To solve the problem of finding the number of triangles whose vertices are at the vertices of a regular octagon, but none of whose sides is a side of the octagon, we can follow these steps:
First, calculate the total number of triangles that can be formed by choosing any three vertices from the octagon. A regular octagon has 8 vertices. The number of ways to choose 3 vertices from 8 is given by the combination formula:
\(^8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\).
Next, subtract the number of triangles where at least one side is a side of the octagon from the total number of triangles.
To determine these triangles, notice that for a triangle to have a side of the octagon, it must include two adjacent vertices:
There are 8 such pairs of adjacent vertices (one pair for each side of the octagon). Thus, the number of unwanted triangles (where one side is a side of the octagon) is:
\(8 \times 5 = 40\).
Finally, subtract the number of unwanted triangles from the total number possible:
\(56 - 40 = 16\).
Thus, the number of triangles where none of the sides is a side of the octagon is 16.
Therefore, the correct answer is 16.
The number of triangles having no side common with an \( n \)-sided polygon is given by:
\[ \text{no. of triangles having no side common with a } n \text{-sided polygon} = \binom{n}{1} \times \binom{n-4}{2} \div 3 \]
Substitute \( n = 8 \):
\[ = \binom{8}{1} \times \binom{4}{2} \div 3 \]
\[ = 8 \times 6 \div 3 \]
\[ = 16. \]
Thus, the number of such triangles is 16.
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}