OWAO 2023, Теория — Задача 7. One Thousand and One Nights

• 0.357

  Official solution: $\eta = 1 - \sin 40^{\circ} = 0.357 = 35.7\,\%$.

First, we determine the inclination $\varepsilon$ of the ecliptic to the equator as the inclination of a great circle passing through the points $(\alpha_1; \delta_1)$, $(\alpha_2; \delta_2)$.

Both the right ascension and the ecliptic longitude are counted from the local point of the vernal equinox $(0; 0)$ — that is the point of intersection of the ecliptic and the celestial equator. Therefore, the inclination can be determined from a spherical triangle with vertices at the vernal equinox point ♈, a point on the ecliptic $M$ and the projection of this point on the celestial equator $M_{\perp}$.

Using the five-part rule and the law of sines we obtain:

Substitution of $\sin\lambda$ yields

Next, we can get an estimate of the inclination from the data on any of the two points,

Another approach. We introduce a Cartesian coordinate system centered at the place of observation. The $x$-axis is directed to the local point of the vernal equinox, the $z$-axis is directed to the north celestial pole, the $y$-axis complements the axes to a right-handed coordinate system. The vectors of two points are as follows:

Next, we determine the vector $\vec{w}$ of the pole of the great circle (namely, the north ecliptic pole) using the vector product

where

$\vec{w}$ points to the north ecliptic pole. $\theta = 90^{\circ} - \varepsilon$ and $\alpha_0$ are the declination and the right ascension of the north ecliptic pole respectively,

Polar nights are possible only at latitudes with a modulus greater than $\varphi_0 = 90^{\circ} - \varepsilon = \theta = 40^{\circ}$. The corresponding points form two spherical caps around the poles. Let us determine the fraction of the planet’s surface with latitudes modulus more than $40^{\circ}$:

where

is the height of the spherical cap.

Finally,

Marking Scheme:

• Inclination of the ecliptic.

  In case of using spherical trigonometry:

  • Distance between two points on a sphere — 2 pt.
  • Choosing a valid spherical triangle — 2 pt.
  • Inclination $\varepsilon$ — 1 pt.

  It is acceptable to write the great circle equation for the ecliptic in polar coordinates without derivation.

  In case of using Cartesian coordinates:

  • Conversion from ecliptic coordinates to Cartesian coordinates: formula + evaluation — 1 pt + 1 pt.
  • Vector multiplication — 2 pt.
  • Inclination $\varepsilon$ — 1 pt.

• Fraction of the surface with polar nights:

  • Limiting latitudes ($|\varphi| = 90^{\circ} - \varepsilon$) — 2 pt.
  • Area of spherical caps — 2 pt.
  • Fraction $\eta$ — 1 pt.