We will draw the diagonals from one vertex of a regular pentagon, then construct a regular triangle outward over the side between the endpoints of these diagonals. By lifting the triangles along the diagonals and the mentioned side, until their free vertices meet, we get a three-sided pyramid. What is the volume of the pyramid?
Let the diagonals of the regular pentagon $A D_{1} B C D_{2}$ be $A B$ and $A C$, and the new vertex of the equilateral triangle constructed over the side $B C$ be $D_{3}$. The free vertices $D_{1}, D_{2}, D_{3}$ indeed meet at a point $D$, forming the pyramid. Indeed, each of the three raised triangles is an isosceles triangle, and their bases are equal to the side $a$ of the given pentagon; furthermore, the free vertex is above the perpendicular bisector of the fold line in the base plane, e.g., $D_{1}$ at most up to the reflection of $D_{1}$ over $A B$, denoted as $D_{1}^{\prime}$. These three perpendicular bisectors have a common point, the center $O$ of the given pentagon, and this point lies on all three segments $D_{i} D_{i}^{\prime}$ $(i=1,2,3)$, since $O A B \varangle = O A C \varangle = B A C \varangle / 2 = 18^{\circ} < 36^{\circ} = B A D_{1}$, and $O B C \varangle = 54^{\circ} < 60^{\circ} = D_{3} B C$.

![](https://cdn.mathpix.com/cropped/2024_05_02_4b112edc53284d4c10eag-1.jpg?height=664&width=751&top_left_y=454&top_left_x=665)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_05_02_4b112edc53284d4c10eag-1.jpg?height=480&width=290&top_left_y=1213&top_left_x=901)

Figure 2

The height $m$ of the base triangle corresponding to the base $B C = a$ is also the symmetry axis of the pentagon, from which the area $t$ of the base is

$$
m = \frac{a}{2} \tan 72^{\circ}, \quad t = \left(\frac{a}{2}\right)^{2} \cdot \tan 72^{\circ} = \frac{a^{2}}{4} \sqrt{5 + 2 \sqrt{5}}
$$

Let the diagonal $A B$ of the pentagon be $d$, and its midpoint be $F$. Then the height of the side face $A B D$, its projection on the base plane, and the height of the pyramid are

$$
\begin{gathered}
D F = \frac{d}{2} \tan 36^{\circ} = \frac{d}{2} \sqrt{5 - 2 \sqrt{5}}, \quad O F = \frac{d}{2} \tan 18^{\circ} = \frac{d}{2} \sqrt{\frac{5 - 2 \sqrt{5}}{5}} \\
D O = \sqrt{D F^{2} - O F^{2}} = \frac{d}{\sqrt{5}} \sqrt{5 - 2 \sqrt{5}}
\end{gathered}
$$

Finally, the diagonal $d$ and the sought volume $V$ are

$$
\begin{gathered}
d = \frac{a}{2 \cos 72^{\circ}} = \frac{a}{2} (\sqrt{5} + 1) \\
V = \frac{1}{3} \cdot \frac{a^{2}}{4} \sqrt{5 + 2 \sqrt{5}} \cdot \frac{a}{2} \cdot \frac{\sqrt{5} + 1}{\sqrt{5}} \cdot \sqrt{5 - 2 \sqrt{5}} = \frac{a^{3}}{24} (\sqrt{5} + 1) \quad \left(\approx a^{3} \cdot 0.135\right)
\end{gathered}
$$

Lajos Birkner (Budapest, Konyves Kalman Gymnasium III. grade)

Remarks. 1. Each used trigonometric value can be derived from $\sin 18^{\circ}$, which in turn follows from the fact that if the base of an isosceles triangle with base $a$ and equal sides $b$ is $36^{\circ}$, then the bisector of one of the angles at the base divides the triangle into two isosceles triangles, and the triangle resting on the base is similar to the original: $(b - a) : a = a : b$, and from this

$$
\left(\frac{a}{b}\right)^{2} + \left(\frac{a}{b}\right) - 1 = 0, \quad \sin 18^{\circ} = \frac{a}{2 b} = \frac{\sqrt{5} - 1}{4}
$$

(the other root is negative). For example,

$$
\begin{aligned}
& \sin 72^{\circ} = \cos 18^{\circ} = \sqrt{1 - \sin^{2} 18^{\circ}} = \frac{\sqrt{10 + 2 \sqrt{5}}}{4} \\
& \tan 72^{\circ} = \frac{\sqrt{10 + 2 \sqrt{5}}}{\sqrt{5} - 1} = \sqrt{\frac{5 + \sqrt{5}}{3 - \sqrt{5}}} = \sqrt{5 + 2 \sqrt{5}}
\end{aligned}
$$

2. Those familiar with the so-called Ptolemy-Dürer method for constructing the side of a regular decagon and pentagon inscribed in a given circle will notice that the height of our pyramid is equal to the side $a_{10}$ of a regular decagon inscribed in the circle $k$ circumscribing the initial pentagon.

![](https://cdn.mathpix.com/cropped/2024_05_02_4b112edc53284d4c10eag-2.jpg?height=545&width=590&top_left_y=790&top_left_x=754)

Figure 3

Indeed, in the said method, we take the diameter $L M$ perpendicular to the radius $O A$ of $k$, and the midpoint $N$ of the radius $O M$ and the circle with radius $N A$ intersect the radius $O L$ at $P$. Then $O N = r / 2, A N = r \sqrt{5} / 2$ and thus $P O = r (\sqrt{5} - 1) / 2 = 2 r \sin 18^{\circ} = a_{10}$, and $A P = a_{5} = 2 r \sin 36^{\circ} = 4 r \sin 18^{\circ} \cdot \cos 18^{\circ}$, indeed, from the previous remark,

$$
a_{5}^{2} - a_{10}^{2} = 4 r^{2} \sin^{2} 18^{\circ} \left(4 \cos^{2} 18^{\circ} - 1\right) = r^{2} \frac{3 - \sqrt{5}}{2} \cdot \frac{3 + \sqrt{5}}{2} = r^{2}
$$

thus indeed $a_{5}^{2} = a_{10}^{2} + r^{2} = P O^{2} + O A^{2} = A P^{2}$, and therefore the right triangle $P A O$ is congruent to the triangle $D A O$ of our pyramid: $m_{0} = O D = O P = a_{10}$.

Based on our insight, a simplification arises in the calculation of the volume:

$$
V = \frac{1}{3} t m_{0} = \frac{1}{6} \left(a d \sin 72^{\circ}\right) \cdot a_{10} = \frac{a d}{12} \cdot \left(2 a_{10} \sin 72^{\circ}\right) = \frac{a^{2} d}{12} = \frac{a^{3}}{24} (\sqrt{5} + 1)
$$

if, indeed, the three consecutive vertices of the regular decagon inscribed in $k$ are $B, A^{\prime}, C$, then $B A^{\prime} C \varangle / 2 = 72^{\circ}$, and as the shortest diagonal of the decagon, $a_{5} = B C = 2 B A^{\prime} \cdot \sin \left(B A^{\prime} C \varangle / 2\right) = 2 a_{10} \sin 72^{\circ}$.

3. Similarly to the above, it can also be shown that if the other intersection point of the circle $N(N A)$ and the line $L M$ is denoted by $Q$, then $A Q = A B$ is the diagonal $d$ of the pentagon, and at the same time the length of the diagonal spanning four sides of the decagon, and $O Q$ is the length of the diagonal spanning three sides. [^0]

[^0]: ${ }^{1}$ See problem 1472, K. M. L. 34 (1967) 19. p.; problem 1213, 27 (1963) 20. p.