Consider all convex, equilateral pentagons whose angles are not greater than $120^{\circ}$ and which have an axis of symmetry. What are the limits within which the angles of such pentagons lie?
Let there be a pentagon $A B C D E$, which is supposed to be symmetric. The axis of symmetry $t$ passes through one of the vertices, and the reflection either pairs the vertices or, if a vertex lies on $t$, it maps this vertex to itself. One of the vertices of the pentagon is of the latter type because 5 is an odd number.

Let's choose the labeling such that $t$ passes through $A$, so $t$ perpendicularly bisects the side $C D$ and the diagonal $B E$. It is sufficient to examine the limits of the angles $\gamma$ at $C$, $\beta$ at $B$, and $\alpha$ at $A$.

In the practice problem Gy. 1526 (H), we proved that each angle of the pentagon in question is obtuse, regardless of whether it has an axis of symmetry or not, so $90^{\circ}$ is an unattainable lower limit for all three angles. Therefore, in the symmetric trapezoid $B C D E$, $B E > C D$.

It is easy to see that by choosing the position of $A$ on $t$, and $A B$ and $\alpha$, we can construct $B$, and then $C$ on the line that is at a distance of $A B / 2$ from $t$, meaning that the pentagon (and thus $\beta$ and $\gamma$) is essentially determined by $\alpha$. Similarly, by choosing $t$, the symmetric pair of points $C, D$ with respect to $t$, and $\gamma (>90^{\circ})$, we can determine $B$, and then $A$. Finally, choosing $\beta$ also determines $\alpha$ and $\gamma$: after constructing the triangle $B A C$, we draw a tangent from $C$ to the circle centered at $A$ with radius $A B / 2$ (such that it separates $B$ from the circle), which gives the direction of $t$.

Let's introduce the notations $A B E \varangle=\beta_{1}$, $E B C \varangle=\beta_{2}$, $B A C \varangle=\alpha_{1}$, and $C A F \varangle=\alpha_{2}$, where $F$ is the point on the side $C D$ that lies on $t$.

![](https://cdn.mathpix.com/cropped/2024_05_02_e7afc0ca5545c994ed22g-1.jpg?height=740&width=1179&top_left_y=801&top_left_x=468)

Due to convexity, $\beta=\beta_{1}+\beta_{2}$ and $\alpha=2\left(\alpha_{1}+\alpha_{2}\right)$. Then the following relationships hold:

$$
\begin{gathered}
\beta_{1}=90^{\circ}-\frac{\alpha}{2} \\
\cos \beta_{2}=\frac{B E-C D}{2 \cdot B C}=\frac{2 \cdot B A \sin \frac{\alpha}{2}-B A}{2 \cdot B A}=\sin \frac{\alpha}{2}-\frac{1}{2} \\
\cos \gamma=\cos \left(180^{\circ}-\beta_{2}\right)=\frac{1}{2}-\sin \frac{\alpha}{2}
\end{gathered}
$$

The angle $\alpha$ can take its maximum allowed value, $120^{\circ}$, because then $\beta_{1}=30^{\circ}$, $\beta_{2}=68^{\circ} 32'$, $\beta=98^{\circ} 32'$, and $\gamma=111^{\circ} 28'$, which are allowed values.

By decreasing $\alpha$ in this value range, $\beta_{1}$ strictly increases, and $\beta_{2}$ also strictly increases because $\cos \beta_{2}$ strictly decreases. On the other hand, $\gamma$ strictly decreases because $\cos \gamma$ strictly increases. Therefore,

$$
\begin{aligned}
& \alpha_{\max }=120^{\circ}, \quad \beta_{\min }=98^{\circ} 32' \quad(\text { rounded up }), \\
& \gamma_{\max }=111^{\circ} 28' \quad \text { (rounded down) } \\
& \left(\cos \beta_{\min }=(3-\sqrt{3}-\sqrt[4]{12}) 4\right)
\end{aligned}
$$

$\alpha$ can decrease until $\beta$ reaches $120^{\circ}$ in the observed increase, if it ever does. The value $\beta=120^{\circ}$ is achievable because then $\alpha_{1}=30^{\circ}$, $A C=\sqrt{3} A B$, and $\sin \alpha_{2}=\frac{C D}{2 \cdot A C}=\frac{1}{2 \sqrt{3}}$ gives $\alpha_{2}=16^{\circ} 47'$, from which $\alpha=93^{\circ} 34'$, and $\gamma=103^{\circ} 13'$, which are allowed values. Therefore,

$$
\beta_{\max }=120^{\circ} \quad \text { and } \quad \alpha_{\min }=93^{\circ} 34'
$$

since the above can be reversed, $\alpha$ reaches this value with $\alpha_{1}$ and $\alpha_{2}$ strictly decreasing (which can also be read from the construction based on $\beta$); furthermore, the found value of $\gamma$ gives the minimum value of $\gamma$, since $\alpha$ and $\gamma$ vary in the same direction: $\gamma_{\min }=103^{\circ} 13'$.

According to all of this, our answer is:

$$
93^{\circ} 34' \leq \alpha \leq 120^{\circ}, \quad 120^{\circ} \geq \beta \geq 98^{\circ} 32', \quad 103^{\circ} 13' \leq \gamma \leq 111^{\circ} 28'
$$

Our diagrams illustrate a mechanical model of the pentagon in the two extreme positions: the vertices $A, \ldots, E$ are connected by hinges of equal length, the axis is realized by a plane on which $A$ is fixed, and $F$ slides, keeping $C D$ always perpendicular.

![](https://cdn.mathpix.com/cropped/2024_05_02_e7afc0ca5545c994ed22g-2.jpg?height=569&width=1104&top_left_y=594&top_left_x=497)

Our graph shows the variation of the angles as a function of the height $A F=m$; we see that $\alpha$, $\beta$, and $\gamma$ become equal simultaneously in the case of a regular pentagon.

[^0]:    ${ }^{1}$ See the solution in K.M.L. 49 (1974) 145.