Let $A, B, C, D$ be consecutive vertices of a regular polygon, and the following holds

$$
\frac{1}{A B}=\frac{1}{A C}+\frac{1}{A D}
$$

How many sides does the polygon have?
I. solution. (1) can be transformed as follows:

$$
A B: A C=A D:(A C+A D)
$$

Measure the diagonal $A D$ on the extension of $A C$ beyond $C$, and let the endpoint be $V$. Then

$$
A V=A C+C V=A C+A D
$$

is the last term of the ratio pair (2). Since on the other hand $\angle B A C = \angle D A C$, therefore, due to (2), $\triangle A B C$ and $\triangle A D V$ are similar triangles. But in the first one, $B C = A C$, so in the second one, $D V = D A = C V$, thus $\triangle C D V$ is also an isosceles triangle, and $\angle C V D = \angle C A D$.

Accordingly, the reflection of $V$ over the line $C D$, denoted as $U$, lies on the circumcircle $k$ of the polygon $S$, and it is the last vertex of $S$ when traversing $A B C D \ldots$, because the reflection makes the quadrilateral $C V D U$ a rhombus; $U D$ is a chord of $k$ parallel to $A C$, so $A U = C D$, as the chords corresponding to parallel arcs. Furthermore, $D U = C V = D A$, so the perpendicular bisector of the side $A U$ of the polygon passes through $D$ and the center of $k$. Thus, reflecting the broken line $A B C D$ over $f$, the vertex $A$ maps to $U$, $D$ remains in place, and the vertices $B, C$ map to the missing vertices of $S$. We have thus constructed all the vertices of $S$, obtaining 7 vertices, so the number of sides of $S$ can only be 7. Based on the above, it is clear that a regular heptagon satisfies (1).

Based on the work of József Berács (Győr, Czuczor G. Gymnasium, 4th grade), with the elimination of angle calculations

![](https://cdn.mathpix.com/cropped/2024_05_02_dd3d852f1168693044fbg-1.jpg?height=634&width=957&top_left_y=949&top_left_x=584)

II. solution. Let the number of sides of the polygon $S$ be $n$. The problem mentions 4 vertices, so $n \geq 4$. For a regular quadrilateral, $A B = A D$, so (1) cannot hold, hence $n > 4$. Let the vertex of $S$ following $D$ be $E$, the intersection of the lines $A B$ and $C D$ be $M$ (which exists because $n > 4$); and let the lengths of the segments $A B, A C, A D, A E$ be denoted by $b, c, d, e$ respectively. Since $S$ is symmetric with respect to the perpendicular bisector of $B C$, the triangle $M B C$ is isosceles: $M B = M C = m$. The angles at $A$ in triangles $A M C$ and $A D E$ are inscribed angles corresponding to equal chords, and $\angle A C M$ is the exterior angle of the cyclic quadrilateral $A C D E$, so it is equal to $\angle A E D$, thus triangles $A M C$ and $A D E$ are similar. Therefore, on the one hand, $A M: M C = A D: D E$, and on the other hand, $M C: A C = D E: A E$, that is,

$$
(b+m): m = d: b, \quad m: c = b: e
$$

From the first, $\frac{b}{m} = \frac{d}{b} - 1$, so from the second,

$$
e = c \frac{b}{m} = c\left(\frac{d}{b} - 1\right) = c d\left(\frac{1}{b} - \frac{1}{d}\right)
$$

We note that so far we have not used (1), the derived relationship is valid for any $n > 4$. According to our notation, (1) implies that $1 / c$ stands in the parentheses, so $e = d$. Thus, $\triangle A D E$ is an isosceles triangle. - The completion of the solution is identical to the one above.

István Siklósi (Budapest, Berzsenyi D. Gymnasium, 4th grade)

Remark. The trigonometric equivalent of our solution is as follows.

Let the center of $S$ be $O$, the diameter of the circumcircle be 1, and $\angle A O B = 2 \omega$. Then $b = \sin \omega, c = \sin 2 \omega, d = \sin 3 \omega, e = \sin 4 \omega$, so (1) leads to the equation

$$
\frac{1}{\sin \omega} = \frac{1}{\sin 2 \omega} + \frac{1}{\sin 3 \omega}
$$

To solve this, we use the identity corresponding to (3)

$$
\sin 4 \omega = \sin 2 \omega \sin 3 \omega \left(\frac{1}{\sin \omega} - \frac{1}{\sin 3 \omega}\right)
$$

This identity can also be verified independently: multiplying and rearranging, we get the identity

$$
\sin 4 \omega + \sin 2 \omega = 2 \cos \omega \sin 3 \omega
$$

which is correct based on the known identity for the sum of sines of two angles. 

Now, (5) leads to (4) being the equation $\sin 4 \omega = \sin 3 \omega$. According to the problem, assuming that $S$ is a convex polygon, $0 < 3 \omega < 4 \omega < 180^\circ$, so $3 \omega$ and $4 \omega$ are supplementary angles: their sum is $180^\circ$, $\omega = \frac{180^\circ}{7}$, thus the number of sides of $S$ is 7.