OPhO 2026, Единствен кръг — Задача 6. Stars and Deltas

2026 Online Physics Olympiad · 21 август 2026 г.

Last updated

Last updated: 22nd August 03:35 UTC

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Corrections (all times are in UTC)
  • Q8: Cleared up wording (21/08 04:40 and 04:52)
  • Q3: improved wording (21/08 08:49)
  • Q19: Improved wording (21/08 09:42)
  • Q16: Attempts reset (21/08 22:00), Error bound increased (22/8 02:30)
  • Q1: Improved wording (22/08 03:35)
Указания

Format. The OPhO is a team-based competition (3 members at most per team). Unlike previous years, the 2026 OPhO consists of a single round held from August 21, 12 am (UTC), to August 23, 11:59 pm (UTC). The contest is made up of three parts, all running concurrently over the full contest window:

  • Short Answer — problems requiring numerical answers, submitted on the OPhO portal;
  • Long Answer — problems requiring full written solutions, submitted on the OPhO portal;
  • Simulation — computational problems requiring teams to run a provided executable locally, with solutions submitted on the OPhO portal.

All teams have access to all three parts for the entire duration of the contest. Submissions are made on the OPhO portal https://opho.physoly.org/ using the team login, and may be made at any time within the contest window.

Short Answer. Teams submit numerical answers on the portal, with up to three attempts per problem. Every answer submission must be accompanied by a file upload, on the portal, of the working used to obtain that answer. This working will not be graded, but submissions without it will not be accepted. The base score for each problem depends on the number of teams that successfully answer it, each incorrect attempt multiplies the base score by a factor of 0.8, and there is no time factor. The score for an individual problem follows:

w(N)=(0.8)i[e−x(N)200+e−x(N)2100+1],w(N) = (0.8)^i\left[e^{-\frac{x(N)}{200}} + e^{-\frac{x(N)^2}{100}} + 1\right],

where i∈{0,1,2}i \in \{0, 1, 2\} represents whether a team got it on the first, second, or third try, NN represents the number of teams that answer the problem correctly, and x(N)x(N) is the sum of (0.8)i(0.8)^i across all teams that have solved the problem. The base score is independent of problem number, but expect questions to be ordered roughly by difficulty. A team’s raw Short Answer total is the sum of these values, and all raw totals are then scaled linearly such that the top team receives 55 points.

Long Answer. Teams submit full solutions on the portal. Rubrics are predetermined, and solutions will be reviewed to determine the proper number of points to give. The points breakdown is given per problem, for a total of 35 points. It is recommended that participants write their solutions in LaTeX. Handwritten solutions (or a combination of both) are accepted too; if participants have more than one photo of a handwritten solution (jpg, png, etc.), it is required that they be organized in the correct order in a pdf before submitting.

Simulation. Teams will need to be able to run a provided executable locally. Solutions are submitted on the portal and will be graded against predetermined rubrics. The points breakdown is given per problem, for a total of 10 points.

Scoring. A team’s cumulative score is the sum of its scaled Short Answer score (out of 55 points), Long Answer score (out of 35 points), and Simulation score (out of 10 points), for a maximum score of 100 points.

Permitted Resources. Use of the internet is allowed, including textbooks, published papers, calculators, and computer algebra systems. However, the use of AI tools of any kind is strictly prohibited. This includes, but is not limited to, large language models (e.g. ChatGPT, Claude, Gemini), AI coding assistants, AI-generated search summaries, and any other AI-powered assistance.

Dishonest Behavior. Please refrain from cheating in any form. This includes, but is not limited to, creating alternate accounts to gain additional attempts, sharing answers with other teams, posting problems online, utilizing AI tools as described above, or employing any other method to gain an unfair advantage.

We have previously caught numerous teams engaging in such practices, and pursuing dishonest tactics provides no genuine sense of security. Any teams found to be acting maliciously will be notified via email, along with the opportunity to appeal. Failing to provide a valid appeal will result in a ban from participating in any future Physoly competitions. Our utmost priority is to ensure an enjoyable contest experience for all participants, and we remain committed to preventing any disruption to the fun for others.

Awards and Results. The top competitors will be recognized on the website. No cash prizes will be awarded this year.

List of Constants
  • Proton mass, mp=1.67⋅10−27 kgm_p = 1.67 \cdot 10^{-27}\ \mathrm{kg}
  • Neutron mass, mn=1.67⋅10−27 kgm_n = 1.67 \cdot 10^{-27}\ \mathrm{kg}
  • Electron mass, me=9.11⋅10−31 kgm_e = 9.11 \cdot 10^{-31}\ \mathrm{kg}
  • Avogadro’s constant, N0=6.02⋅1023 mol−1N_0 = 6.02 \cdot 10^{23}\ \mathrm{mol^{-1}}
  • Universal gas constant, R=8.31 J/(mol⋅K)R = 8.31\ \mathrm{J/(mol \cdot K)}
  • Boltzmann’s constant, kB=1.38⋅10−23 J/Kk_B = 1.38 \cdot 10^{-23}\ \mathrm{J/K}
  • Electron charge magnitude, e=1.60⋅10−19 Ce = 1.60 \cdot 10^{-19}\ \mathrm{C}
  • 1 electron volt, 1 eV=1.60⋅10−19 J1\ \mathrm{eV} = 1.60 \cdot 10^{-19}\ \mathrm{J}
  • Speed of light, c=3.00⋅108 m/sc = 3.00 \cdot 10^{8}\ \mathrm{m/s}
  • Universal Gravitational constant,
    G=6.67⋅10−11 (N⋅m2)/kg2G = 6.67 \cdot 10^{-11}\ \mathrm{(N \cdot m^2)/kg^2}
  • Solar Mass
    M⊙=1.988⋅1030 kgM_\odot = 1.988 \cdot 10^{30}\ \mathrm{kg}
  • Acceleration due to gravity, g=9.8 m/s2g = 9.8\ \mathrm{m/s^2}
  • 1 unified atomic mass unit,
    1 u=1.66⋅10−27 kg=931 MeV/c21\ \mathrm{u} = 1.66 \cdot 10^{-27}\ \mathrm{kg} = 931\ \mathrm{MeV/c^2}
  • Planck’s constant,
    h=6.63⋅10−34 J⋅s=4.41⋅10−15 eV⋅sh = 6.63 \cdot 10^{-34}\ \mathrm{J \cdot s} = 4.41 \cdot 10^{-15}\ \mathrm{eV \cdot s}
  • Permittivity of free space,
    ϵ0=8.85⋅10−12 C2/(N⋅m2)\epsilon_0 = 8.85 \cdot 10^{-12}\ \mathrm{C^2/(N \cdot m^2)}
  • Coulomb’s law constant,
    k=14πϵ0=8.99⋅109 (N⋅m2)/C2k = \frac{1}{4\pi\epsilon_0} = 8.99 \cdot 10^{9}\ \mathrm{(N \cdot m^2)/C^2}
  • Permeability of free space,
    μ0=4π⋅10−7 T⋅m/A\mu_0 = 4\pi \cdot 10^{-7}\ \mathrm{T \cdot m/A}
  • Magnetic constant,
    μ04π=1⋅10−7 (T⋅m)/A\frac{\mu_0}{4\pi} = 1 \cdot 10^{-7}\ \mathrm{(T \cdot m)/A}
  • 1 atmospheric pressure,
    1 atm=1.01⋅105 N/m2=1.01⋅105 Pa1\ \mathrm{atm} = 1.01 \cdot 10^{5}\ \mathrm{N/m^2} = 1.01 \cdot 10^{5}\ \mathrm{Pa}
  • Wien’s displacement constant, b=2.9⋅10−3 m⋅Kb = 2.9 \cdot 10^{-3}\ \mathrm{m \cdot K}
  • Stefan-Boltzmann constant,
    σ=5.67⋅10−8 W/m2/K4\sigma = 5.67 \cdot 10^{-8}\ \mathrm{W/m^2/K^4}

Условие

Consider two infinite planar resistive networks.

Circuit 1. The network is obtained by tiling the plane with rhombi having interior angles 60° and 120°, as depicted. Each edge of the tiling is occupied by a resistor. Edges are partitioned into three classes by orientation, indicated by colour: yellow edges each carry resistance R1R_1, blue edges each carry resistance R2R_2, and green edges each carry resistance R3R_3. Let RABR_{AB} denote the equivalent resistance measured between the terminals AA and BB marked in the figure.

Circuit 2. The network is obtained by tiling the plane with regular hexagons and equilateral triangles (the tri-hexagonal tiling), as depicted, with a resistor on every edge. Grey edges each carry resistance R4R_4, red edges each carry resistance R5R_5, and magenta edges each carry resistance R6R_6. Let RCDR_{CD} denote the equivalent resistance between the terminals CC and DD marked in the figure.

Two network drawings with dashed lines indicating they continue to infinity. Left, labelled 'Circuit 1': a tiling of 60°/120° rhombi with yellow horizontal edges and blue and green slanted edges; two nodes are marked by hand in red as A (lower) and B (upper). Right, labelled 'Circuit 2': a tri-hexagonal tiling inside a hexagonal outline with grey vertical edges, red edges rising to the right and magenta edges falling to the right; two nearby nodes are marked by hand in red as C and D.

In both networks, all mutually parallel edges (i.e. all edges of the same orientation) belong to the same colour class and hence carry equal resistance. Note that there are no direct resistor connections between nodes A and B, nor between the nodes C and D.

Given:R1=1 ΩR_1 = 1\,\Omega, R2=2 ΩR_2 = 2\,\Omega, R3=3 ΩR_3 = 3\,\Omega, R4=3 ΩR_4 = 3\,\Omega, R5=1.5 ΩR_5 = 1.5\,\Omega, R6=1 ΩR_6 = 1\,\Omega.

Determine the product RAB×RCDR_{AB} \times R_{CD}, expressed in Ω2\Omega^2.

Отговори

Решение


Оригинал в Архива: problems.pdf · официални решения: solutions.pdf