+0  
 
0
77
2
avatar+188 

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 \(1 \le i < j \le 12\)

 Jul 28, 2020
 #1
avatar
+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.

 Jul 29, 2020
 #2
avatar+188 
+2

Hi Guest! Thank you so much for the response!! The answer is actually 144, and you will see why in the solution below, 

 Jul 31, 2020

27 Online Users

avatar
avatar