Съдържание

Условие 1. Mathematical introduction (3 т.)2. Wire frame in the shape of a prism (7 т.)Отговори Решение

IZhO 2017, Теория — Задача 3. Problem 3. Resistance of a prism

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Всички материали от IZhO 2017

XIII International Zhautykov Olympiad/Theoretical Competition · 10 т.

Внимание: Бележка към темата

Problem 1C refers to a separate answer sheet with the points S1, S2, A1, A2 and a highlighted oval area; that sheet is not in the archive. The grid drawing in the official solution shows its layout.

THEORETICAL COMPETITION. January 14, 2017. Please read this first:

The time available for the theoretical competition is 4 hours. There are three questions.
Use only the pen provided.
You can use your own calculator for numerical calculations. If you don’t have one, please ask for it from Olympiad organizers.
You are provided with Writing sheet and additional paper. You can use the additional paper for drafts of your solutions but these papers will not be checked. Your final solutions which will be evaluated should be on the Writing sheets. Please use as little text as possible. You should mostly use equations, numbers, figures and plots.
Use only the front side of Writing sheets. Write only inside the boxed area.
Begin each question on a separate sheet of paper.
Fill the boxes at the top of each sheet of paper with your country (Country), your student code (Student Code), the question number (Question Number), the progressive number of each sheet (Page Number), and the total number of Writing sheets used (Total Number of Pages). If you use some blank Writing sheets for notes that you do not wish to be evaluated, put a large X across the entire sheet and do not include it in your numbering.
At the end of the exam, arrange all sheets for each problem in the following order:
- Used Writing sheets in order;
- The sheets you do not wish to be evaluated
- Unused sheets and the printed question.
Place the papers inside the envelope and leave everything on your desk. You are not allowed to take any paper out of the room.

SOLUTIONS TO THE PROBLEMS OF THE THEORETICAL COMPETITION

Attention. Points in grading are not divided!

Условие

1. Mathematical introduction (3 т.)

By definition, it is believed that terms of the numerical sequence obey the recurrence relation if each successive term is expressed through the previous ones. For example, for a well known geometric progression we have

x_k = \lambda x_{k-1}, \qquad (1)

where $k = 1, 2, 3, ...$ , $\lambda$ stands for a fixed number and zeroth term of the numerical sequence has some value of $A$ , i.e. $x_0 = A$ .

1.1 Obtain an explicit formula for an arbitrary term of the sequence $x_k$ , i.e. express it through the successive number $k$ , the initial value $A$ and $\lambda$ . [0,2 т.]

Let us consider the number $\lambda = 2 + \sqrt{3}$ . Taking its natural power of $k$ , the result can be presented in the following form

\lambda^k = p_k + q_k\sqrt{3}, \qquad (2)

where $p_k, q_k$ denote some integer numbers.

1.2 Find the recurrence relations, expressing the values of $p_k, q_k$ through the previous values $p_{k-1}, q_{k-1}$ . Find also the inverse relations, expressing $p_{k-1}, q_{k-1}$ through $p_k, q_k$ . [0,4 т.]

1.3 Calculate the numerical values of the coefficients $p_k, q_k$ for $k = 1, 2, 3, 4, 5$ . [0,7 т.]

1.4 Express the number $\lambda^{-k} = \left(2 + \sqrt{3}\right)^{-k}$ in terms of $p_k, q_k$ . [0,2 т.]

Let the terms of a certain numerical sequence obey the recurrence relation

x_{k+1} = 4x_k - x_{k-1},\quad k = 1, ..., N - 1, \qquad (3)

where it is known that $N$ is some integer number, $x_0 = A$ and $x_N = B$ , $A, B$ designate some values.

1.5 Obtain an explicit formula for an arbitrary term $x_k$ of the sequence (3), i.e. express it in terms of the number $k$ and values $A, B, N$ . [1 т.]

1.6 Obtain an explicit formula for an arbitrary term $x_k$ of the sequence (3) through $p_k, q_k$ , found in.1.2-1.3. [0,5 т.]

Hint. The solution to the recurrence relation (3) must be sought in the form $x_k = C\lambda^k$ , where $C$ is a constant. Determine at what values of $\lambda$ it is possible and construct an exact solution that satisfies all above stated conditions.

2. Wire frame in the shape of a prism (7 т.)

Problems are widely known in which you are asked to find the electrical resistance of a simple wire frame. An example of such a frame shaped in the form of a cube is shown below. Let the electric resistance of each edge be equal to $R_0$ .

Wire-frame cube with dots at the vertices; one edge is labelled R0, and a battery is connected by wires to the two ends of one vertical edge on the left (two adjacent vertices).

Wire frame of a regular prism with many side faces drawn in perspective, with dots at the vertices; a lateral edge and a base edge are labelled R0, and two lead wires go to the two ends of the front lateral edge. — Wire frame in the shape of a prism

Part of the prism frame with the upper vertices labelled x with indices N−1, 0, 1, 2 and the lower vertices labelled y with indices N−1, 0, 1, 2; a gap with dots indicates the omitted faces, and a battery is connected to the vertices x0 and y0. — Numbering and denoting vertices' potentials

Ladder circuit: upper nodes x with indices k−1, k, k+1 and lower nodes y with the same indices; horizontal resistors R0 between neighbouring nodes in each row and vertical resistors R0 between x and y at each k.

2.1 Find the total resistance of the cube when the source is connected to the two adjacent cube vertices as shown on the right. [0,8 т.]

Let us now consider a more general case of the wire frame in the form of the regular prism with an arbitrary number $N$ of side faces and determine its electrical resistance when the source is connected to the adjacent vertices of the side edge, as shown in the figure below. The resistance of each frame edge is equal to $R_0$ .

For convenience, the vertices of the prism and their electric potentials on the upper and the bottom sides are consequently numbered and denoted, as shown in the figure below. DC voltage source is applied to zeroth vertices such that the source sets the potentials of the vertices equal to $x_0 = +\varphi_0$ and $y_0 = -\varphi_0$ , respectively.

2.2 Find the relation between the potentials $x_k$ and $y_k$ . Find the relation between the potentials $x_k$ and $x_{N-k}$ [0,2 т.]

Consider an arbitrary lateral edge, except zeroth ( $k = 0$ ) and the last ( $k = N - 1$ ) ones. The corresponding circuit diagram is shown below.

2.3 Find an recurrence relation for the potential $x_k$ as expressed in terms of the neighboring vertices potentials for $k = 1,2...N - 2$ . [1 т.]

2.4 Find boundary conditions at $k = 0$ and $k = N - 1$ necessary for unambiguous determination of the potential $x_k$ . [0,2 т.]

2.5 Find explicit expressions for the potentials $x_k$ and $y_k$ for all possible numbers $k = 0,1,2...N - 1$ . [0,2 т.]

2.6 Express the source current in terms of $\varphi_0, R_0, N$ . Use the appropriate numbers $p_k, q_k$ .obtained in the Mathematical introduction to this problem. [0,4 т.]

2.7 Derive an explicit formula for the resistance $R_N$ of the wire frame, expressed in terms of $R_0, p_N, q_N$ . [0,2 т.]

2.8 Find and tabulate the exact values of the frame resistances for $N = 1,2,3,4,5$ . [1 т.]

2.9 Draw the equivalent circuits for the exotic prisms with $N = 1$ and $N = 2$ . [0,5 т.]

2.10 Find the resistance $R_\infty$ of the wire frame at $N \to \infty$ . [1 т.]

2.11 Find the minimum value of $N$ at which the prism resistance deviation from $R_\infty$ does not exceed 2%. [1,5 т.]

Отговори

1. Mathematical introduction, 1.1 $x_k = A\lambda^k$
1. Mathematical introduction, 1.2 $p_k = 2p_{k-1} + 3q_{k-1},\ q_k = p_{k-1} + 2q_{k-1};\quad p_{k-1} = 2p_k - 3q_k,\ q_{k-1} = 2q_k - p_k$
1. Mathematical introduction, 1.3 p: 2, 7, 26, 97, 362; q: 1, 4, 15, 56, 209 (k = 1…5; p0 = 1, q0 = 0)
1. Mathematical introduction, 1.4 $\lambda^{-k} = p_k - q_k\sqrt{3}$
1. Mathematical introduction, 1.5 $x_k = \frac{A\left(\lambda_1^{N-k} - \lambda_2^{N-k}\right) + B\left(\lambda_1^k - \lambda_2^k\right)}{\lambda_1^N - \lambda_2^N},\quad \lambda_{1,2} = 2\pm\sqrt{3}$
1. Mathematical introduction, 1.6 $x_k = \frac{Aq_{N-k} + Bq_k}{q_N}$
2. Wire frame in the shape of a prism, 2.1 $R = \frac{7}{12}R_0$
2. Wire frame in the shape of a prism, 2.2 $y_k = -x_k,\quad x_{N-k} = x_k$
2. Wire frame in the shape of a prism, 2.3 $x_{k+1} - 4x_k + x_{k-1} = 0$
2. Wire frame in the shape of a prism, 2.4 $x_0 = \varphi_0,\quad x_N = x_0$
2. Wire frame in the shape of a prism, 2.5 $x_k = \varphi_0\frac{q_{N-k} + q_k}{q_N},\quad y_k = -x_k$
2. Wire frame in the shape of a prism, 2.6 $I_0 = \frac{2\varphi_0}{R_0}\frac{p_N - 1}{q_N}$
2. Wire frame in the shape of a prism, 2.7 $R_N = R_0\frac{q_N}{p_N - 1}$
2. Wire frame in the shape of a prism, 2.8 R1 = R0, R2 = 2/3 R0, R3 = 3/4 R0, R4 = 7/12 R0, R5 = 11/19 R0
2. Wire frame in the shape of a prism, 2.10 $R_\infty = \frac{R_0}{\sqrt{3}}$
2. Wire frame in the shape of a prism, 2.11 4

The solution states that already at $N = 4$ the relative error is 1% ( $\infty\approx 4$ ).

Решение

1. Mathematical introduction (3 т.)

1.1 [0.2 points] From the course of school mathematics it is known that geometrical progression terms are explicitly expressed as

x_k = A\lambda^k. \qquad (1)

1.2 [0.4 points] Let us express $\lambda^k$ recurrently in terms of $\lambda^{k-1}$ :

\lambda^k = \lambda^{k-1}\cdot\lambda

and transform it as follows

\begin{aligned} \lambda^k &= \left(p_k + q_k\sqrt{3}\right) = \left(p_{k-1} + q_{k-1}\sqrt{3}\right)\cdot\left(2 + \sqrt{3}\right) = 2p_{k-1} + p_{k-1}\sqrt{3} + 2q_{k-1}\sqrt{3} + 3q_{k-1} = \\ &= \left(2p_{k-1} + 3q_{k-1}\right) + \left(p_{k-1} + 2q_{k-1}\right)\sqrt{3}. \end{aligned} \qquad (2)

This equality implies the required recurrence relations in the form

\begin{aligned} p_k &= 2p_{k-1} + 3q_{k-1} \\ q_k &= p_{k-1} + 2q_{k-1}. \end{aligned} \qquad (3)

Inverse relations are obtained analogously

\begin{aligned} \lambda^{k-1} &= p_{k-1} + q_{k-1} = \lambda^k\cdot\lambda^{-1} = \left(p_k + q_k\sqrt{3}\right)\cdot\left(2 - \sqrt{3}\right) = \\ &= \left(2p_k - 3q_k\right) + \left(2q_k - p_k\right)\sqrt{3}, \end{aligned} \qquad (4)

and, thus,

\begin{aligned} p_{k-1} &= 2p_k - 3q_k, \\ q_{k-1} &= 2q_k - p_k. \end{aligned} \qquad (5)

1.3 [0.7 points] Calculation of the coefficients is much easier to carry out in series, given that $p_0 = 1,\quad q_0 = 0$ . The results are shown in Table 1.

Table 1.

$k$	$p_k$	$q_k$
0	1	0
1	2	1
2	7	4
3	26	15
4	97	56
5	362	209

1.4 [0.2 points] Note that

\lambda^{-1} = \frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}, \qquad (6)

therefore,

\lambda^{-k} = \left(2 - \sqrt{3}\right)^k = p_k - q_k\sqrt{3}. \qquad (7)

1.5 [1.0 points] Using the hint, we substitute $x_k = C\lambda^k$ into the recurrence relation and obtain the equation to determine $\lambda$ in the form

\lambda^{k+1} = 4\lambda^k - \lambda^{k-1}. \qquad (8)

After reduction the following quadratic equation is derived

\lambda^2 - 4\lambda + 1 = 0, \qquad (9)

which has two solutions

\lambda_{1,2} = 2\pm\sqrt{3}. \qquad (10)

Consequently, the general solution to the recurrence relation (3) is explicitly written by

x_k = C_1\lambda_1^k + C_2\lambda_2^k, \qquad (11)

where $C_1, C_2$ are arbitrary constants that are determined by the boundary conditions:

\begin{aligned} x_0 = A &\quad\Rightarrow\quad C_1 + C_2 = A \\ x_0 = B &\quad\Rightarrow\quad C_1\lambda_1^N + C_2\lambda_2^N = B. \end{aligned} \qquad (12)

Solving the linear set of equation yields

\begin{cases} C_1 + C_2 = A, \\ C_1\lambda_1^N + C_2\lambda_2^N = B \end{cases} \Rightarrow \begin{cases} C_1 = \dfrac{B - A\lambda_2^N}{\lambda_1^N - \lambda_2^N}, \\ C_2 = \dfrac{A\lambda_1^N - B}{\lambda_1^N - \lambda_2^N}. \end{cases} \qquad (13)

Substituting this solution into (11), it is possible to rewrite it in the following symmetrical form

\begin{aligned} x_k &= C_1\lambda_1^k + C_2\lambda_2^k = \frac{B - A\lambda_2^N}{\lambda_1^N - \lambda_2^N}\lambda_1^k + \frac{A\lambda_1^N - B}{\lambda_1^N - \lambda_2^N}\lambda_2^k = \\ &= \frac{A\lambda_1^N\lambda_2^k - B\lambda_2^k + B\lambda_1^k - A\lambda_2^N\lambda_1^k}{\lambda_1^N - \lambda_2^N} = \frac{A\left(\lambda_1^{N-k} - \lambda_2^{N-k}\right) + B\left(\lambda_1^k - \lambda_2^k\right)}{\lambda_1^N - \lambda_2^N}. \end{aligned} \qquad (14)

The derivation of the last relation takes into account that according to the Vieta theorem $\lambda_2 = \lambda_1^{-1}$ .

1.6 [0.5 points] In view of the above formulas for the $\lambda_{1,2}^k$ , we find that

\lambda_1^k - \lambda_2^k = \lambda_1^k - \lambda_1^{-k} = \left(p_k + q_k\sqrt{3}\right) - \left(p_k - q_k\sqrt{3}\right) = 2q_k\sqrt{3}, \qquad (15)

and, finally,

x_k = \frac{A\left(\lambda_1^{N-k} - \lambda_2^{N-k}\right) + B\left(\lambda_1^k - \lambda_2^k\right)}{\lambda_1^N - \lambda_2^N} = \frac{Aq_{N-k} + Bq_k}{q_N}. \qquad (16)

2. Wire frame in the shape of a prism (7 т.)

2.1 [0.8 points] If the vertices of the cube with the same potentials are connected, then, the following equivalent circuits are obtained

and easily calculated using the standard method as

Ultimately, the cube resistance for the given connection is found as

R = \frac{7}{12}R_0. \qquad (17)

2.2 [0.2 points] Visual symmetry of the circuit and of the initial conditions provides obvious relations

y_k = -x_k, \qquad (18)

x_{N-k} = x_k. \qquad (19)

2.3 [1.0 points] The algebraic sum of the currents entering a node is equal to zero, thus, using Ohm's law, the following equation is obtained for the node $x_k$

\frac{x_{k-1} - x_k}{R_0} + \frac{x_{k+1} - x_k}{R_0} + \frac{y_k - x_k}{R_0} = 0. \qquad (20)

Since $y_k = -x_k$ , the recurrence relation holds

x_{k+1} - 4x_k + x_{k-1} = 0. \qquad (21)

2.4 [0.2 points] For an unambiguous determination of all values $x_k$ , we need to explicitly specify two boundary conditions. One of those is the initial potential defined as

x_0 = \varphi_0, \qquad (22)

whereas the other follows from the symmetry condition (19), which is valid for any $k$ , and, in particular, for $k = 0$ (despite the fact that the node with the number $N$ does not exist in the circuit!)

x_N = x_0. \qquad (23)

2.5 [0.2 points] The recurrence relation (21) has been considered in the Mathematical introduction. Therefore, you can use the obtained solution (16) by setting:

x_k = \frac{Aq_{N-k} + Bq_k}{q_N} = \varphi_0\frac{q_{N-k} + q_k}{q_N}. \qquad (24)

2.6 [0.4 points] The current in the source circuit is found as the sum of the currents flowing from the node $x_0$ :

I = \frac{x_0 - x_1}{R_0} + \frac{x_0 - x_{N-1}}{R_0} + \frac{x_0 - y_0}{R_0} = \frac{4x_0 - 2x_1}{R_0}. \qquad (25)

Here it has been taken into account that $y_0 = -x_0,\quad x_{N-1} = x_1$ . Substituting the values for $x_0, x_1$ , results in

\begin{aligned} I_0 &= \frac{4x_0 - 2x_1}{R_0} = \frac{2}{R_0}\left(2\phi_0 - \phi_0\frac{q_{N-1} + q_1}{q_N}\right) = \frac{2\phi_0}{R_0}\left(2 - \frac{q_{N-1} + 1}{q_N}\right) = \\ &= \frac{2\phi_0}{R_0}\frac{2q_N - q_{N-1} - 1}{q_N} = \frac{2\phi_0}{R_0}\frac{2q_N - \left(2q_N - p_N\right) - 1}{q_N} = \frac{2\phi_0}{R_0}\frac{p_N - 1}{q_N}. \end{aligned} \qquad (26)

At the last step the relation (5) has been used, $q_{N-1} = 2q_N - p_N$ .

2.7 [0.2 points] By formulation, the input voltage for the given circuit is

U_0 = 2\varphi_0, \qquad (27)

consequently, the resistance is found in the following elegant form

R_N = \frac{U_0}{I_0} = R_0\frac{q_N}{p_N - 1}. \qquad (28)

2.8 [1.0 points] Calculations are easily performed using numerical values in Table 1.

Table 2. Resistances of prisms.

$N$	$p_N$	$q_N$	$R_N$
1	2	1	$R_0$
2	7	4	$R_0\frac{4}{7 - 1} = \frac{2}{3}R_0$
3	26	15	$R_0\frac{15}{26 - 1} = \frac{3}{4}R_0$
4	97	56	$R_0\frac{56}{97 - 1} = \frac{7}{12}R_0$
5	362	209	$R_0\frac{209}{362 - 1} = \frac{11}{19}R_0$

Note that for a cubic prism with $N = 4$ the resistance coincides with that previously found in 2.1.

2.9 [0.5 points] For $N = 1$ the circuit is obvious:

but for $N = 2$ the prism should be additionally closed as:

In both cases the corresponding resistances coincide with the values shown in Table 2.

2.10 [1.0 points] The limit of the formula (28) can be found in various ways, for example, expressing

p_N = \frac{1}{2}\left(\lambda^N - \lambda^{-N}\right),\quad q_N = \frac{1}{2\sqrt{3}}\left(\lambda^N + \lambda^{-N}\right), \qquad (29)

where $\lambda = 2 + \sqrt{3} > 1$ .

Then,

R_\infty = \lim_{N\to\infty}R_N = R_0\lim_{N\to\infty}\frac{q_N}{p_N - 1} = R_0\lim_{N\to\infty}\frac{\frac{1}{2\sqrt{3}}\left(\lambda^N + \lambda^{-N}\right)}{\frac{1}{2}\left(\lambda^N - \lambda^{-N}\right) - 1} = \frac{R_0}{\sqrt{3}}. \qquad (30)

2.11 [1.5 points] Evaluation gives rise to

\frac{R_\infty}{R_0} = \frac{1}{\sqrt{3}}\approx 0.577. \qquad (31)

Then, we carry out the calculation of the relative error of the approximate expression for different values of $N$ listed in Table 2.

Table 3.

$N$	$R_N$	$\frac{R_N}{R_0}$	$\varepsilon = \frac{R_\infty - R_N}{R_N}$
1	$R_0$	1.000	-0.423
2	$\frac{2}{3}R_0$	0.667	-0.134
3	$\frac{3}{4}R_0$	0.750	-0.038
4	$\frac{7}{12}R_0$	0.583	-0.010
5	$\frac{11}{19}R_0$	0.579	$\lt$ -0.004

It is seen that already at $N = 4$ the relative error is 1%. Consequently, in this problem four is equal to infinity!

\infty\approx 4. \qquad (32)

	Content	points
1.1	Formula (1) $x_k = A\lambda^k$	0.2	0.2
1.2	Formula (3) $p_k = 2p_{k-1} + 3q_{k-1}\quad q_k = p_{k-1} + 2q_{k-1}$	0.2	0.4
	Formulas (5) $p_{k-1} = 2p_k - 3q_k\quad q_{k-1} = 2q_k - p_k$	0.2
1.3	Correct initial values $p_0 = 1,\quad q_0 = 0$	0.2	0.7
	Correct values in Table 1.	0.5
1.4	Formula (7) $\lambda^{-k} = p_k - q_k\sqrt{3}$	0.2	0.2
1.5	Formula (10) $\lambda_{1,2} = 2\pm\sqrt{3}$	0.2	1.0
	Formula (11) $x_k = C_1\lambda_1^k + C_2\lambda_2^k$	0.2
	Formula (12) $C_1 + C_2 = A\quad C_1\lambda_1^N + C_2\lambda_2^N = B$	0.2
	Solution (13) $C_1 = \frac{B - A\lambda_2^N}{\lambda_1^N - \lambda_2^N}\quad C_2 = \frac{A\lambda_1^N - B}{\lambda_1^N - \lambda_2^N}$	0.2
	Formula (14) $x_k = \frac{A\left(\lambda_1^{N-k} - \lambda_2^{N-k}\right) + B\left(\lambda_1^k - \lambda_2^k\right)}{\lambda_1^N - \lambda_2^N}$	0.2
1.6	Formula (16) $x_k = \frac{Aq_{N-k} + Bq_k}{q_N}$	0.5	0.5
2.1	Equivalent circuit	0.3	0.8
	Formula (17) $R = \frac{7}{12}R_0$	0.5
2.2	Formula (18) $y_k = -x_k$	0.1	0.2
	Formula (19) $x_{N-k} = x_k$	0.1
2.3	Formula (20) $\frac{x_{k-1} - x_k}{R_0} + \frac{x_{k+1} - x_k}{R_0} + \frac{y_k - x_k}{R_0} = 0$	0.5	1.0
	Formula (21) $x_{k+1} - 4x_k + x_{k-1} = 0$	0.5
2.4	Formula (22) $x_0 = \varphi_0$	0.1	0.2
	Formula (23) $x_N = x_0$	0.1
2.5	Formula (24) $x_k = \frac{Aq_{N-k} + Bq_k}{q_N} = \varphi_0\frac{q_{N-k} + q_k}{q_N}$	0.2	0.2
2.6	Formula (25) $I = \frac{4x_0 - 2x_1}{R_0}$	0.2	0.4
	Formula (26) $I_0 = \frac{2\varphi_0}{R_0}\frac{p_N - 1}{q_N}$	0.2
2.7	Formula (27) $U_0 = 2\varphi_0$	0.1	0.2
	Formula (28) $R_N = \frac{U_0}{I_0} = R_0\frac{q_N}{p_N - 1}$	0.1
2.8	Correct values in Table 2.	1.0	1.0
2.9	Equivalent circuit for $N = 1$	0.1	0.5
	Equivalent circuit for $N = 2$	0.4
2.10	Formula (29) $p_N = \frac{1}{2}\left(\lambda^N - \lambda^{-N}\right),\quad q_N = \frac{1}{2\sqrt{3}}\left(\lambda^N + \lambda^{-N}\right)$	0.5	1.0
	Formula (30) $R_\infty = \frac{R_0}{\sqrt{3}}$	0.5
2.11	Formula (31) $\frac{R_\infty}{R_0}\approx 0.577$	0.2	1.5
	Correct values in Table 3.	1.0
	Formula (32) $\infty\approx 4$	0.3
Total			10.0

Cube connected to a battery across two adjacent vertices, with dashed arrows joining vertices of equal potential, an arrow to the equivalent planar ladder circuit: vertical resistors R0, R0/2 and R0 between the two rails, and horizontal resistors R0/2 in the upper and lower rails.

Step-by-step reduction of the ladder circuit, linked by curved arrows: first the ladder with R0/2 horizontal and R0, R0/2, R0 vertical resistors; then R0 with R0/2 in parallel with 2R0; then R0 with 2/5 R0; then R0 in parallel with 7/5 R0; finally a single resistor 7/12 R0.

Prism with N = 1: a single edge R0 connected to a battery, an arrow, and the equivalent circuit of one resistor R0 between two terminals.

Prism with N = 2: two lateral edges connected by curved upper and lower edges R0, joined to a battery, an arrow, and the equivalent circuit with vertical resistors R0 and R0 and horizontal resistors R0/2 in each rail.

Equivalent circuit row of the grading table for 2.1: cube with a battery and dashed lines joining equal-potential vertices, an arrow, and the ladder circuit with R0, R0/2 and R0 vertical resistors and R0/2 horizontal resistors.

Equivalent circuit row of the grading table for N = 1: one edge R0 with a battery, an arrow, and a single resistor R0 between two terminals.

Equivalent circuit row of the grading table for N = 2: prism frame with curved edges R0 and a battery, an arrow, and the circuit with vertical resistors R0, R0 and horizontal resistors R0/2.

Оригинал в Архива: IZhO-2017-Theory_eng.pdf · официални решения: IZhO-2017-Theory_eng_sol.pdf