A convex hexahedron has two faces that are regular pentagons $A B C D E$ and $A B F G H$, and two other faces that are equilateral triangles $A E H$ and $B C F$. Determine the surface area and volume of the solid.

 untranslated portion:
Egy konvex hatlapú test két lapja az $A B C D E$ és $A B F G H$ szabályos ötszög, további két lapja az $A E H$ és $B C F$ szabályos háromszög. Határozzuk meg a test felszínét és térfogatát.
The two pentagonal faces can be rotated into each other around the $AB$ edge. Therefore, on the one hand, they are mirror images of each other on the plane bisecting the angle between their planes. Thus, the edges $EH$, $CF$, and the line $DG$ are perpendicular to this plane, parallel to each other, and thus the two remaining faces of the body are the trapezoids $DGHE$ and $DGFC$, due to convexity, since $DG$ is farther from $AB$ than $CF$ and $EH$.

![](https://cdn.mathpix.com/cropped/2024_05_02_be08c97129971b4242a1g-1.jpg?height=534&width=1282&top_left_y=302&top_left_x=411)

From the rotation of the pentagonal faces, it also follows that the line $DE$ intersects the $AB$ rotation axis at point $M$, and the rotated position of $DE$, the line $GH$, also passes through $M$. On $M$, based on the symmetries of the pentagon, $MH = MA = ME = CE$, the diagonal of the pentagon - whose length, as is known, is $2 \cos 36^\circ = (\sqrt{5} + 1) / 2$, taking $AB$ as the unit length - and similarly for the intersection point $N$ of the edge pair $DC$, $GF$, $NC = NF = NB = CE$.

Accordingly, the body in question can be constructed from the tetrahedron $MNDG$ by cutting off the tetrahedra $MAEH$ and $NBCF$. And since, on the other hand, the body is also symmetric with respect to the perpendicular bisector plane of the edge $AB$, its volume is twice the difference in the volumes of the tetrahedra $MODG$ and $MAEH$, where $O$ is the midpoint of the edge $AB$.

Notice also that since $BD$ is parallel to $AE$, the plane $BDG$ is parallel to $AEH$, so $BDG$ is an equilateral triangle as the image of $AEH$ scaled from the center $M$, thus $DB = BG = AC$, and the two trapezoidal faces of the body are congruent to the trapezoid $ABCE$.

Accordingly, the surface area $f$ of the body is given by four times the area of the trapezoid $ABCE$ and twice the area of the triangles $CDE$ and $AEH$. The height of the trapezoid and the triangle $CDE$ perpendicular to the base $CD$ is $\sin 72^\circ = \sqrt{10 + 2\sqrt{5}} / 4$, so

\[
\begin{aligned}
f & = 4 \cdot \frac{1 + \frac{\sqrt{5} + 1}{2}}{2} \sin 72^\circ + 2 \cdot \frac{1}{2} \sin 72^\circ + 2 \frac{\sqrt{3}}{4} = \\
& = \frac{4 + \sqrt{5}}{4} \cdot \sqrt{10 + 2\sqrt{5}} + \frac{\sqrt{3}}{2} = 6.797 \text{ square units. }
\end{aligned}
\]

In the isosceles triangle $ODG$,

\[
OD = \sqrt{BD^2 - BO^2} = \frac{1}{2} \sqrt{5 + 2\sqrt{5}}
\]

so the height from $O$ is:

\[
OO' = \sqrt{OD^2 - \frac{DG^2}{4}} = \frac{1}{4} \sqrt{14 + 6\sqrt{5}} = \frac{\sqrt{5} + 3}{4}
\]

The height of the tetrahedron $MODG$ perpendicular to the face $ODG$ is:

\[
MO = MA + AO = \frac{\sqrt{5} + 2}{2}
\]

so this volume is:

\[
\frac{1}{3} \cdot \frac{1}{2} \cdot \frac{\sqrt{5} + 1}{2} \cdot \frac{\sqrt{5} + 3}{4} \cdot \frac{\sqrt{5} + 2}{2} = \frac{9 + 4\sqrt{5}}{24}
\]

The radius of the circumcircle of the triangle $AEH$ is $AM' = 1 / \sqrt{3}$, so the height from $M$ in the pyramid $MAEH$ is:

\[
MM' = \sqrt{CE^2 - \frac{1}{3}} = \sqrt{\frac{14 + 6\sqrt{5}}{12}} = \frac{\sqrt{5} + 3}{2\sqrt{3}}
\]

so the volume is:

\[
\frac{1}{3} \cdot \frac{\sqrt{3}}{4} \cdot \frac{\sqrt{5} + 3}{2\sqrt{3}} = \frac{\sqrt{5} + 3}{24}
\]

Finally, the volume of the body in question, based on the above, is

\[
V = \frac{2 + \sqrt{5}}{4} = 1.059 \text{ cubic units. }
\]

György Turán (Budapest, Fazekas M. Gyak. Gimn.)

Remarks. 1. The volume of the body can also be calculated from other decompositions, for example, the plane $FCE$ - which also passes through $H$ - divides the body into two oblique triangular prisms (the directions of their edges are $CE$ and $CF$).

2. Similarly, by decomposing the body into the frustums $DGAEH$ and $DGBFC$ and the pyramid $DGAB$, the volume of each of these three parts is $V / 3$.