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Polya’s method for extremums

Revision as of 10:37, 22 February 2024 by Vvsss (talk | contribs) (→‎The segment of the shortest length)

The segment of the shortest length

Shortest segment.png

The segment $AB$ has the ends on the sides of a right angle and contains a point $C(x_C, y_C).$ Find the shortest length of such a segment.

Solution

Let's imagine that $AB$ is a spring rod that cannot bend, but tends to shorten its length.

The rod is fixed at point $C$ on a hinge without friction. The hinge allows the rod to rotate and slide.

The ends of the rod can slide without friction along the grooves - the sides of the corner.

Let the rod be balanced, and the force pulling it together is equal to $T.$ The grooves can create a force only along the normal, so they act on the rod with forces \[F_B = \frac {T}{\sin \alpha}, F_A = \frac {T}{\cos \alpha}.\] For the rod to be balanced, it is required that the moments of the forces be equal relative to point $C.$ The moments of forces are: \[F_A \cdot AC \cdot \sin \alpha = F_B \cdot BC \cdot \cos \alpha \implies \frac {BC}{AC} = \tan^2 \alpha,\] \[\frac {y_C}{x_C} = \frac {BC \cdot \sin \alpha}{AC \cdot \cos \alpha} = \tan^3 \alpha \implies\] \[AB = \left ( x_C^ {\frac {2}{3}} + y_C^ {\frac {2}{3}}\right ) ^{\frac {3}{2}}.\] vladimir.shelomovskii@gmail.com, vvsss



The circle inside a fixed plane angle

Circle and triangle.png

Let the plane angle with vertex $A$ and the circle inside the angle be given. A straight line is drawn through point M of this circle, tangent to the circle and intersecting the sides of the angle at points $B$ and $C.$ Find the condition under which the area of $\triangle ABC$ is the smallest.

Solution

Let us imagine that $AB$ is a rod that cannot bend, and whose ends slide freely along the sides of the angle, and at point $M$ it rests on a convex curve.

Let $\triangle ABC$ be covered with a soap film, which, as usual, tends to reduce its area proportional to the energy. Let ABC be a section of a triangular prism, two faces of which AB and AC are fixed, and the third (BC) is a piston whose width can be changed so that it fits hermetically to the faces AB and AC and can slide along them. The piston rests on a cylinder whose cross-section is a circle with center O.

Air has been removed from the prism. External air presses on the piston and it comes to an equilibrium position. In this case, the energy of the system will be minimal if the volume of the prism becomes minimal. In the equilibrium position, the moments of forces applied to the piston $BC$ on the segments $BM$ and $MC$ are equal. These moments are proportional to the square of the length of the segment, that is, the equilibrium condition is the equality $BM = MC$ or $M$ is the midpoint $BC$ (the center of mass of a homogeneous segment $BC.$)

vladimir.shelomovskii@gmail.com, vvsss

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