Juquinha marks points on a circumference and draws triangles by connecting 3 of these points. The lengths of the arcs between 2 consecutive points are equal.

a) By marking 4 points on the circumference, how many triangles can he draw?

b) By marking 5 points on the circumference, how many equilateral triangles can he draw?

c) How many right triangles can he draw if he marks 6 points?
Solution

a) 4. We need to choose 3 points out of a total of 4. In situations like this, when we choose any triangle, for example, triangle $A B C$, all permutations of points $A, B$, and $C$ still form triangle $A B C$. Therefore, when using the fundamental principle of counting, we must divide by the number of permutations of 3 points. Thus, the total number of triangles we can draw, using four points on a circle as vertices, is $\frac{4 \cdot 3 \cdot 2}{3 \cdot 2 \cdot 1}=4$.

b) None. Since the lengths of the arcs delimited by 2 consecutive points are equal, each of these arcs measures $\frac{360^{\circ}}{5}=72^{\circ}$. When drawing an equilateral triangle inscribed in a circle, each arc measures $\frac{360^{\circ}}{3}=120^{\circ}$. Therefore, $120^{\circ}$ should be a multiple of $72^{\circ}$, which is absurd. Thus, it is not possible to draw an equilateral triangle with vertices at 3 of the 5 points symmetrically drawn on a circle.

c) 12. A right triangle inscribed in a circle must have one of its sides (the hypotenuse) on the diameter. Since 6 points are symmetrically drawn on the circle, the measure of each arc delimited by 2 consecutive points is $\frac{360^{\circ}}{6}=60^{\circ}$, meaning we need to choose 2 points with 3 arcs on each side. This is the same as drawing a regular hexagon using these points and taking the longest diagonal as the hypotenuse. Thus, there are 3 of these diagonals (diameters), and for each of them, there are 4 points to complete the triangle. Therefore, the total number of right triangles is $3 \cdot 4=12$.