In an equilateral triangle of side length 6 , pegs are placed at the vertices and also evenly along each side at a distance of 1 from each other. Four distinct pegs are chosen from the 15 interior pegs on the sides (that is, the chosen ones are not vertices of the triangle) and each peg is joined to the respective opposite vertex by a line segment. If N denotes the number of ways we can choose the pegs such that the

Step-by-step solution

Case I: To divide the triangle into 9 regions. Two pegs must be selected from a side and the other two from a different side. This can be done in ^3 C₂ ^5 C₂ ^5 C₂=300 ways Case II: Now we are choosing 3 points on three sides, such that three lines from those points are concurrent. By using Ceva's theorem in which product of three different ratio leads to 1 . Possible ratio on side A B, B C and C A will be of the form m n , n m and 1 i.e. m n n m 1=1 Ratio 1: 1 can be choosen in 3 ways for all three sides other rat