number of right angle triangles in $n$-sided regular polygon

Question

In a regular polygon of $12$ sides , three vertices are selected at random to form a triangle. Then no. of right angle triangles formed.

And also find probability that triangle formed is equilateral given that triangle formed is obtuse.

Solution I try:

Total number of triangles is $\displaystyle \binom{12}{3}$ and number of equilateral triangle is $4$ .how i find right angle triangle and obtuse angle triangle.

Answer 1

For the right triangle question . . .

Pick a vertex $P$, and consider triangle $APB$, where $A,B$ are two other vertices.

Then triangle $APB$ has a right angle at $P$ if and only if $AB$ is a diameter.

For each $P$, there are $5$ diameters which don't meet $P$.

Since there are $12$ choices for $P$, it follows that there are $(12)(5) = 60$ right triangles, hence the probability of getting a right triangle is $$\frac{60}{\binom{12}{3}}=\frac{60}{220}=\frac{3}{11}$$

For the non-obtuse triangle question . . .

Define the vertex distance between two vertices as the least number of vertex steps to get from one vertex to the other.

For distinct vertices $A,P,B$ of the $12$-gon, triangle $APB$ has an obtuse angle at $P$ if and only if $A,B$ are not diametrically opposite, and $P$ is strictly between $A$ and $B$ on minor arc $AB$.

Suppose distinct vertices $A,B$ of the $12$-gon are not diametrically opposite.

Then the vertex distance from $A$ to $B$ is at least $2$, and at most $5$.

For each $k \in \{2,3,4,5\}$, there are $12$ choices for a chord $AB$ for which the vertex distance from $A$ to $B$ is $k$, and $k-1$ choices for the point $P$, hence the number of obtuse triangles is $$\sum_{k=2}^5 (12)(k-1) = 12(1+2+3+4)=120$$

It follows that of the ${\large{\binom{12}{3}}}$ possible triangles, $${\small{\binom{12}{3}}}-120=220-120=100$$ of them are non-obtuse, and of those, $4$ are equilateral, so the probability of getting an equilateral triangle, given that the chosen triangle is non-obtuse is $$\frac{4}{100}=\frac{1}{25}$$

If you want "acute" instead of "non-obtuse", a minor adjustment is needed . . .

Of the ${\large{\binom{12}{3}}}$ possible triangles, $${\small{\binom{12}{3}}}-(60+120)=220-180=40$$ of them are acute, and of those, $4$ are equilateral, so the probability of getting an equilateral triangle, given that the chosen triangle is acute is $$\frac{4}{40}=\frac{1}{10}$$

Answer 2

Hint:

Denote $O$ the centre of the regular dodecahedron, and $A_1,\dots, A_{12}$ its vertices. Computing the angles in triangles $OA_iA_{i+1}$, you can check the only way to obtain a right triangle $A_iA_jA_k$ is to take three consecutive vertices.