IAO 2007, Теория, Група α, английски — Задача 1. Earth and Moon.

Let us find the distance from the satellite to the Moon.

Let $\rho$, $R$ and $L$ are angular radii as seen from the satellite, (radii and distances to the Earth (E) and Moon (M)). From the equations

one can find

or

The close positions of the Earth and the Moon on the picture shows us that the satellite, the Earth and the Moon are approximately on one line and therefore we can approximately say

or

where $L_{EM}$ is the distance from the Earth to the Moon.

From the picture we see that for the observer on the satellite the phase of the Moon is full moon. If $m_0$ is the magnitude of the full moon for the observer on the Earth (the distance to the Moon is $L_{EM}$), the magnitude for the observer on the satellite is

We may measure values $\rho_E$ and $\rho_M$ on the picture and take values $R_M$, $R_E$ and $m_0$ (and, by the way, $L_{EM}$) from the table of the Solar System,

Note 1 for Jury members. Due to difference in printers the scales of the pictures were different in different copies of the pictures. Therefore the values of the lengths may be different from those written here. But the ratios, which are necessary for the solutions, of the lengths should be about the same. Nevertheless the problem is estimational. Measuring of $\rho_E/\rho_M$ and correct calculations have to result to answers from $-10.6^{\mathrm m}$ to $-11.0^{\mathrm m}$.

Note 2 for Jury members. By the way, the short approximate solution that also should be regarded as a full solution (the so called solution in integer values):

After using the formula $L_M=L_{EM}/(1-(R_E/R_M)(\rho_M/\rho_E))$ (or an analogous formula) and calculations we may find that $L_M\approx2.5L_{EM}$, enlarging distance every 2.5 times (more exactly 2.512) enlarges the magnitude on $2^{\mathrm m}$, so