In a tetrahedron, external normal vectors are drawn to all faces. The length of each vector is numerically equal to the area of ​​the corresponding face. Prove that the sum of vectors is 0

Let us pump gas at pressure p into the tetrahedron. Then the force the gas pushes on the face is perpendicular to the face and proportional to its area: F = pS. From the equilibrium condition, the sum of the forces is 0, and then the sum of our verctors is also 0. This proof can easily be generalized to an arbitrary polyhedron.

Well-known

A right triangle is defined by the hypotenuse c and the vertex angle α. Then, from dimensional considerations, its area S is equal to f(α) * c^2, where f(x) — some function of the angle (in fact, f(x) = 0.5 * sin(x)). Let's draw the height to the hypotenuse, dividing the triangle into two small triangles with areas S1 and S2. Both triangles have an angle α. Summing up their areas using the formula, we reduce them by f(α) and get the equality.

Inside an acute triangle ABC, a point P is marked, minimizing the sum of the distances PA+PB+PC. Prove that the sides of the triangle ABC are visible from point P at equal angles.

Let's fasten three threads at one point (point P), which can slide freely along the plane. Let's make holes in the vertices of the triangle and pass the threads through them. We attach 3 equal masses on the lower ends of the threads. The minimum potential energy corresponds to the maximum total length of the vertical segments of the threads, that is, the minimum total length of segments PA, PB, PC. Point P is in a state of equilibrium, and three identical forces pull it. Then, from symmetry, the angles between each pair of threads are equal to 120 degrees.