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A regular dodecahedron \(P_1 P_2 P_3 \dotsb P_{12}\) 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 \(1 \le i < j \le 12.\)

There was another post with the same question but had the wrong answer

 Mar 22, 2020
 #1
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Hint; " Sums of squares of lengths should make you think of what?"

 Mar 22, 2020
 #2
avatar+2497 
+1

Nice, Guest! Another hint would be:

 

Since a dodecahedron is inscribed in a circle with radius one, \((P_1 P_2)^2 + (P_1 P_3)^2 + \dots + (P_{11} P_{12})^2.\) would be TWO dodecahedrons.

 Mar 22, 2020
 #3
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Would the answer just be equal to the perimeter of the dodecahedron times two?

Guest Mar 22, 2020
 #4
avatar+2497 
+1

Squared, so technically two dodecahedrons, I think.

CalTheGreat  Mar 22, 2020
 #8
avatar+663 
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I calculated the perimeter of the dodecagon. The answer is not the sum of two dodecagons, but nice try Cal!

AnExtremelyLongName  Mar 30, 2020
 #5
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still not sure how to find the perimeter of the dodecahedron xd

 Mar 22, 2020
 #6
avatar+21955 
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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

Find this value and multiply by 12.

 Mar 23, 2020
 #7
avatar+25574 
+2

A regular dodecahedron \(P_1 P_2 P_3 \dotsb P_{12}\) 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 \(1 \le i < j \le 12\).

 

\(\small{ \begin{array}{|lrcll|} \hline \mathbf{\text{Compute}} \\ \hline &s=(P_1 P_2)^2 + (P_1 P_3)^2+(P_1 P_4)^2+(P_1 P_5)^2+(P_1 P_6)^2+(P_1 P_7)^2+(P_1 P_8)^2+(P_1 P_9)^2+(P_1 P_{10})^2+(P_1 P_{11})^2+(P_1 P_{12})^2 \\ & +(P_2 P_3)^2+(P_2 P_4)^2+(P_2 P_5)^2+(P_2 P_6)^2+(P_2 P_7)^2+(P_2 P_8)^2+(P_2 P_9)^2+(P_2 P_{10})^2+(P_2 P_{11})^2+(P_2 P_{12})^2 \\ & +(P_3 P_4)^2+(P_3 P_5)^2+(P_3 P_6)^2+(P_3 P_7)^2+(P_3 P_8)^2+(P_3 P_9)^2+(P_3 P_{10})^2+(P_3 P_{11})^2+(P_3 P_{12})^2 \\ & +(P_4 P_5)^2+(P_4 P_6)^2+(P_4 P_7)^2+(P_4 P_8)^2+(P_4 P_9)^2+(P_4 P_{10})^2+(P_4 P_{11})^2+(P_4 P_{12})^2 \\ & +(P_5 P_6)^2+(P_5 P_7)^2+(P_5 P_8)^2+(P_5 P_9)^2+(P_5 P_{10})^2+(P_5 P_{11})^2+(P_5 P_{12})^2 \\ & +(P_6 P_7)^2+(P_6 P_8)^2+(P_6 P_9)^2+(P_6 P_{10})^2+(P_6 P_{11})^2+(P_6 P_{12})^2 \\ & +(P_7 P_8)^2+(P_7 P_9)^2+(P_7 P_{10})^2+(P_7 P_{11})^2+(P_7 P_{12})^2 \\ & +(P_8 P_9)^2+(P_8 P_{10})^2+(P_8 P_{11})^2+(P_8 P_{12})^2 \\ & +(P_9 P_{10})^2+(P_9 P_{11})^2+(P_9 P_{12})^2 \\ & +(P_{10} P_{11})^2+(P_{10} P_{12})^2 \\ & +(P_{11} P_{12})^2 \\ \hline \end{array} }\)

 

\(\begin{array}{|rcll|} \hline (P_1 P_2)=(P_2 P_3)=(P_3 P_4)=(P_4 P_5)=(P_5 P_6)=(P_6 P_7)=(P_7 P_8) \\ =(P_8 P_9)=(P_9 P_{10})=(P_{10} P_{11})=(P_{11} P_{12})=(P_1 P_{12})= 2*\sin(15^\circ) \\\\ (P_1 P_2)=(P_2 P_4)=(P_3 P_5)=(P_4 P_6)=(P_5 P_7)=(P_6 P_8)=(P_7 P_9) \\ =(P_8 P_{10})=(P_9 P_{11})=(P_{10} P_{12})=(P_{1} P_{11})=(P_2 P_{12})= 2*\sin(30^\circ) \\\\ (P_1 P_4)=(P_2 P_5)=(P_3 P_6)=(P_4 P_7)=(P_5 P_8)=(P_6 P_9)=(P_7 P_{10}) \\ =(P_8 P_{11})=(P_9 P_{12})=(P_{1} P_{10})=(P_{2} P_{11})=(P_3 P_{12})= 2*\sin(45^\circ) \\\\ (P_1 P_5)=(P_2 P_6)=(P_3 P_7)=(P_4 P_8)=(P_5 P_9)=(P_6 P_{10})=(P_7 P_{11}) \\ =(P_8 P_{12})=(P_1 P_{9})=(P_{2} P_{10})=(P_{3} P_{11})=(P_5 P_{12})= 2*\sin(60^\circ) \\\\ (P_1 P_6)=(P_2 P_7)=(P_3 P_8)=(P_4 P_9)=(P_5 P_{10})=(P_6 P_{11})=(P_7 P_{12}) \\ =(P_1 P_{8})=(P_2 P_{9})=(P_{3} P_{10})=(P_{4} P_{11})=(P_5 P_{12})= 2*\sin(75^\circ) \\\\ (P_1 P_7)=(P_2 P_8)=(P_3 P_9)=(P_4 P_{10})=(P_5 P_{11})=(P_6 P_{12})= 2*\sin(90^\circ)=2 \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline s&=& 12*(2*\sin(15^\circ))^2 + 12*(2*\sin(30^\circ))^2+12*(2*\sin(45^\circ))^2 \\ && +12*(2*\sin(60^\circ))^2 +12*(2*\sin(75^\circ))^2+6*(2*\sin(90^\circ))^2 \\\\ s&=& 48\sin^2(15^\circ) + 48\sin^2(30^\circ)+48\sin^2(45^\circ) \\ && +48\sin^2(60^\circ) + 48\sin^2(75^\circ)+24 \\\\ s&=& 48\left( \sin^2(15^\circ) + \sin^2(30^\circ)+\sin^2(45^\circ)+\sin^2(60^\circ) + \sin^2(75^\circ) \right) +24 \\\\ s&=& 48\left( \sin^2(15^\circ) + \sin^2(30^\circ)+\sin^2(45^\circ)+\cos^2(30^\circ) + \cos^2(15^\circ) \right) +24 \\\\ s&=& 48\left( \sin^2(15^\circ) + \cos^2(15^\circ) + \sin^2(30^\circ)+\cos^2(30^\circ)+\sin^2(45^\circ)\right) +24 \\\\ s&=& 48\left( 2+\sin^2(45^\circ)\right) +24 \quad | \quad \sin(45^\circ)=\dfrac{\sqrt{2}}{2} \\\\ s&=& 48\left( 2+\dfrac{1}{2}\right) +24 \\\\ s&=& \dfrac{48*5}{2} +24 \\\\ s&=& 24*5 +24 \\\\ s&=& 24*6 \\\\ \mathbf{s}&=& \mathbf{144} \\\\ \mathbf{s} &=& \mathbf{12^2} \\ \hline \end{array}\)

 

The sum is \(\mathbf{12^2 = 144}\)

 

laugh

 Mar 24, 2020

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