In a regular pentagon $A B C D E$, an equilateral triangle $A B M$ is contained. Determine the size of the angle $B C M$.

(L. Hozová)

Hint. What are the sizes of the interior angles of a regular pentagon?
The size of the internal angles of an equilateral triangle is $60^{\circ}$, the size of the internal angles of a regular pentagon is $108^{\circ}$. At vertex $B$, we find that the size of angle $C B M$ is $108^{\circ}-60^{\circ}=48^{\circ}$.

Segments $A B, B C$, and $B M$ are congruent, so triangle $C B M$ is isosceles with base $C M$. The internal angles at the base are congruent and their sum is a right angle. The size of angle $B C M$ is therefore $\frac{1}{2}\left(180^{\circ}-48^{\circ}\right)=66^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_e57040b3fafc3d67d9eag-4.jpg?height=603&width=629&top_left_y=792&top_left_x=722)

Note. A general $n$-sided polygon can be divided into $n-2$ triangles (whose vertices are the vertices of the $n$-sided polygon), so the sum of the sizes of its internal angles is $(n-2) \cdot 180^{\circ}$. A regular $n$-sided polygon has all internal angles congruent, so the size of each is $\frac{n-2}{n} \cdot 180^{\circ}$. This explains the initial relationships for $n=3$ and $n=5$.

A regular $n$-sided polygon can also be divided into $n$ congruent isosceles triangles with a common vertex at the center of the $n$-sided polygon. For $n=5$, we get that angle $A S B$ has a size of $\frac{1}{5} \cdot 360^{\circ}=72^{\circ}$, so angles $S A B, S B A$, etc., have a size of $\frac{1}{2}\left(180^{\circ}-72^{\circ}\right)=54^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_e57040b3fafc3d67d9eag-4.jpg?height=597&width=631&top_left_y=1846&top_left_x=718)