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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

 Jun 17, 2020
 #1
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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.

 Jun 17, 2020
 #2
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That answer is incorrect sorry...

Guest Jun 17, 2020

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