So seen some answers with explnations but they are all incorrect
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 \(1 \le i < j \le 12.\))
The correct answer is not, 432, 42, or 320
I have check them all and none are correct.
thx in advance
A regular dodecahedron has 12 congruent sides.
If the center of the circle is C, draw all the radii from C to each of the points P1, P2, ... P12.
Each of the triangles formed will be congruent to each other.
Consider the triangle, triangle(CP1P2).
Side CP1 = 1 and side CP2 = 1.
The central angle angle(P1CP2) = 30o. (360o / 12o = 30o)
Use the Law of Cosines on this triangle to find side P1P2:
(P1P2)2 = (CP1)2 + (CP2)2 - 2·(CP1)·(CP2)·cos( angle(P1CP2) )
(P1P2)2 = 12 + 12 - 2·1·1·sqrt(3)/2
This makes the sum work out to 252.
