Is there a plane that intersects a regular pentagonal pyramid

a) axially,

b) centrally

in a hexagon?
Solution. We will show that the answer is yes in both cases. We will prove that for any regular pentagonal pyramid, there exists a plane that intersects the pyramid in a hexagon that is both axially and centrally symmetric.

Let the base of the pyramid be the unit side length regular pentagon $ABCDE$, and let the sixth vertex be $F$. We know that each angle of the regular pentagon is $108^\circ$, and the length of the diagonals is $(\sqrt{5}+1)/2$ times the length of the sides. Let the intersection of the lines $AE$ and $CD$ be $M$, and the intersection of the lines $BC$ and $ED$ be $N$ (Figure 1). Then $\angle DEM = \angle EDM = 180^\circ - 108^\circ = 72^\circ$. Therefore,

\[
\begin{aligned}
\angle EMD & = 180^\circ - 2 \cdot 72^\circ = 36^\circ = \\
& = \frac{180^\circ - 108^\circ}{2} = \angle ECD,
\end{aligned}
\]

so the triangle $CEM$ is isosceles, and thus $EM = EC = (\sqrt{5}+1)/2$. By the symmetry of the pentagon, it follows that $CN = DN = DM = (\sqrt{5}+1)/2$.

![](https://cdn.mathpix.com/cropped/2024_05_02_c5329fa8b8a3ab1c3e51g-1.jpg?height=567&width=572&top_left_y=725&top_left_x=760)

The plane $\mathcal{T}$, which is the perpendicular bisector of the segment $AB$, contains $D$ and $F$, and $C$ and $E$, as well as $N$ and $M$, are reflections of each other with respect to $\mathcal{T}$.

Let $0 \leq \lambda \leq 1$ be a real number, and let $U_\lambda$ and $V_\lambda$ be the points on the segments $CD$ and $ED$ respectively, such that $CU_\lambda = \lambda = EV_\lambda$. Then $U_\lambda$ and $V_\lambda$ are reflections of each other with respect to $\mathcal{T}$. Let $\mathcal{S}_\lambda$ be the plane containing the line $U_\lambda V_\lambda$ and parallel to the plane $FNM$. Since $U_\lambda V_\lambda$ is perpendicular to $\mathcal{T}$, $\mathcal{S}_\lambda$ is also perpendicular to $\mathcal{T}$. Let the intersection points of $\mathcal{S}_\lambda$ with the edges $FE$, $FA$, $FB$, and $FC$ of the pyramid be $W_\lambda$, $Z_\lambda$, $X_\lambda$, and $Y_\lambda$ respectively (Figure 2). Then $W_\lambda$ and $Y_\lambda$, as well as $Z_\lambda$ and $X_\lambda$, are reflections of each other with respect to $\mathcal{T}$, so the hexagon $U_\lambda V_\lambda W_\lambda Z_\lambda X_\lambda Y_\lambda$ is symmetric with respect to $\mathcal{T}$, and therefore axially symmetric with respect to the line $\mathcal{T} \cap \mathcal{S}_\lambda$. Thus, the plane $\mathcal{S}_\lambda$ intersects the pyramid in a hexagon that is axially symmetric.

If we consider the angles $\angle FAE = \alpha$ and $\angle FAB = \beta$, we get that

\[
\angle AZ_1W_1 = 180^\circ - \beta - 2\alpha > \beta = \angle Z_1AW_1.
\]

Since in any triangle, the side opposite the larger angle is longer, it follows that $W_1Z_1 < AW_1$, and since $AW_1 = EZ_0 = W_0Z_0 = V_1W_1$, we have $W_1Z_1 < V_1W_1$.

We have already seen that $0 = V_0W_0 < V_1W_1 = W_0Z_0$. If $\lambda$ increases continuously from 0 to 1, then by the intercept theorem, $V_\lambda W_\lambda = \lambda \cdot V_1W_1$ also increases continuously from 0 to $V_1W_1$, and $W_\lambda Z_\lambda$ decreases continuously from $W_0Z_0$ to $W_1Z_1$. Since $W_1Z_1 < V_1W_1$, there will be a value of $\lambda$ for which $V_\lambda W_\lambda = W_\lambda Z_\lambda$. (By repeatedly applying the intercept theorem, it can be calculated that $\lambda = \frac{5+\sqrt{5}}{10}$ satisfies this equality, see the Remark after the solution.)

If $V_\lambda W_\lambda = W_\lambda Z_\lambda$, then by the axial symmetry of the hexagon, we have $V_\lambda W_\lambda = Y_\lambda U_\lambda$ and $W_\lambda Z_\lambda = X_\lambda Y_\lambda$. Therefore, the two pairs of opposite sides of the hexagon, $V_\lambda W_\lambda$ and $X_\lambda Y_\lambda$, as well as $Y_\lambda U_\lambda$ and $W_\lambda Z_\lambda$, are not only parallel but also equal in length. This means that the quadrilaterals $V_\lambda W_\lambda X_\lambda Y_\lambda$ and $Y_\lambda U_\lambda W_\lambda Z_\lambda$ are parallelograms. Therefore, their diagonals bisect each other. If the midpoint of the segment $W_\lambda Y_\lambda$ is $O$, then the midpoints of the segments $V_\lambda X_\lambda$ and $U_\lambda Z_\lambda$ are also $O$. This means that the hexagon $U_\lambda V_\lambda W_\lambda Z_\lambda X_\lambda Y_\lambda$ is centrally symmetric with respect to $O$.

This completes the proof of our statement.

Remark. To calculate the exact value of $\lambda$, we use the notation in Figure 4. Let the intersection of the plane $\mathcal{S}_\lambda$ and the line $AM$ be $Q$, and let the distance $EQ$ be $x$. Then, on the one hand,

\[
\begin{aligned}
W_\lambda Z_\lambda & = QZ_\lambda - QW_\lambda = \frac{AQ}{AM} \cdot MF - \frac{EQ}{EM} \cdot MF \\
& = \left( \frac{1+x}{\frac{\sqrt{5}+3}{2}} - \frac{x}{\frac{\sqrt{5}+1}{2}} \right) MF = \frac{3-\sqrt{5} + (4-2\sqrt{5})x}{2} MF
\end{aligned}
\]

On the other hand, the line $U_\lambda V_\lambda$, which is parallel to $EC$, also passes through $Q$, so $\angle EV_\lambda Q = 36^\circ$, and the triangles $EV_\lambda Q$ and $EMD$ are similar, so $EV_\lambda = \frac{\sqrt{5}+1}{2} EQ$, and thus

\[
W_\lambda V_\lambda = \frac{EV_\lambda}{EN} \cdot NF = \frac{\frac{\sqrt{5}+1}{2} x}{\frac{\sqrt{5}+3}{2}} NF = \frac{(\sqrt{5}-1)x}{2} NF
\]

Since $MF = NF$, $W_\lambda Z_\lambda = W_\lambda V_\lambda$ holds exactly when

\[
\frac{3-\sqrt{5} + (4-2\sqrt{5})x}{2} = \frac{(\sqrt{5}-1)x}{2}
\]

which simplifies to $x = \sqrt{5}/5$. From this, we get

\[
\lambda = EV_\lambda = \frac{\sqrt{5}+1}{2} x = \frac{\sqrt{5}+1}{2} \cdot \frac{\sqrt{5}}{5} = \frac{5+\sqrt{5}}{10} \approx 0.7236
\]

![](https://cdn.mathpix.com/cropped/2024_05_02_c5329fa8b8a3ab1c3e51g-3.jpg?height=491&width=588&top_left_y=302&top_left_x=758)