+624 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.)
+394 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 \(15-5\sqrt{5}\).
+394 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 \(15-5\sqrt{5}\).
