I identified my position $\bar{A}$ on a hillside with a point on the map, but I cannot read its elevation. The shadow of a single (small) cloud $F$ in the sky is seen exactly at the lookout point of a mountain peak $C$, while the reflection of $F$ is at a point $P$ on the shore of a lake that can be marked on the map. According to the map, $\bar{A} C=2200 \mathrm{~m}, \bar{A} P=400 \mathrm{~m},$ and the height of $C$ above the lake is $550 \mathrm{~m}$; my compass indicates that the direction of $\bar{A} C$ deviates from north towards west by $71^{\circ}$, and the direction of $\bar{A} P$ deviates from south towards west by $64^{\circ}$. I know that the magnetic declination is $2^{\circ}$ towards west. It is September 23, 11:36 AM; according to my sundial, on this day the Sun is visible in the sky from 5:31 AM to 5:41 PM in city $V$ - which is on the same meridian as I am - and my watch also runs according to this time. The map shows that I am at a latitude of $49^{\circ}$ north. How high am I above the surface of the lake?
![](https://cdn.mathpix.com/cropped/2024_05_02_847990d7d81d7ce5a39dg-1.jpg?height=526&width=602&top_left_y=122&top_left_x=751)

Notice that the detection time in hours and minutes is the arithmetic mean of the sunrise and sunset times of the day: (5:31 + 17:41)/2 = 11:36. From this, it can be approximated that at the time of detection, the Sun is directly south in A, meaning the sun rays are in the plane of A's meridian. Considering the relatively short distance of AC, this conclusion can also be applied to C and F, so F is in the plane of C's meridian and on the north-south line through C on the map (projection). On the other hand, F is also in the plane perpendicular to the lake's surface (the map's plane) and passing through A and P, so the projection of F is on A'P.

According to the date of September 23, on the day of detection, the plane of the Earth's Equator passes through the Sun, and at noon, the sun rays fall perpendicularly to the horizontal in the points of the Equator, and simultaneously to the Earth's axis of rotation. Therefore, on the 49° latitude circle, the sun rays form a 90° - 49° = 41° angle with the horizontal.

From these, we can determine the position of F on the map (its distance from A and C) and its height above the lake. Denoting the projections of the points on the plane of the lake's surface with 0-index, the angles at C0 and F0 in the A0C0F0 triangle and the angles between A0F0 and the geographic north-south line at A0 are alternate angles:

$$
\hat{A}_{0} C_{0} F_{0} \varangle=2^{\circ}+71^{\circ}=73^{\circ}, \quad \hat{A}_{0} F_{0} C_{0} \varangle=-2^{\circ}+64^{\circ}=62^{\circ},
$$

thus \( C_{0} \hat{A}_{0} F_{0} \varangle=45^{\circ} \), so by the sine rule, \( \hat{A}_{0} F_{0}=2200 \sin 73^{\circ} / \sin 62^{\circ} \approx 2383 \, \text{m} \), and similarly \( C_{0} F_{0} \approx 1762 \, \text{m} \). - Based on these, in the vertical plane passing through C and F, the height difference between C and F is \( C_{0} F_{0} \cdot \tan 41^{\circ} \approx 1532 \, \text{m} \), so the height of F and the depth of its reflection F* above the lake surface is \( 1532 + 550 = 2082 \, \text{m} \). Finally, in the vertical plane passing through A and F, from the similar right-angled triangles \( \hat{A} \hat{A}_{0} P \) and \( F^{*} F_{0} P \), the height of A above the lake is \( m_{\hat{A}} \approx 2082 \cdot \hat{A}_{0} P / P F_{0} \approx 420 \, \text{m} \), since \( P F_{0}=\hat{A}_{0} F_{0}-\hat{A}_{0} P_{0} \approx 1983 \, \text{m} \).

Miklóssy Endre (Szolnok, Verseghy F. g. IV. o. t.)

Remarks. 1. Several solutions gave a height of 419.9 m, or a slightly different value. Excessive precision in this case, even to the centimeter, is unrealistic. Indeed, F cannot be a point but must be larger than the meter range if the facts given in the problem are indeed observable (the shadow of F is more than 2 km away at C, and its observation from A, more than 2 km away). Furthermore, distances taken from the map, angles measured with a compass, and declination data, as well as time data, are rounded (to the nearest 100 m, at most 10 m, or to the nearest whole degree, whole minute).

The rounding justifies the simplifying assumption that the Sun is at the same meridian in A and C, as a rough estimate suggests that this difference is about 6 seconds, which is much smaller than the possible error in the time data. Those who claimed that the Sun is exactly at the meridian at the time of detection are slightly overestimating; because north of the Tropic of Cancer in summer and autumn, the morning half of the day is slightly longer than the afternoon half (almanacs show that the length of the day decreases by about 10 minutes every three days, or nearly 1 minute every quarter day), and finally, the date only ensures that the sun rays are perpendicular to the Earth's axis at some point during the detection (on September 23 and March 21, the angle between the line connecting the centers of the Earth and the Sun and the Earth's axis changes by about 0.4° per day).

Although we cannot give a precise estimate of the error, these considerations sufficiently justify not overestimating the accuracy of our result.

2. It may seem contradictory that on the "equinox" day, the Sun is visible in the sky for 12 hours and 10 minutes; however, this can be explained by refraction of light.

Csikor Ferenc (Budapest, I. István g. IV. o. t.)