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A regular dodecagon P1P2P3P12 is inscribed in a circle with radius 1. Compute (P1P2)2+(P1P3)2++(P11P12)2.(The sum includes all terms of the form (PiPj)2, where 1i<j12.)

 Aug 6, 2020
 #1
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There are 12 digaonals that have a length of P_1 P_2, which from the Sine Law, is sin (15 degrees).  There are 12 diagonals that have a legnth of P_1 P_3, which from the Sine Law, is sin (30 degrees).  We can appy the same reasoning to the other diagonals, which gives us a total sum of

 

(12 sin 15)^2 + (12 sin 30)^2 + (12 sin 45)^2 + (12 sin 60)^2 + (12 sin 75)^2 + (12 sin 90)^2 + (12 sin 105)^2 + (12 sin 120)^2 + (12 sin 135)^2 + (12 sin 150)^2 + (12 sin 175)^2 = 440.

 Aug 6, 2020
 #2
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Wrong

jimkey17  Aug 6, 2020

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