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USAAAO 2025, I кръг — Задача 25

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Всички материали от USAAAO 2025

USAAAO 2025 - First Round · 8 февруари 2025 г.

Внимание: Бележка към темата

All 30 separately numbered original first-round multiple-choice questions and their matching official explanations/answer letters are retained. The printed explanations for Questions 2 and 3 stop before a promised formula; Question 26 refers to a following triangle absent from its native solution page. No missing expression or diagram is reconstructed. Native scientific discrepancies and ambiguous formula notation remain literal and are explicitly attributed in local source notes. A separate two-page common USAAAO archive constants/reference table is included with its own source identity; it is undated in the catalogue, and its use as the exact 2025 attachment is not established. No separate dated instructions, answer sheet or constants companion is present in the 2025 archive folder. No point total or time limit is printed in these two exam booklets.

Original first-round date and writers

USAAAO 2025 - First Round

February 8th, 2025

Writers: Abhay Bestrapalli, Sidhant Chaliha, Ferdinand, Ved Gund, Hagan Hensley, Yehong Jiang, Alexander Li, Joe McCarty, Lucas Carrit Delgado Pinheiro, Sahil Pontula.

Common archive reference — Table of Physical Constants

This separate common USAAAO archive reference is undated in the catalogue. Its use as the exact 2025 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: The common table prints $g=9.807\ \mathrm{m\ s^{-2}}$ and $R_E=6.3781\times10^6\ \mathrm m$ . The 2025 question14/key15 use 9.81 and key7 uses 6371 km; those original exam values are preserved separately.

Условие

Evan uses a telescope of focal distance $f=1000\ \mathrm{mm}$ to observe a star known to have a parallax $p=50\ \mathrm{mas}$ . Observing the CCD of the telescope, Evan notices that the star has a diameter $D=0.1\ \mu\mathrm{m}$ . Furthermore, he measures its apparent magnitude to be $m=1$ . What is the approximate temperature of this star? Please, neglect the atmospheric seeing in your calculations.

(a) 3000 K

(b) 3500 K

(d) 4500 K

(e) 5000 K

Отговори

Решение

The distance of the star in parsecs is

d=\frac1{p('')}=\frac1{50\times10^{-3}}=20\ \mathrm{pc}

Converting to meters:

d=3.086\times10^{16}\times20=6.172\times10^{17}\ \mathrm m

The angular radius of the star in radians is

\theta=\frac1f\cdot\frac D2=\frac11\cdot\frac{0.1\times10^{-6}}2=5\times10^{-8}

Converting to meters:

R=\theta d=6.172\times10^{17}\times5\times10^{-8}=3.086\times10^{10}\ \mathrm m

Calculating the apparent magnitude of the Sun:

m_{\odot}-M_{\odot}=5\log a_{\oplus}-5

m_{\odot}=4.83+5\log\frac1{206265}-5=-26.74

Comparing the star to the Sun using Pogson’s equation:

m-m_{\odot}=-2.5\log\left(\frac F{F_{\odot}}\right)

F=F_{\odot}\cdot10^{0.4(m_{\odot}-m)}

\frac{4\pi R^2\sigma T^4}{4\pi d^2}=\frac{L_{\odot}}{4\pi a_{\oplus}^2}\cdot10^{0.4(m_{\odot}-m)}

T=\left(\frac{d^2L_{\odot}}{4\pi\sigma R^2a_{\oplus}^2}\cdot10^{0.4(m_{\odot}-m)}\right)^{1/4}

T=\left(\frac{(6.172\times10^{17})^2\times3.85\times10^{26}}{4\pi\times5.671\times10^{-8}\times(3.086\times10^{10})^2\times(1.496\times10^{11})^2}\cdot10^{0.4\times(-26.74-1)}\right)^{1/4}

\boxed{T\approx3000\ \mathrm K}

Answer: A

Теми и трудност

Оценена трудност: 3/5.

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Предпоставки: Гимназиална подготовка, Разширена олимпиадна подготовка

Оригинал в Архива: exam_2025.pdf · официални решения: sol_2025.pdf · допълнителни данни: physical-constants.pdf