Variant 132. $[-5 ;-2018]$. Variant 133. $[4 ; 2016]$. Variant 134. $[-5 ;-2017]$.

Grading criteria: 20 points - correct (not necessarily the same as above) solution and correct answer; 15 points - mostly correct solution and correct answer, but there are defects (for example: the transition from (1) to (2) is made without justification or with incorrect justification); $\mathbf{1 0}$ points - it is not proven that the equation has no other solutions except $x \in(4 ; 5)$ (note that the proof must be clear and neat; arguments like "it is obvious that this function is better, and therefore there can be no intersections" do not qualify); 5 points - the transition to equation (2) is justified, but there is no further progress.

## Problem 6 (Variant 131)

There are a lot of homogeneous circular disks of the same thickness and density. These disks are placed on a horizontal board such that the planes of the disks are vertical. Two adjacent disks touch each other, and the radii of two adjacent disks are in the ratio $2: 1$. The disk on the left is always larger than the disk on the right (see figure). The radius of the largest disk is 2 m. Determine the distance from the center of the largest disk to the center of mass of the system.
Solution. The radii of the disks form a geometric progression with a common ratio $q<1$. Here $q=\frac{1}{2}, R=2$. Therefore, the masses of the disks also form a geometric progression with a common ratio $q^{2}<1$. The total mass of the system is finite, as is its size, so the position of the center of mass is defined.

We find the height at which the center of mass is located above the board:

$$
M_{i}=M q^{2(i-1)}, i=1,2,3, \ldots ; R_{i}=R q^{i-1}, i=1,2,3, \ldots
$$

$$
Y=\frac{\sum_{i=1}^{\infty}\left(M_{i} R_{i}\right)}{\sum_{i=1}^{\infty} M_{i}}=R \frac{1-q^{2}}{1-q^{3}}
$$

We connect the centers of two adjacent disks with a segment, and from the right triangle with this hypotenuse and legs parallel and perpendicular to the board, we find the angle between this segment and the board: $\sin \alpha=\frac{1-q}{1+q}$. Since this angle does not depend on the disk number, all the centers of the disks, and thus the center of mass, lie on a line passing through the center of the first disk and having the found inclination to the board.

From the right triangle with legs parallel and perpendicular to the board and the hypotenuse connecting the center of the first disk with the center of mass of the system, we find the desired distance: $d=\frac{R-Y}{\sin \alpha}=R \frac{q^{2}(1+q)}{1-q^{3}}$.

Answer: $\frac{6}{7}$ m.

## Other answers: