A regular dodecagon \($P_1 P_2 P_3 \dotsb P_{12}$\) is inscribed in a circle with radius 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 \(1 \le i < j \le 12\)

Noori Jul 28, 2020

#1**+2 **

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.

Guest Jul 29, 2020