Question

A regular dodecagon
`P_1 P_2 P_3 \dotsb P_(12)` is inscribed in a circle with radius 1. Compute


`\[(P_1 P_2)^2 + (P_1 P_3)^2 + \dots + (P_(11) P_(12))^2.\]`
(The sum includes all terms of the form
`(P_i P_j)^2,` where
`1 \le i \ \textless \ j \le 12.)`

Answer 1

Let
`p_i` denote the vector starting at the origin and ending at the vertex
`P_i` of the 12-gon. There is an angle of (360/12)º = 30º between consecutive vectors.

Recall that for any two vectors
`u,v`, we have

`u\cdot v=\|u\|\|v\|\cos\theta`

with
`\theta` the angle between the two vectors. Also recall that

`\|v\|^2=v\cdot v`

For
`1\le i<j\le12`,
`P_iP_j` is the length of the vector
`p_i-p_j`. So

`(P_iP_j)^2=(p_i-p_j)\cdot(p_i-p_j)=p_i\cdot p_i-2p_i\cdot p_j+p_j\cdot p_j`

The 12-gon is inscribed in a circle of radius 1, which means each vector
`p_i` has length 1, and from this we have

`(P_iP_j)^2=2-2p_i\cdot p_j=2-2\cos\theta_(i,j)`

where
`\theta_(i,j)` is the angle between vectors
`p_i` and
`p_j` with
`i\\eq j`, and these angles are multiples of 30º.

There are
`\binom{12}2=66` terms in the sum (from 12 total vertices, you take 2 at a time).

• 11 of these terms are the squared distances between consecutive vertices and separated by 30º, equal to
  `(P_1P_2)^2`;
• 10 of them are the squared distances between vertices that are two vertices apart, separated by 60º, equal to
  `(P_1P_3)^2`;
• 9 of them are the squared distances between vertices that are three vertices apart, separated by 90º, equal to
  `(P_1P_4)^2`;
• and so on, down to the 1 remaining uncounted squared distance between vertices that are ten vertices apart, separated by 330º,
  `(P_1P_(11))^2`.

So we have

`\displaystyle\sum_(1\le i<j\le12)(P_iP_j)^2=\sum_(n=1)^(11)(12-n)(P_1P_n)^2`

`=\displaystyle2\sum_(n=1)^(11)(12-n)(1-\cos(30n)^\circ)=\boxed{144}`

Answer 2

There are 12 digaonals that have a length of P_1 P_2, which from the Sine Law, is sin (15 degrees). There are 12 diagonals that have a legnth of P_1 P_3, which from the Sine Law, is sin (30 degrees). We can appy the same reasoning to the other diagonals, which gives us a total sum of

(12 sin 15)^2 + (12 sin 30)^2 + (12 sin 45)^2 + (12 sin 60)^2 + (12 sin 75)^2 + (12 sin 90)^2 + (12 sin 105)^2 + (12 sin 120)^2 + (12 sin 135)^2 + (12 sin 150)^2 + (12 sin 175)^2 = 440.