A city $A$ is located in the southern hemisphere of the Earth, and a tower stands in its mirror image on the Equatorial plane. The two towers are of equal height. On May 2nd at noon, when the sun's rays are inclined at an angle of $15^{\circ}$ to the Equatorial plane, the shadow of the tower in $A$ is 3 times as long as the shadow of the tower in $B$. Both shadows were measured on a horizontal plane. - Where could the city $A$ be?
In May, the Sun travels over points with a latitude $23.5^{\circ}$ less than that of the Northern Hemisphere, which is why the shadow of the tower in $A$ definitely points south. Let the latitude of $A$ be $-\varphi$ (south), the common height of the towers be $m$, then the length of the shadow is $m \operatorname{tg}\left(15^{\circ}+\varphi\right)$, because the sun rays form an angle of $15^{\circ}+\varphi$ with the vertical in $A$.

In $B$, the shadow can point north if the place's latitude $+\varphi$ (north) is greater than $15^{\circ}$ (1st figure), and it can point south if $\varphi<15^{\circ}$ (2nd figure).
![](https://cdn.mathpix.com/cropped/2024_05_02_c198df2dd76990d49d47g-1.jpg?height=1224&width=718&top_left_y=404&top_left_x=693)

2nd figure

The right-angled triangle defined by the tower and its shadow has an angle at the top of the tower of $\varphi_{1}-15^{\circ}$ in the first case, and $15^{\circ}-\varphi_{2}$ in the second case, so the equation for determining $\varphi$ is

$$
\begin{aligned}
& m \operatorname{tg}\left(\varphi_{1}+15^{\circ}\right)=3 m \operatorname{tg}\left(\varphi_{1}-15^{\circ}\right), \quad \text { or } \\
& m \operatorname{tg}\left(\varphi_{2}+15^{\circ}\right)=3 m \operatorname{tg}\left(15^{\circ}-\varphi_{2}\right)
\end{aligned}
$$

Using the addition theorem and noting that $\operatorname{tg} 15^{\circ}=2-\sqrt{3}$, from (1):

$$
2 \operatorname{tg} 15^{\circ} \cdot \operatorname{tg}^{2} \varphi_{1}-\left(1+\operatorname{tg}^{2} 15^{\circ}\right) \operatorname{tg} \varphi_{1}+2 \operatorname{tg} 15^{\circ}=0
$$

The discriminant

$$
D_{1}=\left(1+\operatorname{tg}^{2} 15^{\circ}\right)^{2}-16 \operatorname{tg}^{2} 15^{\circ}=(8-4 \sqrt{3})^{2}-16(7-4 \sqrt{3})=0
$$

thus

$$
\operatorname{tg} \varphi_{1}=\frac{1+\operatorname{tg}^{2} 15^{\circ}}{4 \operatorname{tg} 15^{\circ}}=1, \quad \varphi_{1}=45^{\circ}
$$

Similarly from (2):

$$
\begin{gathered}
\operatorname{tg} 15^{\circ} \cdot \operatorname{tg}^{2} \varphi_{2}-2\left(1+\operatorname{tg}^{2} 15^{\circ}\right) \operatorname{tg} \varphi_{2}+\operatorname{tg} 15^{\circ}=0, \\
D_{2} / 4=\left(1+\operatorname{tg}^{2} 15^{\circ}\right)^{2}-\operatorname{tg}^{2} 15^{\circ}=(112-64 \sqrt{3})-(7-4 \sqrt{3})= \\
=105-60 \sqrt{3}=15(7-4 \sqrt{3})=[\sqrt{15}(2-\sqrt{3})]^{2} \\
\operatorname{tg} \varphi_{2}=\frac{(8-4 \sqrt{3}) \pm \sqrt{15}(2-\sqrt{3})}{2-\sqrt{3}}=4 \pm \sqrt{15}
\end{gathered}
$$

and since the equation was set up with the condition $\varphi_{2}<15^{\circ}$, $\operatorname{tg} \varphi_{2}=4-\sqrt{15}=0.1270, \varphi_{2}=7^{\circ} 14^{\prime}$.

There is little land at the $45^{\circ}$ south latitude: only in Chile and Argentina, between the longitudes of $66^{\circ}$ and $74^{\circ}$. This is where $A$ can be, because the reflection of this segment on the Equatorial plane is in Canada and the United States, slightly east of Montreal, also on land.

Similarly, by comparing the land arcs of the $7^{\circ} 14^{\prime}$ south and north latitudes, $A$ can be on the arc between the longitudes of $58^{\circ}$ and $78^{\circ}$ west, from the western coast of Peru to the Tapajós River in Brazil, and in Africa, between the longitudes of $13^{\circ}$ and $39^{\circ}$ east, roughly between Luanda and Dar es Salaam (the capitals of Angola and Tanzania).

Robert Kovalszky (Budapest, Landler J. Gymnasium)