Polya’s method for extremums

Polya’s method

People involved in spirituality believe that we are created and placed in the best of all possible worlds.

Therefore, the laws governing our world are based on the principles of optimality.

Natural scientists have found a number of empirical laws and methods that describe nature and therefore optimal laws.

Pólya proposed a method in which the researcher replaces the geometric problem of finding the optimum with a mechanical or optical model and applies empirical dependencies to this model.

The segment of the shortest length

The segment has the ends on the sides of a right angle and contains a point Find the shortest length of such a segment.

Solution

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

The rod is fixed at point 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 The grooves can create a force only along the normal, so they act on the rod with forces For the rod to be balanced, it is required that the moments of the forces be equal relative to point The moments of forces are: vladimir.shelomovskii@gmail.com, vvsss

The circle inside a fixed plane angle

Let the plane angle with vertex 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 and Find the condition under which the area of is the smallest.

Solution

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

Let 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 on the segments and are equal. These moments are proportional to the square of the length of the segment, that is, the equilibrium condition is the equality or is the midpoint (the center of mass of a homogeneous segment )

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The sphere inside a fixed trihedral angle

There is a sphere inside a fixed trihedral angle with vertex A plane is drawn through point of this sphere, tangent to the sphere and intersecting the edges of the angle at points and

Find the condition under which the volume of the pyramid is the smallest.

Solution

Let us imagine the plane in the form of the piston covering the evacuated volume There is no friction of the piston against the walls.

In the equilibrium position, the volume of the evacuated part of the system is minimal. The equilibrium condition is the equality of the moments of pressure forces, which is equivalent to the equality of the moments of gravity for a homogeneous plate So the point of the contact with the sphere (or any convex solid) must be at the center of mass of the polygon

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The smallest inscribed equilateral triangle

A right triangle with sides and be given. Find the area of the smallest regular triangle that can be inscribed in it. All vertices of the required triangle must be located on different sides of this triangle.

Solution

Let's imagine the desired triangle as a cross-section of a drill drilling the plane of a given triangle. The axis of rotation does not have to be in the center of the desired triangle.

Let's turn the drill at a small angle clockwise so that the points of the drill touching the sides of this triangle move along the sides.

The required triangle is minimal, which means that all its vertices cannot lie on its sides. One or more will go inside. When turning counterclockwise the situation is similar.

This means that there is an instantaneous axis of rotation of the drill for such turns. Since the required regular triangle is pedal for a point on the instantaneous axis of rotation, this point is the first Apollonius point of given triangle.

Using the properties of this point, we find that the required area is vladimir.shelomovskii@gmail.com, vvsss

Curve of the smallest length

Find the length of the curve of the smallest length dividing a given equilateral triangle with side into two equal parts.

Solution

Note that the shortest (continuous) curve has ends and on two sides of the triangle, which we will call blue and red. Let us denote the common vertex of these two sides as

Let us consider a set of six identical regular triangles in each of which the desired (identical!) curve is drawn.

We place the first of them in an arbitrary way in a certain plane.

We apply the second triangle to the first in the same plane so that the two blue sides and common vertices of these triangles coincide. In this case, the ends of the curve on the blue line will coincide, that is, the set of these two curves is also a continuous curve.

We will apply the third triangle to the second so that the red sides, vertices and points are coincide. We will similarly apply the fourth, fifth and sixth triangles.

As a result, we obtain a regular hexagon formed by sides opposite to vertex inside which a continuous curve limits an area whose area is equal to half the area of the hexagon. It is equal to the area of the three given triangles.

It is clear that this is a circle, its area (and radius) are known. The required length of the curve is a sixth of the length of its circumference.

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Triangle of the biggest area

A fixed point in the plane be given. The distances from to the vertices of triangle are Find the condition that the area of triangle ABC is maximum.

Solution

Let's imagine that at points there are rods fixed on hinges that cannot bend or move out of a predetermined plane, but with minimal force they lengthen or shorten. The ends of the rods are connected in pairs in the form of a triangle and slide freely together by changing the length of the rods.

Let there be a soap film located in the plane outside the triangle, which, as usual, tends to reduce its area proportional to the energy.

The forces applied to the rod are directed perpendicular to therefore

In the equilibrium position, the moments of forces applied to the rod on segments and are equal, which means that В is the midpoint of

This means that the area is maximum if is the center of the circumcircle

The area is times larger than the area respectively, the area is maximum if is the orthocenter of

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Clever point elephant in hollow tetrahedron

Let the hollow regular unit tetrahedron be given. A clever point elephant wants to fly along the shortest route that has points belonging to each face of the tetrahedron.

Find the length of this path.

Solution

Denote

Let and be the medians of the faces and

Let

It is known that the common perpendicular to these medians is

1. Note that

Let be symmetrical to with respect to

Then

We similarly prove that

This means that all the vertices of the tetrahedron are located on circles of equal radius with centers on straight line and the distances between the centers of the circles are equal.

Consequently, the tetrahedron is inscribed in a cylinder of radius with axis

2. Let be the angle between and

Let be the center of the face,

Let point be symmetrical to with respect to is symmetrical to with respect to

is located in the plane containing perpendicular to the straight line that is, this point lies on and the tetrahedron

Similarly, we will construct an arbitrarily long chain of equal tetrahedra inscribed in a cylinder whose median faces contain a sequence of points

It follows that the shortest path of the elephant cannot be shorter than There are some such ways inside the tetrahedron. One of them is shown in the diagram.

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