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 $1 \le i < j \le 12.$ )

Please, I want hints, not a solution!

Pangolin14 Oct 3, 2020

#1**0 **

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 = 864.

Since we have double-counted, the answer is 864/2 = 432.

Guest Oct 3, 2020

#2**+1 **

I already know that's incorrect, and also, please don't give me the answer, read the problem GUEST!

Pangolin14
Oct 4, 2020