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 \(1 \le i < j \le 12.\))
We can use the sine law to find P1 P2
By the sine law, P1 P2 = sin 15. We can also find P1 P_3 = sin 60, P1 P4 = sin 90, etc.
So (P1 P2)^2 + (P1 P3)^2 + … + (P11 P12)^2 = 12 (sin 15)^2 + 12 (sin 30)^2 + 12 (sin 45)^2 + … + 12 (sin 90)^2 = 42.
+118608 Guest, my answer, that I have not finished, would be vey similar to yours but I think it would be a bit different, probably giving a different answer. I do like you idea though.
I can think of other ways to do this but they all involve at least some small amount of trigonometry.
Maybe you could use similar triangles based on 30 and 60 degree triangle. I suppose you could say that is not trignometry...
But even that I do not think would be quite enough on its own. You'd also need to know a ittle about angles of any magnitude ...
I suppose yo could do it using co-ordinate geommetry but it would be very lon winded.
This video could be helpful.
