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 drawn line segments divide the interior of the triangle into exactly nine regions, find the sum of the squares of the digits of N.

Step 1: To divide the triangle into 9 regions. Two pegs must be selected from $a$ side and the other two from $a$ different

Step 2: This can be done in 3C2 \times 5C2 \times 5C2 = 300 ways

Step 3: Now we are choosing 3 points on three sides, such that three lines from those points are concurrent.

Step 4: By using Ceva’s theorem in which product of three different ratio leads to 1.

Step 5: Possible ratio on side AB, BC and CA will be of the form , and 1.

Step 6: i.e. \times \times1 = 1

Step 7: Ratio 1 : 1 can be choosen in 3 ways for all three sides other ratio can be choosen in 4 ways for other two sides

Step 8: i.e. there are 3 × 4 + 1 = 13 ways

Step 9: Fourth point can be choosen in 12C1 , ways

Step 10: Total such possibilities = 12 × 13 = 156 ways

Step 11: Total ways = 300 + 156 = 456

Step 12: Sum of squares of digit = 42 + 52 + 62 = 16 + 25 + 36 = 77