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2026 National Astronomy Competition 1 Instructions (Please Read Carefully)Table of Physical Constants Astrophysical Data Other data and conversion factors Trigonometrical identities Prefixes Условие Решение

USAAAO 2026, II кръг — Задача 4. Solar Observations

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Всички материали от USAAAO 2026

2026 National Astronomy Competition · 20 т.

Внимание: Бележка към темата

Questions 3 and 9 refer to designated answer sheets, and question 6 to provided grid paper; these sheets were not obtained in the two native 2026-II PDFs or their two-row same-folder archive catalogue. The separately attributed physical-constants.pdf is an undated common USAAAO archive reference; its identity as the table supplied for this exam is unproven.

Instructions (Please Read Carefully)

2026 National Astronomy Competition

1 Instructions (Please Read Carefully)

The top 15 eligible scorers on the NAC will be invited to the training camp. A third round will be conducted during the camp to select the national team for the IOAA 2026. In order to qualify for camp, you must be a high school student with US citizenship or permanent residency.

The test must be completed within 2.5 hours (150 minutes) and the maximum number of points is 300 points. The point breakdown per problem is as follows.

Problem	P1	P2	P3	P4	P5	P6	P7	P8	P9	Total
Points	10	15	15	20	20	40	40	55	85	300

Please solve each problem on a blank piece of paper and mark the number of the problem at the top of the page. Some problems may require a designated answer sheet; these will be provided. 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 calculator (non-programmable and non-graphing) 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 20, 2026. 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. Students must be proctored during the entire duration of the exam.

This exam sheet and your answer sheets (including scratch papers) should be returned to your proctors once the exam ended. Your proctors should upload your answer sheets to the Google Form provided to them.

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

Srihari (Hari) Balaji, Feodor Yevtushenko, Ferdinand, Hagan Hensley, Joe McCarty, Lucas Pinheiro, Vincent Bian, Elizabeth Lee

After reading the instructions, please make sure to sign, affirming that:

All work on this exam has been completed by me.
I took this exam under the supervision of a proctor.
I did not receive any external help beyond the materials provided.
I will not discuss the contents of this exam with anyone until April 20, 2026.
Failure to follow these rules will result in disqualification from the exam.

Student Signature: ____________________ Date: ____________________

Undated common archive reference · Table of Physical Constants

Table of Physical Constants

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 ( $\vphantom{C}^{12}\mathrm C$ 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

Undated common archive reference · Astrophysical Data and Conversion Factors

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	$1\,\mathrm{N\,m^{-2}}$
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

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

Условие

An astronomer who enjoys solar observations has the habit of recording the Sun’s right ascension and declination every day. These coordinates are always recorded at the same civil time (the time on the clock). Between two consecutive observations, which happened when the Sun was in the Northern Celestial Hemisphere, the declination of the Sun increased by $0.3650^\circ$ . Estimate the days in which these two observations occurred.

Neglect the eccentricity of Earth’s orbit.

Решение

Solution:

First Approach:

In order to estimate the days of the observations, the first step is to determine the declination of the Sun at the point in which its rate of variation in declination is $0.3650^\circ/\text{day}$ . Note that there are two such points in the celestial sphere, but since the problem statement specifies that the Sun was in the Northern Celestial Hemisphere, it is possible to obtain an unique solution.

Using the figure below, it is possible to obtain an expression for angle $x$ , which can be used to break down the angular velocity of the Sun into a right ascension component and a declination component.

Original spherical geometry diagram with celestial equator, ecliptic, delta, epsilon, lambda, x and both right-angle marks.

Using the spherical law of sines:

\frac{\sin x}{\sin(90^\circ-\varepsilon)}=\frac{\sin90^\circ}{\sin(90^\circ-\delta)}

\sin x=\frac{\cos\varepsilon}{\cos\delta}

Using the Pythagorean trigonometric identity:

\cos x=\sqrt{1-\left(\frac{\cos\varepsilon}{\cos\delta}\right)^2}

Neglecting the eccentricity of the Earth’s orbit, the angular velocity of the Sun on the Ecliptic corresponds to $\omega=\frac{360^\circ}{365.2564\text{ days}}=0.98561^\circ/\text{day}$ .

Original angular-velocity component diagram, both full arrows, omega, omega-delta and x.

The declination component of the angular velocity ( $\omega_\delta$ ) can be obtained through the following expression:

\begin{aligned}\omega_\delta&=\omega\cos x\\ \frac{\omega_\delta}{\omega}&=\sqrt{1-\left(\frac{\cos\varepsilon}{\cos\delta}\right)^2}\\ \frac{\cos\varepsilon}{\cos\delta}&=\pm\sqrt{1-\left(\frac{\omega_\delta}{\omega}\right)^2}\\ \cos\delta&=\pm\frac{\cos\varepsilon}{\sqrt{1-\left(\frac{\omega_\delta}{\omega}\right)^2}}\\ \cos\delta&=\pm\frac{\cos23.44^\circ}{\sqrt{1-\left(\frac{0.365}{0.98561}\right)^2}}\\ \cos\delta&=\pm0.9877\end{aligned}

Since the Sun is in the Northern Celestial Hemisphere and the change in declination is positive, the Sun must be in the first quadrant of the Ecliptic. Therefore:

\cos\delta=0.9877

\delta=8.995^\circ

The next step is to find the ecliptic longitude of the Sun. Using the spherical law of sines:

\begin{aligned}\frac{\sin\lambda}{\sin90^\circ}&=\frac{\sin\delta}{\sin\varepsilon}\\ \sin\lambda&=\frac{\sin8.995^\circ}{\sin23.44^\circ}\\ \sin\lambda&=0.3930\\ \lambda&=23.14^\circ\end{aligned}

Since the angular velocity of the Sun throughout the ecliptic is constant if Earth’s orbital eccentricity is constant, it is possible to obtain the number of days since the vernal equinox:

\begin{aligned}\lambda&=\omega d\\ d&=\frac\lambda\omega\\ d&=\frac{23.14^\circ}{0.98561^\circ/\text{day}}\\ d&=23.48\text{ days}\end{aligned}

Therefore, the observation corresponds to days 23 and 24 after the vernal equinox. Since the vernal equinox occurs around March $20^{\text{th}}$ , the observations correspond to April $12^{\text{th}}$ and April $13^{\text{th}}$ .

Since the equinox does not necessarily happen on March $20^{\text{th}}$ , pairs of days between April $10^{\text{th}}$ and April $15^{\text{th}}$ are acceptable without any point deductions.

Second Approach:

It is also possible to solve this problem in an alternative way. Consider the following expression, which can be obtained through the law of sines for spherical triangles:

\sin\delta=\sin\varepsilon\sin\lambda

Deriving this expression with respect to time:

\begin{aligned}\cos\delta\frac{d\delta}{dt}&=\sin\varepsilon\cos\lambda\frac{d\lambda}{dt}\\ \sqrt{1-\sin^2\delta}\frac{d\delta}{dt}&=\sin\varepsilon\cos\lambda\frac{d\lambda}{dt}\\ \sqrt{1-(\sin\varepsilon\sin\lambda)^2}\frac{d\delta}{dt}&=\sin\varepsilon\cos\lambda\frac{d\lambda}{dt}\\ [1-(\sin\varepsilon\sin\lambda)^2]\left(\frac{d\delta}{dt}\right)^2&=\sin^2\varepsilon\cos^2\lambda\left(\frac{d\lambda}{dt}\right)^2\end{aligned}

The rate $\frac{d\lambda}{dt}$ is constant and corresponds to $\frac{360^\circ}{365.2564\text{ days}}=0.98561^\circ/\text{day}$ . Based on the problem statement, $\frac{d\delta}{dt}=0.3650^\circ/\text{day}$ . In order to simplify the expression, it is convenient to define a variable $k=\left(\frac{d\lambda}{dt}\right)^2/\left(\frac{d\delta}{dt}\right)^2\approx7.29$ .

Therefore:

\begin{aligned}1-\sin^2\varepsilon\sin^2\lambda&=k\sin^2\varepsilon\cos^2\lambda\\ 1&=\sin^2\varepsilon(\sin^2\lambda+k\cos^2\lambda)\\ \frac1{\sin^2\varepsilon}&=\sin^2\lambda+k(1-\sin^2\lambda)\\ \frac1{\sin^2\varepsilon}&=\sin^2\lambda+k-k\sin^2\lambda\\ \frac1{\sin^2\varepsilon}&=(1-k)\sin^2\lambda+k\\ \sin^2\lambda&=\frac1{1-k}\left(\frac1{\sin^2\varepsilon}-k\right)\end{aligned}

Since the declination increased between the two observations and the Sun was in the Northern Celestial Hemisphere, $\lambda$ must be in the first quadrant. Therefore:

\sin^2\lambda=0.1545

\sin\lambda=0.3930

\lambda=23.14^\circ

From this point onward, the solution follows the same steps as the first approach.

Rubrics - First Approach:

(7 points) correctly drawing or describing the geometry of the problem.
(4 points) finding an expression for the cosine of $x$ .
(1 point) correctly breaking down the angular velocity of the Sun to find the declination component.
(2 points) determining the declination of the Sun.
(2 points) determining the ecliptic longitude of the Sun.
(2 points) determining the number of days since the vernal equinox.
(1 point) estimating a reasonable day for the vernal equinox.
(1 point) correct final answer.

Rubrics - Second Approach:

(1 point) writing down the expression that correlates $\sin\delta$ , $\sin\varepsilon$ and $\sin\lambda$ .
(3 points) correctly deriving the expression with respect to the time.
(12 points) correctly simplifying the expression to obtain the value of $\lambda$ .
(2 points) determining the number of days since the vernal equinox.
(1 point) estimating a reasonable day for the vernal equinox.
(1 point) correct final answer.

Оригинал в Архива: usaaao2026.pdf · официални решения: usaaao2026_sol.pdf · допълнителни данни: physical-constants.pdf