IAO 2011, Теория, Група α — Задача 1. Observation of a star

Latitude of observation

The difference is $7^{\mathrm h}59^{\mathrm m}$ between the two observations. It is almost exactly $1/3$ of a sidereal day (the exact value is $7^{\mathrm h}58^{\mathrm m}41^{\mathrm s}$). During this time the star moves $120^\circ$ of its daily path. And its zenith distance has changed by $\zeta=90^\circ-87^\circ12'=2^\circ48'$. This value is small, so it is possible to work with the approximation of plane geometry. If $\beta$ is the angular distance from the star to the pole P, we may conclude (see figure) that during this time the star has moved from the zenith point Z to point Y, and changed the zenith distance by

therefore

Hence, the declination of the star is equal to

Accordingly, this star culminates at zenith only at latitude $\varphi=\delta$, that is the latitude also equals $88^\circ23'$.

But it is not a complete solution. However, the value of $88^\circ23'$ is only the absolute value of the unknown value. In principle, both the values: $+88^\circ23'$ (Northern latitude, the star is near the North Pole) and $-88^\circ28'$ (Southern latitude, the star near the South Pole) can meet the condition of the more general problem. However, according to the condition of our problem the observations were made with the naked eye on 16 June, when polar day appears at such northern latitudes. Therefore, correct answer is only one: $-88^\circ23'$ ($88^\circ23'$ South latitude).

Transcription note: the intermediate southern value is printed $-88^\circ28'$; the calculation and final answer print $-88^\circ23'$. Both are retained.