Three mountains lie on the same meridian. The spherical distance between the first two is $a^{0}$, and between the second two is $b^{0}$; their heights are $h_{1}, h_{2}, h_{3}$ respectively. Does the second mountain obscure the third mountain when viewed from the first mountain?

Numerical values: $a=1^{\circ}, b_{0}=1^{\circ} 30^{\prime}, h_{1}=2500 \mathrm{~m}, h_{2}=3000 \mathrm{~m}, h_{3}=8800 \mathrm{~m}$ and the radius of the Earth is $6366 \mathrm{~km}$.
According to our diagram, $h_{1}=A B, h_{2}=C D, h_{3}=E F$. The ray from the peak $B$ of mountain $A B$ through point $D$ intersects $E F$ at point $G$: the mountain $C D$ obscures the part $E G$ of mountain $E F$. We need to calculate this. To do this, we first need to determine $O G$. $O G$ can be calculated from $\triangle O D G$ or $\triangle O B G$; however, we first need to determine the angles on the side $B D$ in $\triangle O B D$. Indeed, in $\triangle O B D$, two sides, $O B$ and $O D$, and the angle between them, $\alpha^{\circ}$, are known. Therefore,

\[
\begin{gathered}
(O D+O B):(O D-O B)=\tan \frac{180^{\circ}-\alpha^{\circ}}{2}: \tan \frac{\beta-\delta_{1}}{2} \\
12737.5: 0.5=\tan 89^{\circ} 30': \tan \frac{\beta-\delta_{1}}{2} \\
\tan \frac{\beta-\delta_{1}}{2}=\frac{\tan 89^{\circ} 30'}{25475}
\end{gathered}
\]

From this,

\[
\frac{\beta-\delta_{1}}{2}=0^{\circ} 15' 30''
\]

However,

\[
\frac{\beta+\delta_{1}}{2}=89^{\circ} 30'
\]

and thus $\beta=89^{\circ} 45' 30''$, $\delta_{1}=89^{\circ} 14' 30''$, $\delta_{2}=90^{\circ} 45' 30''$, and $\gamma=87^{\circ} 44' 30''$. Now, from $\triangle O D G$,

\[
O G=\frac{O D \sin \delta_{2}}{\sin \gamma}=\frac{6369 \sin 90^{\circ} 45' 30''}{\sin 87^{\circ} 44' 30''}=6373.3 \text{ km}
\]

Therefore,

\[
E G=O G-O E=6373.3-6366=7.3 \text{ km}
\]

Accordingly, the part of $E F$ that is still visible from $A$ is $G F=8800-7300=1500 \text{ m}$.

Miklós Kemény (Berzsenyi Dániel Grammar School, 6th grade, Budapest)

Note. Without considering the curvature of the Earth, a much larger part of $E F$ would be visible from the peak $A$ (approximately 5000 m).