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Условие Отговори Решение

OPhO 2026, Единствен кръг — Задача 21. Teo II

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2026 Online Physics Olympiad · 21 август 2026 г.

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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\left[e^{-\frac{x(N)}{200}} + e^{-\frac{x(N)^2}{100}} + 1\right],

where $i \in \{0, 1, 2\}$ represents whether a team got it on the first, second, or third try, $N$ represents the number of teams that answer the problem correctly, and $x(N)$ is the sum of $(0.8)^i$ 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, $m_p = 1.67 \cdot 10^{-27}\ \mathrm{kg}$
Neutron mass, $m_n = 1.67 \cdot 10^{-27}\ \mathrm{kg}$
Electron mass, $m_e = 9.11 \cdot 10^{-31}\ \mathrm{kg}$
Avogadro’s constant, $N_0 = 6.02 \cdot 10^{23}\ \mathrm{mol^{-1}}$
Universal gas constant, $R = 8.31\ \mathrm{J/(mol \cdot K)}$
Boltzmann’s constant, $k_B = 1.38 \cdot 10^{-23}\ \mathrm{J/K}$
Electron charge magnitude, $e = 1.60 \cdot 10^{-19}\ \mathrm{C}$
1 electron volt, $1\ \mathrm{eV} = 1.60 \cdot 10^{-19}\ \mathrm{J}$
Speed of light, $c = 3.00 \cdot 10^{8}\ \mathrm{m/s}$
Universal Gravitational constant, $G = 6.67 \cdot 10^{-11}\ \mathrm{(N \cdot m^2)/kg^2}$
Solar Mass $M_\odot = 1.988 \cdot 10^{30}\ \mathrm{kg}$
Acceleration due to gravity, $g = 9.8\ \mathrm{m/s^2}$
1 unified atomic mass unit, $1\ \mathrm{u} = 1.66 \cdot 10^{-27}\ \mathrm{kg} = 931\ \mathrm{MeV/c^2}$
Planck’s constant, $h = 6.63 \cdot 10^{-34}\ \mathrm{J \cdot s} = 4.41 \cdot 10^{-15}\ \mathrm{eV \cdot s}$
Permittivity of free space, $\epsilon_0 = 8.85 \cdot 10^{-12}\ \mathrm{C^2/(N \cdot m^2)}$
Coulomb’s law constant, $k = \frac{1}{4\pi\epsilon_0} = 8.99 \cdot 10^{9}\ \mathrm{(N \cdot m^2)/C^2}$
Permeability of free space, $\mu_0 = 4\pi \cdot 10^{-7}\ \mathrm{T \cdot m/A}$
Magnetic constant, $\frac{\mu_0}{4\pi} = 1 \cdot 10^{-7}\ \mathrm{(T \cdot m)/A}$
1 atmospheric pressure, $1\ \mathrm{atm} = 1.01 \cdot 10^{5}\ \mathrm{N/m^2} = 1.01 \cdot 10^{5}\ \mathrm{Pa}$
Wien’s displacement constant, $b = 2.9 \cdot 10^{-3}\ \mathrm{m \cdot K}$
Stefan-Boltzmann constant, $\sigma = 5.67 \cdot 10^{-8}\ \mathrm{W/m^2/K^4}$

Условие

Kai Wen has discovered another planet: Teo II, a small, dense core with mass $M$ surrounded by an atmosphere with negligible mass. The atmosphere is isothermal with temperature $T$ , and is composed of two gases with equal adiabatic constants but different molecular masses $\alpha m, (1 - \alpha)m$ where $\alpha < 1/2$ . Define the dimensionless radial coordinate $\eta = \dfrac{rkT}{GMm}$ , and let the mole fraction of the lighter gas be $\chi(\eta)$ .

Kai Wen is still angry about his decisive poker loss to the inhabitants of Teo I. He detonates a large bomb at radius $\eta_0$ , sending sound waves through the atmosphere. After a long time, he notices that the sound intensity is large at radius $\eta_0$ .

(a) Find $\eta_0$ in terms of $\chi_0 = \chi(\eta_0)$ . [3.5 pts]

(b) What are the possible values for $\eta_0$ , if $\alpha$ is fixed but the atmospheric composition can vary? [3.5 pts]

(c) Inspecting more closely, he sees that in general, the locations of locally maximum intensity on the sphere of radius $\eta_0$ form rings. What is the smallest possible $\eta_0$ such that there are no rings? [3.0 pts]

Отговори

Решение

(a) If the speed of sound were constant, the sound would not be refracted and all of it would escape. Thus, the trapping of the sound must be due to a varying speed of sound $c(\eta)$ . Considering a ray at angle $\theta$ , divide the atmosphere into slices of height $d\eta$ , each with constant speed of sound:

Two concentric circular arcs (curving downwards to the right): the lower one labelled η with speed of sound c, the upper one labelled η + dη with speed of sound c + dc. A blue ray leaves the lower arc at angle θ to it, crosses to the upper arc where it makes angle θ′ with the arc, and continues as an arrow at angle θ + dθ.

Snell's law gives $\dfrac{1}{c}\cos(\theta') = \dfrac{1}{c + dc}\cos(\theta + d\theta)$ . Furthermore, note that $\theta' = \theta + \dfrac{d\eta}{\eta\tan(\theta)}$ . Altogether, we obtain $\tan(\theta)\dfrac{d\theta}{d\eta} = -\dfrac{1}{c}\dfrac{dc}{d\eta} + \dfrac{1}{r} = -\dfrac{d\ln(c)}{d\eta} + \dfrac{1}{\eta}$ , where the first term is due to refraction and the second is due to the curvature of the sphere. We expect the intensity to be maximized at the radius where rays can propagate indefinitely; thus, we need $\left.\dfrac{d\ln(c)}{d\eta}\right|_{\eta_0} = \dfrac{1}{\eta_0}$ .
The cause of the changing speed of sound is the changing density of the atmosphere. Note that the density is $\rho = mn(\alpha\chi + (1 - \alpha)(1 - \chi))$ where $n$ is the total number density, while the pressure is $P = nkT$ . Thus, since the composite gas has constant adiabatic index $\gamma$ , we have:

c = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma}{\alpha\chi + (1 - \alpha)(1 - \chi)}\frac{kT}{m}}

\Rightarrow \frac{d\ln(c)}{d\eta} = -\frac{1}{2}\frac{d}{d\eta}\ln(\alpha\chi + (1 - \alpha)(1 - \chi)) = \frac{1 - 2\alpha}{2}\frac{\chi'}{\alpha\chi + (1 - \alpha)(1 - \chi)}

Letting the number densities of each gas be $n_i$ , thermal equilibrium implies:

n_i(\eta) \propto \exp\left(-\frac{m_iV(\eta)}{kT}\right) = \exp\left(\frac{m_i}{m\eta}\right)

Giving $n_i' = -\dfrac{m_i}{m\eta^2}n_i$ . Then, we have:

\chi' = \frac{d}{d\eta}\frac{n_1}{n_1 + n_2} = \frac{n_1'n_2 - n_1n_2'}{(n_1 + n_2)^2} = (n_1'/n_1 - n_2'/n_2)\frac{n_1n_2}{(n_1 + n_2)^2} = (1 - 2\alpha)\chi(1 - \chi)\frac{1}{\eta^2}

\Rightarrow \frac{1}{\eta_0} = \left.\frac{d\ln(c)}{d\eta}\right|_{\eta_0} = \frac{(1 - 2\alpha)^2}{2}\frac{\chi_0(1 - \chi_0)}{\alpha\chi_0 + (1 - \alpha)(1 - \chi_0)}\frac{1}{\eta_0^2}

\Rightarrow \boxed{\eta_0 = \frac{(1 - 2\alpha)^2}{2}\frac{\chi_0(1 - \chi_0)}{\alpha\chi_0 + (1 - \alpha)(1 - \chi_0)}}

(b) One might think that any $\eta_0$ given by some $\chi_0 \in (0, 1)$ is possible. However, our only requirement above was that $\eta_0$ is a stationary point; we still need to show stability, so that nearby rays oscillate around $\eta_0$ .
For this, note that $\tan(\theta)\dfrac{d\theta}{d\eta} = \dfrac{d\theta}{d\xi} \approx \dfrac{d^2\eta}{d\xi^2}$ where $\xi$ is the tangential distance, giving $\dfrac{d^2\eta}{d\xi^2} \approx -\dfrac{d\ln(c/\eta)}{d\eta}$ . Thus, we can consider $\ln(c/\eta)$ as an effective potential, whose local minima are stable equilibria for $\eta$ . We compute:

\begin{aligned} \frac{d^2\ln(c/\eta)}{d\eta^2} &= \frac{d}{d\eta}\left(\frac{1 - 2\alpha}{2}\frac{\chi'}{\alpha\chi + (1 - \alpha)(1 - \chi)} - \frac{1}{\eta}\right) \\ &= \frac{1 - 2\alpha}{2}\left(\frac{\chi''}{\alpha\chi + (1 - \alpha)(1 - \chi)} - \frac{(2\alpha - 1)\chi'^2}{(\alpha\chi + (1 - \alpha)(1 - \chi))^2}\right) + \frac{1}{\eta^2} \end{aligned}

Note that:

\chi'' = \frac{d}{d\eta}\frac{(1 - 2\alpha)\chi(1 - \chi)}{\eta^2} = \left(\frac{(1 - 2\alpha)(1 - 2\chi)}{\eta^2} - \frac{2}{\eta}\right)\chi'

\begin{aligned} \Rightarrow \left.\frac{d^2\ln(c/\eta)}{d\eta^2}\right|_{\eta_0} &= \left(\frac{(1 - 2\alpha)(1 - 2\chi_0)}{\eta_0^2} - \frac{2}{\eta_0}\right)\cdot\frac{1}{\eta_0} + \frac{2}{\eta_0^2} + \frac{1}{\eta_0^2} \\ &= \frac{(1 - 2\alpha)(1 - 2\chi_0)}{\eta_0^3} + \frac{1}{\eta_0^2} \\ &= \frac{1 - 2\alpha}{2\eta_0^3(\alpha\chi_0 + (1 - \alpha)(1 - \chi_0))}(3(1 - 2\alpha)\chi_0^2 + (6\alpha - 5)\chi_0 + 2(1 - \alpha)) \end{aligned}

Thus, $\left.\dfrac{d^2\ln(c/\eta)}{d\eta^2}\right|_{\eta_0} > 0$ iff $\chi_0 \in \left(0, \dfrac{5 - 6\alpha - \sqrt{1 + 12\alpha - 12\alpha^2}}{6(1 - 2\alpha)}\right)$ . One can verify that $\eta_0$ is an increasing function of $\chi_0$ on this interval (the local maximum is at $\chi_0 = \dfrac{\sqrt{1 - \alpha}}{\sqrt{\alpha} + \sqrt{1 - \alpha}}$ ), showing that $\boxed{\eta_0 \in \left(0, \dfrac{2 - \sqrt{1 + 12\alpha - 12\alpha^2}}{3}\right)}$ .
(c) Since near-horizontal rays oscillate around $\eta_0$ with the same period (and all originate at the detonation point), the rings are the locations where the rays converge, and occur at spacing $\lambda/2$ where $\lambda$ is the wavelength of the oscillation. Using $\dfrac{d^2\eta}{d\xi^2} \approx -\left.\dfrac{d^2\ln(c/\eta)}{d\eta^2}\right|_{\eta_0}\eta$ , we have $\left(\dfrac{2\pi}{\lambda}\right)^2 = \left.\dfrac{d^2\ln(c/\eta)}{d\eta^2}\right|_{\eta_0}$ .
For there to be no rings, the only location of convergence can be at the poles, meaning $\lambda/\eta_0 \in 2\pi, 4\pi, \ldots$ . Note that $\eta_0^2\left.\dfrac{d^2\ln(c/\eta)}{d\eta^2}\right|_{\eta_0} = \dfrac{(1 - 2\alpha)(1 - 2\chi_0)}{\eta_0} + 1$ is a decreasing function of $\eta_0$ , so to minimize $\eta_0$ we should take $\lambda/\eta_0 = 2\pi$ . Thus, we need $\left.\dfrac{d^2\ln(c/\eta)}{d\eta^2}\right|_{\eta_0} = \dfrac{1}{\eta_0^2}$ , implying $\chi_0 = 1/2$ (independent of $\alpha$ !) and $\boxed{\eta_0 = (1 - 2\alpha)^2/4}$ .

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