Count me if you can Part 2

Probability Level 5

You are given a 20 -sided regular polygon. Find the number of pairs of diagonals that intersect either in the interior of the polygon or the exterior of the polygon. Let the answer be $N$ . What is the sum of digits of $N$ ?

Remarks : Diagonals are not edges.

Note: This would be a lot easier if you have solved Count me if you can.

Check it out here

5 6 7 8

1 solution

Mark Hennings
Feb 3, 2020

There are $\tfrac 12 \times 20 \times (20-3) = 170$ diagonals, and hence a total of $\binom{170}{2}=14365$ pairs of diagonals. A total of $10\binom{9}{2} + 10\binom{8}{2} = 640$ pairs of diagonals are parallel (so do not intersect at all), and a total of $20\binom{17}{2} = 2720$ pairs of diagonals meet at a vertex, so the desired count of diagonal pairs is $14365-640-2720=11005$ , making the answer $1+1+0+0+5=\boxed{7}$ .

I did not understand how you got 640 pairs of parallel diagonals... Please explain

Tattwa shiwani - 1 year, 3 months ago

There are $8$ diagonals that are parallel to any given edge, and $9$ diagonals perpendicular to any diagonally opposite pair of vertices. Now count the pairs of diagonals...

Mark Hennings - 1 year, 3 months ago

why do you have (20-3) over there? Where did that 3 come from and also why you have 1/2? 1/2 means that the polygon looks like a circle??? alsom 170/2 = 14365 but i calculated it, it goes 85. I am really confused lol

Slayer white - 1 year, 3 months ago

20 choices of one vertex, and then 17 choices of a non-adjacent vertex. Since we are only interested in vertex pairs, and not the order in which the vertices are chosen, we divide by 2.

Mark Hennings - 1 year, 3 months ago

@Mark Hennings , Brilliant approach.

Nikola Alfredi - 11 months, 3 weeks ago

Count me if you can Part 2

Note: This would be a lot easier if you have solved Count me if you can.

1 solution

0 pending reports