USAAAO 2025, I кръг — Задача 5
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 | ||
| Proton rest mass | ||
| Electronic charge | ||
| Speed of light in free space | ||
| Permeability of free space | ||
| Permittivity of free space | ||
| Planck’s constant | ||
| Reduced Planck’s constant | ||
| Boltzmann’s constant | ||
| Gas constant | ||
| Molar gas constant | ||
| Avogadro’s number | ||
| Standard molar volume | ||
| Unified atomic mass unit ( scale) | ||
| Mass of hydrogen atom | ||
| Bohr magneton | or | |
| Nuclear magneton | or | |
| Proton magnetic moment | ||
| Neutron magnetic moment | ||
| Bohr radius | ||
| Fine structure constant | ||
| Compton wavelength of electron | ||
| Rydberg’s constant | ||
| Stefan-Boltzmann constant | ||
| Radiation density constant | ||
| Gravitational constant |
Rest masses of some leptons and hadrons in :
0.5110, 105.66, 1777, 134.98, 139.57, 493.7, 497.7, 547, 1865, 1869, p 938.3, n 939.6, 1115.7, 1189, 1193, 1197, 1315, 1321, 1672, , .
| Quark | Charge | Mass () | |||||
|---|---|---|---|---|---|---|---|
| u | 0 | 0 | 0 | 0 | |||
| d | 0 | 0 | 0 | 0 | |||
| c | 0 | 0 | +1 | 0 | 0 | ||
| s | 0 | -1 | 0 | 0 | 0 | ||
| t | 0 | 0 | 0 | 0 | +1 | 174.3 | |
| b | 0 | 0 | 0 | -1 | 0 | 4.2 |
Common archive reference — Astrophysical Data and identities
Astrophysical Data
| Quantity | Symbol | Value |
|---|---|---|
| 1 astronomical unit | AU | |
| 1 parsec | pc | |
| Luminosity of Sun | ||
| Mass of Sun | ||
| Radius of Sun | ||
| Mass of Earth | ||
| Radius of Earth |
Other data and conversion factors
| Quantity | Symbol | Value |
|---|---|---|
| 1 ångstrom | Å | |
| 1 fermi | fm | |
| 1 barn | b | |
| 1 pascal | Pa | |
| 1 standard atmosphere | ||
| Standard acceleration due to gravity | ||
| 1 electron volt | eV | |
| Wavelength of 1 eV photon |
Trigonometrical identities
In a triangle ABC,
and
Prefixes
| Prefix | Name | Factor |
|---|---|---|
| T | tera | |
| G | giga | |
| M | mega | |
| k | kilo | |
| c | centi | |
| m | milli | |
| micro | ||
| n | nano | |
| p | pico | |
| f | femto |
Source note: The common table prints and . The 2025 question14/key15 use 9.81 and key7 uses 6371 km; those original exam values are preserved separately.
Условие
It may be useful to know the semi-latus rectum of an ellipse is the distance between one of its foci and the point on the ellipse immediately above or below it, as shown in the diagram below.

Consider a highly eccentric planet with a semi-latus rectum that is nearly a hundred times smaller than its semi-major axis. What is its eccentricity?
(a) 0.99
(b) 0.995
(c) 0.9999
(d) 0.99995
(e) 0.999999
Отговори
- B
Решение
By constructing a right triangle, using the two foci and the point on the semi-latus rectum, we can find that the semi-latus rectum length is related to the semi-major axis and the eccentricity by
where is the distance between the point on the semi-latus rectum and the other focus (the one not directly below it), i.e, the hypotenuse of the triangle. Since an ellipse is defined such that all points on its circumference have a constant sum of distances from the foci , we also have that
We can now solve these equations, most easily by using difference of squares on the first and then dividing by the second to obtain , and it is now easy to see that . We now rearrange for . Plugging in , we get 0.995.
Answer: B
Source note: The first displayed relation has two equal signs in the source. The following paragraph instead invokes a difference of squares; both printed forms are preserved.
Теми и трудност
Оценена трудност: 2/5.
Небесна механика · Орбитални елементи
Предпоставки: Гимназиална подготовка, Разширена олимпиадна подготовка
Оригинал в Архива: exam_2025.pdf · официални решения: sol_2025.pdf · допълнителни данни: physical-constants.pdf