In a certain polygon, the ratio of the sum of the magnitudes of its interior angles to the sum of the magnitudes of $\mathrm{k}$ supplementary angles is $3: 5$. (To explain: a supplementary angle complements a given angle to a full angle.)

How many vertices does that polygon have?

(I. Jančigová)
The sum of the sizes of the interior angles in every triangle is $180^{\circ}$, in every quadrilateral $360^{\circ}$, etc. Generally, it is true that every $n$-sided polygon can be composed of $n-2$ triangles, so the sum of its interior angles is $(n-2) \cdot 180^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_18c503b3c9e047833e05g-6.jpg?height=486&width=808&top_left_y=288&top_left_x=630)

The sum of the sizes of the exterior angles of a general $n$-sided polygon is

$$
n \cdot 360^{\circ}-(n-2) \cdot 180^{\circ}=(n+2) \cdot 180^{\circ}.
$$

The ratio of these two values is $(n-2):(n+2)$, which is supposed to be $3: 5$. By rearranging, we get:

$$
\begin{aligned}
\frac{n-2}{n+2} & =\frac{3}{5}, \\
5 n-10 & =3 n+6, \\
2 n & =16, \\
n & =8.
\end{aligned}
$$

The unknown polygon is an octagon.

Another solution. The sums of the sizes of the interior and exterior angles are the same in all polygons with the same number of vertices. Therefore, it is sufficient to consider (for example) only regular polygons.

A regular $n$-sided polygon can be composed of $n$ congruent isosceles triangles. The interior angle at the main vertex of the triangle has a size of $\frac{1}{n} 360^{\circ}$, the sum of the interior angles at the base is equal to the interior angle of the regular $n$-sided polygon and has a size of $180^{\circ}-\frac{1}{n} \cdot 360^{\circ}$. The exterior angle of the regular $n$-sided polygon thus has a size of $360^{\circ}-\left(180^{\circ}-\frac{1}{n} \cdot 360^{\circ}\right)=180^{\circ}+$ $+\frac{1}{n} \cdot 360^{\circ}$.
![](https://cdn.mathpix.com/cropped/2024_04_17_18c503b3c9e047833e05g-6.jpg?height=530&width=1588&top_left_y=2006&top_left_x=240)

The sums of the sizes of the interior and exterior angles are, respectively,

$$
\begin{aligned}
& n\left(180^{\circ}-\frac{1}{n} \cdot 360^{\circ}\right)=(n-2) \cdot 180^{\circ}, \\
& n\left(180^{\circ}+\frac{1}{n} \cdot 360^{\circ}\right)=(n+2) \cdot 180^{\circ}.
\end{aligned}
$$

The ratio of these two values is $(n-2):(n+2)$, which is supposed to be $3: 5$. From this, by the same rearrangements as above, we get $n=8$. The unknown polygon is an octagon.

Evaluation. 2 points for preparatory expressions in terms of $n$; 2 points for setting up and solving the equation; 2 points for the quality of the commentary.

Another solution. Similarly to the previous solution, we focus only on regular polygons, or their division into congruent isosceles triangles, see the illustration above.

In a regular polygon, the ratio of the sums of the sizes of the interior and exterior angles is the same as the ratio of the sizes of these angles at each vertex. This ratio is $3: 5$, precisely when the interior angle of the unknown polygon has a size of

$$
\frac{3}{8} \cdot 360^{\circ}=135^{\circ}
$$

$(3+5=8$ parts correspond to a full angle). The angle at the main vertex of the auxiliary isosceles triangle, i.e., the central angle of the polygon, is $45^{\circ}$ (so that the sum of the interior angles of the triangle is $180^{\circ}$). This angle is one-eighth of a full angle. The unknown polygon is an octagon.

Note. The previous idea can be processed by successively expressing the central, interior, and exterior angles of a regular $n$-sided polygon in terms of $n$ and checking the required ratio:

| $n$ | 3 | 4 | 5 | 6 | 7 | 8 | $\ldots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| central | $120^{\circ}$ | $90^{\circ}$ | $72^{\circ}$ | $60^{\circ}$ | $\frac{1}{7} 360^{\circ}$ | $45^{\circ}$ | $\ldots$ |
| interior | $60^{\circ}$ | $90^{\circ}$ | $108^{\circ}$ | $120^{\circ}$ | $\frac{1}{7} 900^{\circ}$ | $135^{\circ}$ | $\ldots$ |
| exterior | $300^{\circ}$ | $270^{\circ}$ | $252^{\circ}$ | $240^{\circ}$ | $\frac{1}{7} 1620^{\circ}$ | $225^{\circ}$ | $\ldots$ |
| ratio | $1: 5$ | $1: 3$ | $3: 7$ | $1: 2$ | $5: 9$ | $\mathbf{3 : 5}$ | $\ldots$ |

From the geometric intuition, it follows that for increasing $n$, the value of the ratio of the interior and exterior angles gradually increases to $1: 1$. Thus, if the problem has a solution, it is the only solution.

Evaluation. 2 points each for the sizes of the interior and central angles of a regular polygon; 2 points for solving and the quality of the commentary.

When checking step by step, consider the completeness of the commentary. A randomly discovered unexplained solution should be rated 2 points.

[^0]:    The district round of the Z9 category takes place on January 25, 2023, starting no later than 10:00 AM, and participants have 4 hours of clean time to solve the problems. Each problem can earn a participant 6 points, and a successful solver is a student who earns 12 or more points. Allowed aids are writing and drawing tools, school mathematical tables. Calculators and other electronic aids are not allowed.

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