Let a_1, a_2, a_12 be twelve equally spaced points on a circle with radius 1. find (a_1 a_2)^2 + (a_1 a_3)^2 + + (a_11 a_12)^2. (the sum includes the square of the distance between any pair of points, so the sum includes 12choose2 = 66 terms.)

Let be a fixed positive integer. Then points can be equally spaced along the boundary of the unit disk in the complex plane, and their positions described in the complex plane are given by , where .

So, for any , we can write



The distance between two points and is given by the norm of the difference of the points. We denote this by



We're interested in all the possible distinct pairings, which means and are to be considered the same, so to accomplish this we assume .

Then the sum we're interested in finding is



where this sum is identical to the one in your original question, except the indices of each point are shifted down by one.

We have








So we can express the sum as



For example, if we take points, we can see the points make up the vertices of a square with inradius 1, which means the distance between adjacent points () is , while the distance between opposite points (, which gets counted only once) is . So we would expect the sum to be



In terms of our sum, we would add






as required.

This was just an example to convince you that the formula works.

For , you would end up with

Yes

Step-by-step explanation:

13 is awesome and that,s logic!!