A well-known pilot told a story: "In one of my take-offs, I started heading north, then after flying $10000 \mathrm{~km}$, I flew east for the same $10000 \mathrm{~km}$. Finally, I turned north again and after another $10000 \mathrm{~km}$ flight, I landed at my starting point."

A colleague replied: "Your story is surprising, yet I believe it. In fact, if you answer my question, I will calculate the latitude from which you took off, and you will calculate the starting latitude of my somewhat more believable story. - Once, I also returned to where I took off after a $30000 \mathrm{~km}$ flight. In each of the 5 consecutive segments of my journey, I flew $6000 \mathrm{~km}$. I started heading east, then turned north, east, south, and finally east again. About $8 \%$ of my journey was already flown over. So, did you fly over already traveled paths during the third part of your journey, or the half of it? Or perhaps even more than that?"

The first one replied: "The half of it."

Let's calculate the geographical latitude of the take-off point of both pilots with an accuracy of $0.1^{\circ}$.
a) The first pilot - let's call him Elek - can explain his surprisingly strange words upon first hearing based on the question of the second pilot - let's call him Mátyás. Mátyás knew that Elek had flown at least 10,000 km. This is understandable regarding the first and third segments of the route taken south of the equator, as heading north after crossing the North Pole, we start heading south (our direction changes). If we now stop somewhere and then head north again, covering the same distance without changing direction, we indeed return to our starting point. - The coincidence of the starting point of the third segment with the end point of the first segment is possible if, in between, Elek flew along a parallel (always perpendicular to the north-south direction), and the circumference of this parallel is an integer multiple of the 10,000 km length of the second segment. Mátyás indirectly asked Elek about the circumference of this parallel, and from the answer, he calculated the geographic latitude of the parallel, and from that, the latitude of the takeoff point.

According to Elek's answer, he started flying on a segment that was 15,000 km long, so the circumference of the parallel is 5,000 km. On the other hand, the circumference of a parallel at latitude \(\varphi\) in kilometers is \(40,000 \cos \varphi\), because for its radius \(r\) and the radius \(R\) of the great circle, \(r = R \cos \varphi\), and \(2 R \pi = 40,000 \) km, so \(\cos \varphi = 1/8\), \(\varphi = 82.8^\circ\). Thus, after crossing the North Pole, Elek flew 7.2 degrees south, and previously 82.8 degrees north, so his takeoff point was at 7.2 degrees north latitude (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_30b130758e95d9b68970g-1.jpg?height=350&width=1369&top_left_y=763&top_left_x=389)

b) Mátyás flew along the equator in his second and fourth segments, and on parallels in the other three, specifically on the same \(\varphi\) latitude parallel in the first and fifth segments, otherwise he would not have returned to his takeoff point.

If, contrary to custom, we only speak of eastern longitude and do not subtract 360 degrees when crossing the Greenwich meridian in an easterly direction, then Mátyás's aircraft's geographic longitude increased in the 1st, 3rd, and 5th segments, remained unchanged in the 2nd and 4th segments, and increased by 360 degrees, or an integer multiple thereof, over the entire route.

Mátyás did not reach the North Pole in the second segment, because if he had crossed it, he would have ended up on a more southern parallel in the fourth segment by heading south. The 6,000 km segments on the equator result in a 54-degree change in geographic latitude, so Mátyás's takeoff point's latitude is 54 degrees less than the latitude of the third segment.

![](https://cdn.mathpix.com/cropped/2024_05_02_30b130758e95d9b68970g-1.jpg?height=740&width=1100&top_left_y=1587&top_left_x=445)

The projection of the first pilot's route on the Egrenlitô plane

Figure 2

Mátyás once flew nearly 2,400 km on a route. This could not have been in the fourth segment, otherwise the entire segment would have been a single route, which is already 20%. - Suppose the 2,400 km mentioned was in the third segment of the flight. Thus, the circumference of the parallel in this segment is approximately \(6,000 - 2,400 = 3,600\) km (slightly more than this). Thus, for the latitude \(\varphi\) of this parallel, similarly to the previous calculation, \(\cos \varphi \approx 3,600 / 40,000 = 0.09\), and thus about \(85^\circ\) is obtained. [2 On this parallel]

Mátyás once flew around it, and another time covered nearly 2/3 of it, so his geographic longitude increased by nearly \(360^\circ + 240^\circ = 600^\circ\) in the third segment. On the other hand, at this time, the 1st and 5th segments were slightly below the \(31^\circ\) latitude, whose circumference is \(40,000 \cos 31^\circ\), so on the two 6,000 km segments, an increase of \(360^\circ \cdot 12,000 / 40,000 \cos 31^\circ = 108^\circ / \cos 31^\circ \approx 126^\circ\) in longitude was achieved. This adds up to about 726 degrees, which is roughly two full circles, from which a refined solution can be expected.

Expressing the part of the longitude increase in the third segment similarly, we get the following equation for \(\varphi\):

\[
\begin{gathered}
2 \cdot \frac{360^\circ \cdot 6,000}{40,000 \cos (\varphi - 54^\circ)} + \frac{360^\circ \cdot 6,000}{40,000 \cos \varphi} = 720^\circ \\
\frac{6}{\cos (\varphi - 54^\circ)} + \frac{3}{\cos \varphi} = 40
\end{gathered}
\]

Transforming this in the usual way, we arrive at an irreducible quartic equation; therefore, we seek an approximate solution by trial and error.

For \(\varphi = 85^\circ\), the value of the left side is \(7.00 + 34.42 = 41.42\), which is 1.42 more than the desired value. As \(\varphi\) decreases, both denominators increase, and thus the left side decreases. Indeed, for \(\varphi = 84.5^\circ\), a deficit is already shown: \(6.96 + 31.30 = 38.26\), with a deficit of 1.74. Thus, a root is expected between the two \(\varphi\) values used, and more closely to \(85^\circ\) than to \(84.5^\circ\). Indeed, for \(\varphi = 84.8^\circ\), the value of the left side is \(6.98 + 33.10 = 40.08\). - Thus, Mátyás took off from the parallel at \(30.8^\circ\) north latitude, the circumference of the parallel in the third segment is about 3,626 km, so the double-covered part of the route is 2,374 km, which is \(7.9\%\).

To verify this, we will also show that there could not have been a double-covered part of Mátyás's route in the 1st and 5th segments. Indeed, if this were the case, the circumference of the relevant parallel would be slightly more than \(2 \cdot 6,000 - 2,400 = 9,600\) km. Thus, for its latitude, \(\cos \varphi \approx 0.24\), from which, on the one hand, only a point on the \(76^\circ\) south latitude parallel could be the takeoff point, and on the other hand, the \(22^\circ\) south latitude parallel for the third segment. On the other hand, the longitude increase achieved in the 1st and 5th segments is only about \(360^\circ \cdot 12,000 / 9,600 = 450^\circ\), and this could not be increased to 510 degrees by the 6,000 km covered on the approximately \(37,000\) km circumference \(22^\circ\) parallel.

Simplified and supplemented from the following works:

András Patkós (Budapest, Radnóti M. g. III. o. t.)

Aranka Siket (Makó, József A. g. III. o. t.)

Csaba Sükösd (Budapest, József A. g. III. o. t.)

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