Squared rays

[bonanova]

Show that the sum of the squares of the lengths of all sides and diagonals

emanating from a vertex of a regular n-gon inscribed in the unit circle is 2n.
 

[gavinksong]

The square lengths of the xth ray can be expressed using the law of cosines: 2 - 2 cos(x * 2pi / n).

If we sum this over all the rays, the cosine terms cancel out, so we are left with 2n.

[bonanova]

The square lengths of the xth ray can be expressed using the law of cosines: 2 - 2 cos(x * 2pi / n).

If we sum this over all the rays, the cosine terms cancel out, so we are left with 2n.

 

I had to think about this point: does it matter there are only n-1 rays?

[gavinksong]

 

The square lengths of the xth ray can be expressed using the law of cosines: 2 - 2 cos(x * 2pi / n).

If we sum this over all the rays, the cosine terms cancel out, so we are left with 2n.

 

I had to think about this point: does it matter there are only n-1 rays?

 

 

That's an excellent question.

 

We include an imaginary 0th ray, the square length of which is 2 - 2 cos(0) = 0.

This is necessary to have all the cosine terms cancel each other out.

[bonanova]

Yeah, I had to think that point through for a moment.

I saw it as having a sum that I didn't know how to evaluate.

So I added and subtracted the term that would make the sum zero.

That step simultaneously evaluated the sum and added 2 to the mix.

 

Your approach is neater, reasoning that the answer is not changed if you add the squared length of the n^th ray.

 

That's a nice insight.