$$

where

$$AB^2 + BC^2 = AC^2$$

This proof can be read in the small brochure by B. Yu. Kogan “The Application of Mechanics to Theometry” (Nauka publishing house, 1965).