USAAAO 2024, I кръг — Задача 1

• B

Solution:

Since the corridor is parallel to the surface of the Earth, we know that the Sun’s altitude is 0 degrees. Define points on the celestial sphere as follows: $A$ is David’s current location, $B$ is the North Celestial Pole, and $C$ is the location of the Sun.

By the spherical law of cosines,

By definition, $A$ is the azimuth angle of $245.81^{\circ}$, $a = 90^{\circ} - \delta$ where $\delta$ is the declination of the Sun, and $c = 90^{\circ} - l$ where $l = 42.36^{\circ}$ is the latitude of Boston. Additionally, $b$ is just 90 degrees minus the altitude of the Sun, so $b = 90^{\circ}$.

Therefore, our equation simplifies to

Calculating this gives $\delta = -17.63^{\circ}$.

Now, we draw a second spherical triangle. Let $A$ be the direction of the vernal equinox, $B$ be the location of the North Celestial Pole, and $C$ be the direction of the Sun.

Since the declination of the Sun is negative but it has been less than 6 months since the last winter solstice, we know that the point $C$ is between the location of the winter solstice and point $A$, and so therefore angle $A = 90^{\circ} + i = 113.44^{\circ}$ where $i = 23.44^{\circ}$ is the Earth’s axial tilt. Also, we know $a = 90^{\circ} - \delta$. And since $A$ is on the equator, we must have that $c = 90^{\circ}$. Using the law of cosines again,

During 365.25 days, the Sun travels $360^{\circ}$ across the celestial sphere, so it would take

to reach the vernal equinox from its current position. Thus all we have to do is count backwards 48-49 days from March 20th to get an answer of January 30.

Answer: B