A lighthouse to the north of a ship is seen from two lighthouses, which are 80 kilometers apart in a straight line to the west. After an hour of drifting, one is seen in the southwest direction, and the other in the south-southwest direction. What is the speed of the ship?
Let $A$ and $B$ be the locations of the lighthouses; $C$ and $D$ be the positions of the ship at the beginning and the end of the hour. Draw a perpendicular from $B$ to the line $AD$ and denote its foot by $E$. From the right-angled isosceles triangle $ABE$, it follows that:

$$
B E=A B: \sqrt{2}
$$

Furthermore, from the congruence of triangles $BED$ and $BCD$, it follows that:

$$
B E=B C
$$

But since the triangle $ACD$ is also isosceles and right-angled, therefore:

$$
\begin{aligned}
& C D=A B+B C=A B+B E= \\
& =A B \frac{1+\sqrt{2}}{\sqrt{2}}=\frac{A B}{2}(2+\sqrt{2})
\end{aligned}
$$

and the sought speed is $CD$ kilometers per hour.

(Aladár Visnya, 8th grade student, Pécs).

# Second solution.

Applying the sine theorem to the triangle $ABD$, we get:

$$
B D=\frac{A B \sin 45^{\circ}}{\sin 22^{\circ} 30^{\prime}}=2 A B \cos 22^{\circ} 30^{\prime}
$$

From the triangle $BCD$, we have:

$$
C D=B D \cos 22^{\circ} 30^{\prime}
$$

and thus

$$
\begin{gathered}
B D=2 A B \cos ^{2} 22^{\circ} 30^{\prime}= \\
160 \cos ^{2} 22^{\circ} 30^{\prime} \\
\log B D=\log 160+2 \log \cos 22^{\circ} 30^{\prime}= \\
=2,20412+9,93124-10= \\
=2,13536 \\
B D=136,57 \mathrm{Km}
\end{gathered}
$$

and the sought speed is $136,57 \mathrm{Km}$. per hour, or 2276 meters per minute, or $37,9 \mathrm{~ms}^{-1}$.

(János Imre, 8th grade student, Nyíregyháza).

The problem was also solved by: Domokos Berzenczey, Sándor Suták, and Aladár Visnya.