+0  
 
0
58
2
avatar

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

 Apr 11, 2020
edited by Guest  Apr 11, 2020
 #1
avatar+20961 
0

If C is the center of the circle, then angle(P1CP2)  =  360o / 12  =  30o.

To find the length of chord P1P2, we can use Heron's formula:  c2  =  a2 + b2 - 2·a·b·cos(C)

--->   c2  =  12 + 12 - 2·1·1·cos(30o)

              =  1 + 1 - 2·sqrt(3)/2

              =  2 - sqrt(3)

 

Since all chords equal each other, the sum is:  12( 2 - sqrt(3) )  =  24 - 12·sqrt(3)

 Apr 11, 2020
 #2
avatar
0

sorry, but the answer is wrong. :(

 Apr 11, 2020

19 Online Users

avatar
avatar
avatar