Josef and Marie discovered a regular pyramid on their vacation, whose base was a square with a side of $230 \mathrm{~m}$ and whose height was equal to the radius of a circle with the same circumference as the base square. Marie marked the vertices of the square as $A B C D$. Josef marked a point $E$ on the line connecting point $B$ with the apex of the pyramid such that the length of the broken line $A E C$ was the shortest possible.

Determine the length of the broken line $A E C$ rounded to the nearest centimeter.

(M. Krejčová, F. Steinhauser)

Hint. What was the relationship between the segments $A E$ and $C E$ with respect to the line connecting point $B$ with the apex of the pyramid?
Due to better clarity, we will illustrate the situation from the problem statement:

![](https://cdn.mathpix.com/cropped/2024_04_17_94af7d7a3fc4841f393fg-08.jpg?height=462&width=714&top_left_y=954&top_left_x=677)

The broken line $AEC$ is the shortest when, after unfolding the pyramid's lateral surface into a plane, it appears as a segment:

![](https://cdn.mathpix.com/cropped/2024_04_17_94af7d7a3fc4841f393fg-08.jpg?height=414&width=663&top_left_y=1689&top_left_x=705)

The segments $AV$ and $CV$ are congruent, as are the segments $AB$ and $CB$. Therefore, after unfolding the lateral surface into a plane, points $A$ and $C$ are symmetric with respect to the line $BV$, and in particular, the segment $AC$ is perpendicular to this line. The length of the shortest possible broken line $AEC$ is equal to twice the height of the triangle $ABV$ from vertex $A$, and we will determine this height.

The triangle $ABV$ is isosceles, and we know the length of its base. The sides are the hypotenuses of right triangles, one leg of which is half the diagonal of the base square, and the other leg is the height of the pyramid:

![](https://cdn.mathpix.com/cropped/2024_04_17_94af7d7a3fc4841f393fg-09.jpg?height=466&width=713&top_left_y=184&top_left_x=680)

Half the diagonal of the base square has (by the Pythagorean theorem) a length of

$$
|AS| = \frac{\sqrt{2}}{2} |AB| \approx 162.635 \text{ m}
$$

The height of the pyramid has (according to the information in the problem about the circumferences) a length of

$$
|SV| = \frac{4}{2\pi} |AB| \approx 146.423 \text{ m}
$$

The edges passing through the apex of the pyramid have (by the Pythagorean theorem) a length of

$$
|AV| = \sqrt{|AS|^2 + |SV|^2} \approx 218.837 \text{ m}
$$

Now we know all the sides of the triangle $ABV$. Its height from vertex $A$ can be expressed using the height from the main vertex $V$ and two expressions for the area of this triangle:

![](https://cdn.mathpix.com/cropped/2024_04_17_94af7d7a3fc4841f393fg-09.jpg?height=414&width=389&top_left_y=1552&top_left_x=842)

The triangle $ABV$ is symmetric with respect to the height passing through the main vertex. This height has (by the Pythagorean theorem) a length of

$$
|VP| = \sqrt{|AV|^2 - \frac{1}{4} |AB|^2} \approx 186.184 \text{ m}
$$

The area of the triangle $ABV$ is equal to $\frac{1}{2} |AB| \cdot |VP| = \frac{1}{2} |BV| \cdot |AE| = \frac{1}{2} |AV| \cdot |AE|$, so the height from vertex $A$ has a length of

$$
|AE| = \frac{|AB| \cdot |VP|}{|AV|} \approx 195.681 \text{ m}
$$

The length of the shortest possible broken line $AEC$ is

$$
|AEC| = 2 |AE| \approx 391.362 \text{ m}
$$

i.e., approximately $391 \text{ m}$ and $36 \text{ cm}$.

Notes. The initial observation with the unfolded lateral surface of the pyramid can be replaced by the following reasoning: The triangles $ABV$ and $BCV$ are congruent, so the segments $AE$ and $EC$ are also congruent, and the length of the broken line $AEC$ is equal to twice the length of the segment $AE$. This is the shortest possible if it is perpendicular to the line $BE$. It is sufficient to determine the height of the triangle $ABV$ from vertex $A$.

In the given solution, rounding can lead to an undesirable accumulation of error. Therefore, we prefer to round to the nearest millimeter. All previous quantities can also be expressed generally in terms of $|AB|$ and substituted only into the final expression. Thus, we get:

$$
|AV| = \sqrt{\frac{\pi^2 + 8}{2\pi^2}} \cdot |AB|, \quad |VP| = \sqrt{\frac{\pi^2 + 16}{4\pi^2}} \cdot |AB|, \quad |AE| = \sqrt{\frac{\pi^2 + 16}{2\pi^2 + 16}} \cdot |AB|.
$$

The commonly used approximate value $\pi \approx \frac{22}{7}$ leads to an error of between 2 and 3 cm in the result. Comparing the overall dimensions and proportions of the pyramid, it appears that Josef and Marie were on vacation in Egypt at the Great Pyramid of Khufu (or Cheops).

The height of the triangle can also be expressed using knowledge of trigonometric functions, their basic relationship $\left((\sin \beta)^2 + (\cos \beta)^2 = 1\right)$, and the cosine rule $\left(|AV|^2 = |AB|^2 + |BV|^2 - 2|AV| \cdot |BV| \cdot \cos \beta\right)$. We cannot assume this knowledge in the given category, but curious solvers can familiarize themselves with it and compare this approach with others.