The foundation of a tent is a square with a side length of $a(=15 \mathrm{~m})$. The boundary of the foundation surrounds this square and consists of 4 arcs and 4 straight line segments. The centers of the arcs are the vertices of the square, with radii of $a, a, a / 2, a / 2$, respectively. The straight line segments are the common external tangents of 2 adjacent arcs. At each vertex of the square, there is a column, and the height of each column is equal to the radius of the arc around its base. The tent's roof consists of 4 parts of conical surfaces, whose base edges are the arcs and whose apexes are the column tops above the arc centers, as well as 5 planar quadrilaterals, whose vertices are the column tops and the endpoints of the tangent segments of the arcs. Calculate the area of the roof, the volume of the tent, and its average height (the ratio of its volume to its base area). Draw a possible, single-piece layout of the roof.
I. Let the base of the tent be the square $ABCD = N$, with the upper endpoints of the columns standing at the vertices being $A^*, B^*, C^*, D^*$ such that $AA^* = BB^* = a$, $CC^* = DD^* = a/2$. The base is bounded by arcs around the vertices of $N$, namely $A_1A_2$, $B_1B_2$, $C_1C_2$, $D_1D_2$, and the line segments $A_2B_1$, $B_2C_1$, $C_2D_1$, and $D_2A_1$. The tent is symmetric with respect to the perpendicular bisector plane of the segment $AB$. The first figure shows the top view and side view of the tent.
![](https://cdn.mathpix.com/cropped/2024_05_02_4c4ec36524d7bd482d0eg-1.jpg?height=734&width=934&top_left_y=346&top_left_x=580)

1. Figure

In the right trapezoid $AA_1D_2D$, $\angle DAA_1 = 60^\circ$ because the perpendicular from $D$ bisects the base $AA_1$, and the distance from the foot of the perpendicular to $A_1$ is equal to $DD_2$, which is $a/2$. This perpendicular thus forms a half of an equilateral triangle with the vertex $A$. Furthermore, $A_2B_1 \parallel AB \parallel D_1C_2$, and $A_2$ and $D_1$ lie on the line $AD$. The central angle of the arc $A_1A_2$ is $120^\circ$, and the central angle of the arc $D_1D_2$ is $60^\circ$, so their lengths are $a \cdot 2\pi / 3$ and $a \cdot \pi / 6$, respectively.

The right triangle $DD_2D^*$ is a $2:1$ scaled-down version of the right triangle $AA_1A^*$. It is evident that $A^*D^*$ and $A_1D_2$ intersect at the reflection of $A$ over $D$, and the planes of the two triangles are parallel because the angles at $A$ and $D$ are corresponding. Therefore, $A^*, A_1, D_2, D^*$ lie in the same plane, $A^*A_1 = a\sqrt{2}$, and $D^*D_2 = a/\sqrt{2}$.

From the right trapezoids $ADD^*A^*$ and $ADD_2A_1$, we get $A^*D^* = a\sqrt{5}/2$ and $A_1D_2 = a\sqrt{3}/2$. Thus, the area of the two symmetric conical sections and the trapezoids along the edges $A^*D^*$ and $B^*C^*$ is

$$
T_1 = 2\left(\frac{1}{2} \cdot a \cdot \frac{2\pi}{3} \cdot a\sqrt{2} + \frac{1}{2} \cdot a \cdot \frac{\pi}{6} \cdot \frac{a}{\sqrt{2}} + \frac{a\sqrt{2} + a/\sqrt{2}}{2} \cdot \frac{a\sqrt{3}}{2}\right) = 3\sqrt{2}(\pi + \sqrt{3}) \cdot \frac{a^2}{4}
$$

On the other hand, the sum of the areas of the rectangles $A_2B_1B^*A^*$, $A^*B^*C^*D^*$, and $D^*C^*C_2D_1$, and the area of the tent $T$ are:

$$
\begin{gathered}
T_2 = a\left(a\sqrt{2} + a\frac{\sqrt{5}}{2} + \frac{a}{\sqrt{2}}\right) = (3\sqrt{2} + \sqrt{5}) \cdot \frac{a^2}{2} \\
T = T_1 + T_2 = (3\sqrt{2}(2 + \sqrt{3} + \pi) + 2\sqrt{5}) \frac{a^2}{4} \approx 8.41a^2
\end{gathered}
$$

The volume of the part of the tent below the trapezoid $A^*D^*D_2A_1$ is a three-sided truncated pyramid. Thus, the sum of the volumes of the parts outside the planes $A^*AD$ and $B^*BC$ is

$$
V_1 = 2\left(\frac{a^2\pi}{3} \cdot \frac{a}{3} + \frac{\pi}{6}\left(\frac{a}{2}\right)^2 \cdot \frac{1}{3} \cdot \frac{a}{2} + \frac{1}{3}\left(\frac{a^2}{2} + \frac{a^2}{8} + \frac{a^2}{4}\right) \cdot \frac{a\sqrt{3}}{2}\right) = (17\pi + 21\sqrt{3}) \cdot \frac{a^3}{72}
$$

The sum of the volumes of the prisms between the two planes is

$$
V_2 = a\left(\frac{a^2}{2} + \frac{3a^2}{4} + \frac{a^2}{8}\right) = \frac{11a^3}{8}
$$

and thus the volume of the tent is

$$
V = (99 + 21\sqrt{3} + 17\pi) \cdot \frac{a^3}{72} \approx 2.62a^3
$$

Finally, the base area, with the parts listed in the order of the two calculations above, is

$$
t = 2\left(\frac{a^2\pi}{3} + \frac{a^2\pi}{6 \cdot 4} + \frac{1}{2} \cdot \frac{3a}{2} \cdot \frac{a\sqrt{3}}{2}\right) + a \frac{5a}{2} = \frac{a^2}{4}(3(\pi + \sqrt{3}) + 10) \approx 6.16a^2
$$

Thus, the average height is

$$
m = \frac{V}{t} = \frac{99 + 21\sqrt{3} + 17\pi}{10 + 3\sqrt{3} + 3\pi} \cdot \frac{a}{18} \approx 0.426a
$$

Substituting $a = 15 \text{ m}$, we get $T = 1892 \text{ m}^2$, $V = 8850 \text{ m}^3$, $t = 1385 \text{ m}^2$, and $m = 6.40 \text{ m}$.

II. For the layout plan of the tent (2. figure), we only need to add the central angles of the conical sections to the previously determined dimensions.

![](https://cdn.mathpix.com/cropped/2024_05_02_4c4ec36524d7bd482d0eg-2.jpg?height=550&width=699&top_left_y=690&top_left_x=694)

Based on the relationship $\alpha = \text{arc}/\text{radius}$, the angle at vertex $A^*$ is $(2\pi a / 3) : (\sqrt{2}a) = \sqrt{2}\pi / 3 \approx 1.481$ radians $= 84.9^\circ$, and the angle at vertex $D^*$ is half of that, $42.4^\circ$.

Sugár László (Budapest, I. István g. IV. o. t.) Körmendi Sándor (Szombathely, Latinka S. Gépip. T. IV. o. t.)