Съдържание

USAAAO 2024, II кръг — Short 3

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Всички материали от USAAAO 2024

2024 National Astronomy Competition · 15 т.

Внимание: Бележка към темата

All eight whole original problems, their native sections, lettered parts, three roman subparts of Long 1(f), full official worked keys and original marking/acceptance notes are retained. Printed scientific discrepancies are preserved in local source notes. The official key adds a Friedmann-variable clarification and a solar apparent magnitude assumption absent from the question booklet, and changes Long1(f)(iii) from a factor to a percentage; these source variants are explicitly attributed. A separate undated common archive constants/reference table is included with its own source identity; its use as the exact 2024 provided attachment is not established, and its constants are not substituted into the printed key. The native cover specifies blank paper rather than a designated answer sheet. April 13th is the end of the examination period, not an established single sitting date.

1 Instructions (Please Read Carefully)

The top 5 eligible scorers on the NAC will be invited to represent USA at the next IOAA. In order to qualify for the national team, you must be a high school student with US citizenship or permanent residency.

This exam consists of 3 parts: Short Questions, Medium Questions and Long Questions.

The maximum number of points is 225 points.

The test must be completed within 2.5 hours (150 minutes).

Please solve each problem on a blank piece of paper and mark the number of the problem at the top of the page. The contestant’s full name in capital letters should appear at the top of each solution page. If the contestant uses scratch papers, those should be labeled with the contestant’s name as well and marked as “scratch paper” at the top of the page. Scratch paper will not be graded. Partial credit will be available given that correct and legible work was displayed in the solution.

This is a written exam. Contestants can only use a scientific or graphing calculator for this exam. A table of physical constants will be provided. Discussing the problems with other people is strictly prohibited in any way until the end of the examination period on April 13th. Receiving any external help during the exam is strictly prohibited. This means that the only allowed items are: a calculator, the provided table of constants, a pencil (or pen), an eraser, blank sheets of papers, and the exam. No books or notes are allowed during the exam. Exam is proctored and recorded. You are expected to have your video on at all times.

We acknowledge the following people for their contributions to this year’s exam:

Abhay Bestrapalli, Orion Foo, Hagan Hensley, David Lee, Sandesh Kalantre, Andrew Liu, Joe McCarty, Leo Yao

Common archive reference — Table of Physical Constants

This separate common USAAAO archive reference is undated in the catalogue. Its use as the exact 2024 exam attachment is not established.

Quantity	Symbol	Value
Electron rest mass	$m_e$	$9.109\times10^{-31}\ \mathrm{kg}$
Proton rest mass	$M_p$	$1.6726\times10^{-27}\ \mathrm{kg}$
Electronic charge	$e$	$1.6022\times10^{-19}\ \mathrm C$
Speed of light in free space	$c$	$2.9979\times10^8\ \mathrm{m\ s^{-1}}$
Permeability of free space	$\mu_0$	$4\pi\times10^{-7}\ \mathrm{H\ m^{-1}}$
Permittivity of free space	$\epsilon_0$	$8.854\times10^{-12}\ \mathrm{F\ m^{-1}}$
Planck’s constant	$h$	$6.626\times10^{-34}\ \mathrm{J\ s}$
Reduced Planck’s constant	$\hbar=h/2\pi$	$1.0546\times10^{-34}\ \mathrm{J\ s}$
	$\hbar c$	$197.33\ \mathrm{MeV\ fm}$
Boltzmann’s constant	$k_B$	$1.3807\times10^{-23}\ \mathrm{J\ K^{-1}}$
Gas constant	$\mathcal R=k_B/m_H$	$8.250\times10^3\ \mathrm{J\ kg^{-1}\ K^{-1}}$
Molar gas constant	$R$	$8.315\ \mathrm{J\ mol^{-1}\ K^{-1}}$
Avogadro’s number	$N_A$	$6.022\times10^{23}\ \mathrm{mol^{-1}}$
Standard molar volume		$22.414\times10^{-3}\ \mathrm{m^3\ mol^{-1}}$
Unified atomic mass unit ( $^{12}\mathrm C$ scale)	$u$	$931.5\ \mathrm{MeV}/c^2=1.660538\times10^{-27}\ \mathrm{kg}$
Mass of hydrogen atom	$m_H$	$1.0078u=1.6735\times10^{-27}\ \mathrm{kg}$
Bohr magneton	$\mu_B$	$9.274\times10^{-24}\ \mathrm{A\ m^2}$ or $\mathrm{J\ T^{-1}}$
Nuclear magneton	$\mu_N$	$5.051\times10^{-27}\ \mathrm{A\ m^2}$ or $\mathrm{J\ T^{-1}}$
Proton magnetic moment	$\mu_p$	$2.7928\mu_N$
Neutron magnetic moment	$\mu_n$	$-1.9130\mu_N$
Bohr radius	$a_0$	$5.292\times10^{-11}\ \mathrm m$
Fine structure constant	$\alpha=e^2/(4\pi\epsilon_0\hbar c)$	$(137.04)^{-1}$
Compton wavelength of electron	$\lambda_C=h/(m_ec)$	$2.4263\times10^{-12}\ \mathrm m$
Rydberg’s constant	$R_{\infty}$	$1.0974\times10^7\ \mathrm{m^{-1}}$
	$R_{\infty}hc$	$13.606\ \mathrm{eV}$
Stefan-Boltzmann constant	$\sigma$	$5.671\times10^{-8}\ \mathrm{W\ m^{-2}\ K^{-4}}$
Radiation density constant	$a=4\sigma/c$	$7.561\times10^{-16}\ \mathrm{J\ m^{-3}\ K^{-4}}$
Gravitational constant	$G$	$6.673\times10^{-11}\ \mathrm{N\ m^2\ kg^{-2}}$

Rest masses of some leptons and hadrons in $\mathrm{MeV}/c^2$ :

$e^{\pm}$ 0.5110, $\mu^{\pm}$ 105.66, $\tau^{\pm}$ 1777, $\pi^0$ 134.98, $\pi^{\pm}$ 139.57, $\mathrm K^{\pm}$ 493.7, $\mathrm K^0$ 497.7, $\eta$ 547, $\mathrm D^0$ 1865, $\mathrm D^{\pm}$ 1869, p 938.3, n 939.6, $\Lambda^0$ 1115.7, $\Sigma^+$ 1189, $\Sigma^0$ 1193, $\Sigma^-$ 1197, $\Xi^0$ 1315, $\Xi^-$ 1321, $\Omega^-$ 1672, $\mathrm Z^0$ $91.187\times10^3$ , $\mathrm W^{\pm}$ $80.41\times10^3$ .

Quark	Charge	$I_3$	$S$	$C$	$B$	$T$	Mass ( $\mathrm{GeV}/c^2$ )
u	$+\frac23$	$\frac12$	0	0	0	0	$\sim0.003$
d	$-\frac13$	$-\frac12$	0	0	0	0	$\sim0.006$
c	$+\frac23$	0	0	+1	0	0	$\sim1.25$
s	$-\frac13$	0	-1	0	0	0	$\sim0.11$
t	$+\frac23$	0	0	0	0	+1	174.3
b	$-\frac13$	0	0	0	-1	0	4.2

Common archive reference — Astrophysical Data and identities

Astrophysical Data

Quantity	Symbol	Value
1 astronomical unit	AU	$1.496\times10^{11}\ \mathrm m$
1 parsec	pc	$3.086\times10^{16}\ \mathrm m$
Luminosity of Sun	$L_{\odot}$	$3.85\times10^{26}\ \mathrm W$
Mass of Sun	$M_{\odot}$	$1.989\times10^{30}\ \mathrm{kg}$
Radius of Sun	$R_{\odot}$	$6.96\times10^8\ \mathrm m$
Mass of Earth	$M_E$	$5.9742\times10^{24}\ \mathrm{kg}$
Radius of Earth	$R_E$	$6.3781\times10^6\ \mathrm m$

Other data and conversion factors

Quantity	Symbol	Value
1 ångstrom	Å	$10^{-10}\ \mathrm m$
1 fermi	fm	$10^{-15}\ \mathrm m$
1 barn	b	$10^{-28}\ \mathrm{m^2}$
1 pascal	Pa	$1\ \mathrm{Nm^{-2}}$
1 standard atmosphere		$1.0132\times10^5\ \mathrm{Pa}$
Standard acceleration due to gravity	$g$	$9.807\ \mathrm{m\ s^{-2}}$
1 electron volt	eV	$1.6022\times10^{-19}\ \mathrm J$
	$\mathrm{eV}/hc$	$8.065\times10^5\ \mathrm{m^{-1}}$
	$\mathrm{eV}/k_B$	$1.1604\times10^4\ \mathrm K$
Wavelength of 1 eV photon		$1.2399\times10^{-6}\ \mathrm m$

Trigonometrical identities

\sin(\theta+\phi)=\sin(\theta)\cos(\phi)+\cos(\theta)\sin(\phi)

\cos(\theta+\phi)=\cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi)

\sin\alpha+\sin\beta=2\sin\frac12(\alpha+\beta)\cos\frac12(\alpha-\beta)

\cos\alpha+\cos\beta=2\cos\frac12(\alpha+\beta)\cos\frac12(\alpha-\beta)

\cos\alpha-\cos\beta=2\sin\frac12(\alpha+\beta)\sin\frac12(\beta-\alpha)

In a triangle ABC, $a/\sin A=b/\sin B=c/\sin C$

and $a^2=b^2+c^2-2bc\cos A$

Prefixes

Prefix	Name	Factor
T	tera	$10^{12}$
G	giga	$10^9$
M	mega	$10^6$
k	kilo	$10^3$
c	centi	$10^{-2}$
m	milli	$10^{-3}$
$\mu$	micro	$10^{-6}$
n	nano	$10^{-9}$
p	pico	$10^{-12}$
f	femto	$10^{-15}$

Source note: This undated common table prints $G=6.673\times10^{-11}\ \mathrm{N\ m^2\ kg^{-2}}$ , $M_E=5.9742\times10^{24}\ \mathrm{kg}$ and $R_E=6.3781\times10^6\ \mathrm m$ . The 2024 eclipse key instead uses $G=6.674\times10^{-11}$ , Earth mass $5.976\times10^{24}\ \mathrm{kg}$ and Earth radius $6.371\times10^6\ \mathrm m$ . Both original sets of values are preserved separately; this reference is not an established exact dated 2024 attachment.

Условие

Astronomer Evan noticed that there is a group of stars $\{X_1,X_2,\ldots,X_{12}\}$ which forms a perfect dodecahedron around the Earth.

Complete unlabelled native dodecahedron with solid visible edges and dotted hidden edges.

Evan went outside to observe some of these stars. At a particular moment in time, he computed the angle

\delta_Z=\max(ZX_{a_1},ZX_{a_2},ZX_{a_3},ZX_{a_4},ZX_{a_5}),

where $Z$ is his zenith and $\{a_1,a_2,\ldots,a_5\}\subset\{1,\ldots,12\}$ are the indices of the five closest stars in angular distance to $Z$ . Compute the minimum possible value of $\delta_Z$ that he could have observed.

Note that $\delta_Z$ is not a declination.

Отговори

Решение

The answer is $\boxed{37.38^\circ}$ .

$\delta_Z$ is minimized when $Z$ points in the center of any polygonal face, call $ABCDE$ . When this is the case, $\delta_Z=ZA=ZB=ZC=ZD=ZE:=X$ . If $Z$ is not located at the center of a face, then the distance of the 5th farthest star from $Z$ is necessarily larger than $X$ .

Complete native dodecahedron with the marked face vertices A, B, C, D and E and dotted hidden edges.

To compute $X$ , consider the spherical triangle $\triangle ZAE$ . When we look at the configuration from a bird’s eye perspective centered at $Z$ , we see that $\angle AZE=72^\circ$ and $ZA=ZE=X$ . Moreover, $\angle ZEA=\angle DEA/2$ . When we look at the configuration from a bird’s eye perspective centered at $E$ , we see that $\angle DEA=120^\circ$ , which implies $\angle ZEA=60^\circ=\angle ZAE$ .

Let $Y=AE$ . Using the spherical law of cosines and spherical law of sines in $\triangle ZEA$ , we have

\cos Y=\sin^2X\cos72^\circ+\cos^2X\qquad\text{and}\qquad\frac{\sin X}{\sin60^\circ}=\frac{\sin Y}{\sin72^\circ},

respectively. From here, there are many ways to solve for $X$ . One way is to substitute $\cos Y=\sqrt{1-\sin^2Y}$ into the first equation. Then, plugging in the expression for $\sin Y$ given by the second equation into the first gives a quadratic in $\sin^2X$ , which can be bashed using the quadratic formula.

If we instead first solve for $Y$ , a slightly cleaner approach is possible. We have

\cos Y=\left(\frac{\sin60^\circ}{\sin72^\circ}\sin Y\right)^2\cos72^\circ+1-\left(\frac{\sin60^\circ}{\sin72^\circ}\sin Y\right)^2,

which implies

1-\cos Y=2\sin^2Y\frac{\sin^260^\circ\sin^236^\circ}{\sin^272^\circ}\quad\Longrightarrow\quad\cos Y=\frac12\left(\frac{\sin72}{\sin36\sin60}\right)^2-1.

Solving for $Y$ and plugging back into our original equations gives $X\approx37.38^\circ$ , as desired.

Source note: The source says 12 stars form a dodecahedron; the unlabelled drawing has the native dodecahedral geometry. The key’s “5th farthest” differs from the question’s five closest stars. The last displayed cosine expression omits degree signs on its sine arguments; all printed forms are retained.

Теми и трудност

Оценена трудност: 4/5.

Небесна сфера и сферична астрономия · Небесна сфера · Небесни координати

Предпоставки: Гимназиална подготовка, Разширена олимпиадна подготовка

Оригинал в Архива: second_exam_2024.pdf · официални решения: second_exam_2024-solutions.pdf · допълнителни данни: physical-constants.pdf