With respect to a fixed globe, at a considerable distance from it, in the plane of its equator, we positioned ourselves so that on it we see the city of Valparaiso (Chile, southern latitude $33^{\circ} 10^{\prime}$, western longitude $71^{\circ} 35^{\prime}$), Szombathely, and the center of the globe in a straight line. At which latitude in the southern hemisphere is our eye then? What angle does the mentioned straight line make with the plane of the globe's equator?

How does the answer change if we take Székesfehérvár instead of Szombathely?
Assuming that our eye and the center $O$ of the globe are in a horizontal plane, and the equator of the globe is also in this plane. Then the perceived image (due to the "sufficiently large" distance) is essentially a front view (second image), where we see the contour of the globe as a circle and the equator as a horizontal diameter.

![](https://cdn.mathpix.com/cropped/2024_05_02_3e6c0d5e5c13011adaa2g-1.jpg?height=862&width=634&top_left_y=285&top_left_x=737)

The tangent of the angle between the image $O''S''z$ of the ray $OSz$ and the image $e''$ of the equator is given by the ratio of the height difference $a$ between $O''$ and $S''z$ and the distance $b$ between the verticals (ordinates) passing through them. Assuming the radius of the globe is unity, $a = |\sin \varphi|$, where $\varphi$ is the latitude of the point $Sz$ (southern latitudes are conventionally negative).

To determine $b$, we use the top view (first image). On this view, the latitudinal circles appear in their true size, their center is the first image $O'$ of $O$, and the meridians (which are actually semicircles) are the radii of the image $e'$ of the equator (counted twice), and the angle between two meridian planes—the difference in longitude—is also seen in its true size between the corresponding two radii. Let the longitude of the meridian in question be $x$, and the longitude of the point $Sz$ be $\lambda$ (western longitudes are conventionally negative), then

$$
b = O'S'z \sin (\lambda - x) = \cos \varphi \sin (\lambda - x)
$$

since the distance between the projections $O'$ and $S'z$ is equal to the radius of the latitudinal circle, which is $\cos \varphi$ according to the second image. Therefore,

$$
\operatorname{tg} \alpha = -\frac{\sin \varphi}{\cos \varphi \sin (\lambda - x)} = \frac{\operatorname{tg} \varphi}{\sin (\lambda - x)}
$$

We see the second images of the two cities, $V''$ and $S''z$, and $O''$, on a straight line if the angles $\alpha$ corresponding to $S''z$ and $V''$ are equal or differ by $180^\circ$, in any case, their tangents are equal. Let's now write the signs of the two coordinates with subscripts 1 and 2, respectively, depending on whether they belong to $Sz$ or $V$, so the requirement is:

$$
\frac{\operatorname{tg} \varphi_1}{\sin (\lambda_1 - x)} = \frac{\operatorname{tg} \varphi_2}{\sin (\lambda_2 - x)}
$$

and from this, $x$ can be calculated.

For the two Hungarian cities, Székesfehérvár (latitude $47^\circ 11'$, longitude $18^\circ 25'$), the calculation is very easy, noticing that its longitude difference relative to Valparaiso is exactly $90^\circ$, so the two denominators are complementary angles in absolute value

$$
\frac{\sin (\lambda_1 - x)}{\sin (\lambda_2 - x)} = \frac{\sin (\lambda_1 - x)}{-\cos (\lambda_1 - x)} = -\operatorname{tg} (\lambda_1 - x) = \frac{\operatorname{tg} \varphi_1}{\operatorname{tg} \varphi_2}
$$

Based on the numerical values, in this case,

$$
\lambda_1 - x = 58^\circ 48', \quad x = -40^\circ 23'
$$

(slightly east of the meridian of Rio de Janeiro), and then $\alpha = 51^\circ 36'$.

In general, from (1), using the addition theorem,

$$
\begin{gathered}
\left(\operatorname{tg} \varphi_2 \cos \lambda_1 - \operatorname{tg} \varphi_1 \cos \lambda_2 \sin x = \right. \\
= \left(\operatorname{tg} \varphi_2 \sin \lambda_1 - \operatorname{tg} \varphi_1 \sin \lambda_2 \right) \cos x \\
\operatorname{tg} x = \frac{\operatorname{tg} \varphi_2 \sin \lambda_1 - \operatorname{tg} \varphi_1 \sin \lambda_2}{\operatorname{tg} \varphi_2 \cos \lambda_1 - \operatorname{tg} \varphi_1 \cos \lambda_2} = \frac{\sin \lambda_1 - k \sin \lambda_2}{\cos \lambda_1 - k \cos \lambda_2}
\end{gathered}
$$

where $k = \frac{\operatorname{tg} \varphi_1}{\operatorname{tg} \varphi_2}$. With the coordinates of Szombathely (latitude $47^\circ 15'$, longitude $16^\circ 37'$), $x = -40^\circ 55'$, $\alpha = 52^\circ 02'$.

László Cserepes (Budapest, Piarist High School, 4th grade)