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Let P1 P2 P3 ... P10 be a regular polygon inscribed in a circle with radius 1. Compute

(P1 P2)^2 + (P2 P3)^2 + ... + (P1 P10)^2.

 

The sum includes all terms of the form (Pi Pi + 1)^2, where 1 <= i <= 9.  We write Pi Pj to mean the length of segment Pi Pj.)

 Mar 10, 2024

Best Answer 

 #1
avatar+394 
+3

Because \(P_1P_2P_3\dots {P}_{10}\)is a regular polygon, all the side lengths are equal.

Therefore we only need to calculate \({(\overline{P_1P_2)}}^{2}*10\).

Suppose the center of the circle is O, then \(\angle P_1OP_2 = {36}^{\circ}\).

Apply law of cosines:

\({(\overline{P_1P_2)}}^{2}={1}^{2}+{1}^{2}-2*\frac{\sqrt{5}+1}{4}=2-\frac{\sqrt{5}+1}{2}\).

\({(\overline{P_1P_2)}}^{2}*10=20-10(\frac{\sqrt{5}+1}{2})=20-5\sqrt{5}-5=15-5\sqrt{5}\).

So our answer is \(15-5\sqrt{5}\).

 Mar 10, 2024
 #1
avatar+394 
+3
Best Answer

Because \(P_1P_2P_3\dots {P}_{10}\)is a regular polygon, all the side lengths are equal.

Therefore we only need to calculate \({(\overline{P_1P_2)}}^{2}*10\).

Suppose the center of the circle is O, then \(\angle P_1OP_2 = {36}^{\circ}\).

Apply law of cosines:

\({(\overline{P_1P_2)}}^{2}={1}^{2}+{1}^{2}-2*\frac{\sqrt{5}+1}{4}=2-\frac{\sqrt{5}+1}{2}\).

\({(\overline{P_1P_2)}}^{2}*10=20-10(\frac{\sqrt{5}+1}{2})=20-5\sqrt{5}-5=15-5\sqrt{5}\).

So our answer is \(15-5\sqrt{5}\).

hairyberry Mar 10, 2024

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