6. A square is made from a wire. Two opposite sides have the same resistance $r$, the other two have the same resistance $2r$. Sliding contacts can move along the two opposite sides with resistance $2r$ (see figure; contacts are shown by arrows). By what factor does the maximum resistance of the square differ from the minimum? How should the contacts be connected to achieve the minimum or maximum resistance of the square? Provide all connection options.
6. Let's find the resistance of a square for some arrangement of contacts and investigate this value for maximum and minimum. Obviously, for any arrangement of contacts on the square, the sum of the resistances of its upper and lower branches is equal to the resistance of the wire from which it is made, i.e.,

![](https://cdn.mathpix.com/cropped/2024_05_06_0aa2685a2c7968513d04g-5.jpg?height=197&width=300&top_left_y=198&top_left_x=1620)
$6 r$. Therefore, if the resistance of the upper branch of the square is $x$, then the resistance of its lower part is $6 r - x$ (see figure). Moreover, the minimum resistance $x$ is $r$ (both contacts at the top), and the maximum is $5 r$ (both contacts at the bottom). Now let's find the resistance of the square connected between sliding contacts. In this case, the upper and lower branches are connected in parallel, and the resistance of the square $R$ will be equal to the ratio of the product of the resistances of the upper and lower segments to their sum

$$
R=\frac{x(6 r-x)}{6 r}
$$

where $x$ in this formula varies from $r$ to $5 r$. The graph of the function $R(x)$ (1) is a parabola with branches directed downwards, defined on the specified interval of values of $x$. Since the values of the function (1) are the same at $x=r$ and $x=5 r$, the function (1) is symmetric with respect to the line $x=3 r$ and

![](https://cdn.mathpix.com/cropped/2024_05_06_0aa2685a2c7968513d04g-5.jpg?height=417&width=731&top_left_y=1039&top_left_x=1208)
reaches its maximum at this value of $x$ (see figure). From this, we conclude that the maximum resistance of the circuit will be at $x=3 r$, and the minimum at $x=r$ and $x=5 r$. The values of the maximum and minimum resistances will be found by substituting these values of $x$ into formula (1):

$$
R_{\max }=\frac{3}{2} r, R_{\min }=\frac{5}{6} r
$$

Let's find the positions of the sliding contacts at which the minimum or maximum resistance of the circuit is achieved. The minimum is achieved if the resistance of the upper segment of the circuit between the contacts $x=r$ or $x=5 r$. Such a position

![](https://cdn.mathpix.com/cropped/2024_05_06_0aa2685a2c7968513d04g-5.jpg?height=191&width=624&top_left_y=1909&top_left_x=1270)

The minimum resistance of the circuit will occur in two cases: if both contacts are at the top, or both contacts are at the bottom. However, the maximum is achieved in many cases. Indeed, the resistance of the upper (and lower) segment between the contacts will be equal to $x=3 r$ if both contacts are at the center of the conductors with resistance $2 r$. But it will remain the same if the left contact is moved up by some amount (relative to the center of the conductor with resistance $2 r$) and the right contact is moved down by the same amount. Therefore, the maximum resistance of the circuit corresponds to any arrangement

![](https://cdn.mathpix.com/cropped/2024_05_06_0aa2685a2c7968513d04g-5.jpg?height=165&width=322&top_left_y=2413&top_left_x=1598)

The minimum resistance of the circuit of contacts, antisymmetric relative to the centers of the conductors with
resistance $2 r$

Criteria for evaluating the problem (maximum score for the problem - 2 points)