Съдържание

OPhO 2020, Открит кръг — Задача 33. AIME

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Всички материали от OPhO 2020

2020 Online Physics Olympiad (OPhO): Open Contest v1.5 · 25 май 2020 г.

2020 Online Physics Olympiad (OPhO): Open Contest v1.5

OPhO Lead Organizers and Writers: Adithya Balachandran, Eddie Chen, Ashmit Dutta, Ethan Hu, Viraj Jayam, Rishab Parthasarathy, Kushal Thaman, Max Wang, QiLin Xue

OPhO Website Developer: Ben Oommen

Advisors: Dr. Branislav Kisačanin, Kye Shi, Kevin Zhou

Указания

If you wish to request a clarification, please use this form. To see all clarifications, see this document.

Use $g = 9.81\ \mathrm{m/s^2}$ in this contest. See the constants sheet on the following page for other constants.
This test contains 55 short answer questions. Each problem will have three possible attempts.
The weight of each question depends on our scoring system found here. Put simply, the later questions are worth more, and the overall amount of points from a certain question decreases with the number of attempts and days that you take to solve a problem as well as the number of teams who solve it. This means that your score decreases with the number of tries and days you take to solve a given problem.
Any team member is able to submit an attempt. Choosing to split up the work or doing each problem together is up to you. Note that after you have submitted an attempt, your teammates must refresh their page before they are able to see it.
Answers should contain at least three significant figures, unless otherwise specified. All answers within the 1% range will be accepted.
When submitting a response using scientific notation, please use exponential form. In other words, if your answer to a problem is $A \times 10^B$ , please type $AeB$ into the submission portal.
A standard scientific or graphing handheld calculator may be used. Technology and computer algebra systems like Wolfram Alpha or the one in the TI nSpire will not be needed. Attempts to use these tools will be classified as cheating.
You are allowed to use Wikipedia or books in this exam. Asking for help on online forums or your teachers will be considered cheating and may result in a possible ban from future competitions.
Top scorers from this contest will qualify to compete in the Online Physics Olympiad Invitational Contest, which is an olympiad-style exam. More information will be provided to invitational qualifiers after the end of the Open Contest.
In general, answer in SI units (meter, second, kilogram, watt) unless otherwise specified. Please answer all questions in degrees unless otherwise specified.
If the question asks to give your answer as a percent and your answer comes out to be “ $x\%$ ”, please input the value $x$ into the submission form.
Do not put letters in your answer on the submission portal! If your answer is “ $x$ meters” , input only the value $x$ into the submission portal.
Do not communicate information to anyone else apart from your team-members before the exam ends on May 29, 2020 at 11:59 PM UTC.

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, $e = 1.60 \cdot 10^{-19}$
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,
$G = 6.67 \cdot 10^{-11}\ \mathrm{(N \cdot m^2)/kg^2}$
Acceleration due to gravity, $g = 9.81\ \mathrm{m/s^2}$
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

Three unit circles, each with radius 1 meter, lie in the same plane such that the center of each circle is one intersection point between the two other circles, as shown below. Mass is uniformly distributed among all area enclosed by at least one circle. The mass of the region enclosed by the triangle shown above is $1\ \mathrm{kg}$ . Let $x$ be the moment of inertia of the area enclosed by all three circles (intersection, not union) about the axis perpendicular to the page and through the center of mass of the triangle. Then, $x$ can be expressed as $\frac{a\pi - b\sqrt{c}}{d\sqrt{e}}$ $\mathrm{kg\ m^2}$ , where $a, b, c, d, e$ are integers such that $\gcd(a, b, d) = 1$ and both $c$ and $e$ are squarefree. Compute $a + b + c + d + e$ .

Three unit circles (red on top, blue lower left, green lower right), each centred at an intersection point of the other two; a black equilateral triangle joins the three centres.

Отговори

Решение

Solution: Define point $O$ as the point in the plane that the axis of rotation passes through. Since moments of inertia simply add about a given axis, we can calculate the moments of inertia of the three ”sectors” whose union forms the given area and subtract twice the moment of inertia of the triangle, so our answer will be $3I_{s,O} - 2I_{t,O}$ .

Claim: The center of mass of a sector is $\frac{2}{\pi}$ away from the vertex of the sector along its axis of symmetry.

Proof: We can divide the sector into arbitrarily small sectors that can be approximated as isosceles triangles. It’s well known that the center of mass of one such isosceles triangle is $\frac{2}{3}$ of the way from the central vertex to the base. Therefore, the center of mass of the sector is the center of mass of the arc with central angle $\frac{\pi}{3}$ and same center with radius $\frac{2}{3}$ contained within the sector. Since the center of mass has to lie on the axis of symmetry, we set that as the x axis with the vertex of the sector being $x = 0$ . Then, the $x$ -coordinate of a point on the arc whose corresponding radius makes an angle of $\theta$ with the axis of symmetry is $\frac{2}{3}\cos(\theta)$ . We can integrate this over all possible angles ( $-\frac{\pi}{6} \le \theta \le \frac{\pi}{6}$ ) and then divide by the range ( $\frac{\pi}{3}$ ) to get the average $x$ -coordinate, or the center of mass.

\frac{\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{2}{3}\cos(\theta)\mathrm{d}\theta}{\frac{\pi}{3}}

\frac{2}{\pi}\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \cos(\theta)\mathrm{d}\theta

\frac{2}{\pi}\left(\sin\left(\frac{\pi}{6}\right) - \sin\left(-\frac{\pi}{6}\right)\right)

\frac{2}{\pi}

This concludes the proof. $\square$

Now define point $X$ as the vertex of a sector and point $M$ as the center of mass of that sector. According to the parallel axis theorem,

I_{s,X} = I_{s,M} + m_s\left(\frac{2}{\pi}\right)^2

. It’s well known that $I_{s,X} = \frac{1}{2}m_s r^2 = \frac{m_s}{2}$ , and so

I_{s,M} = \frac{m_s}{2} - \frac{4m_s}{\pi^2} = m_s\left(\frac{\pi^2 - 8}{2\pi^2}\right)

It’s also well known that $O$ is on the line of symmetry and a distance of $\frac{1}{\sqrt{3}}$ away from $X$ , and so $MX = \frac{2}{\pi} - \frac{1}{\sqrt{3}}$ . Therefore,

I_{s,O} = I_{s,M} + m_s\left(\frac{2}{\pi} - \frac{1}{\sqrt{3}}\right)^2 = m_s\left(\frac{5\pi - 8\sqrt{3}}{6\pi}\right)

It’s well known that, since $O$ is the center of mass of the triangle,

I_{t,O} = \frac{1}{12}

Now we just need to calculate $m_s$ . Since the mass of the triangle is 1 kg, this is equivalent to finding the ratio of the area of a sector to the area of a triangle. Through geometry, this is found to be $\frac{2\pi}{3\sqrt{3}}$ . Finally, we get our answer to be

\left(\frac{2\pi}{\sqrt{3}}\right)\left(\frac{5\pi - 8\sqrt{3}}{6\pi}\right) - \frac{1}{6} = \left(\frac{10\pi - 17\sqrt{3}}{6\sqrt{3}}\right)

and $a + b + c + d + e = 10 + 17 + 3 + 6 + 3 = \boxed{039}$

Оригинал в Архива: open.pdf · официални решения: open_sol.pdf