+109 Let P_1 P_2 P_3 \dotsb P_{10} be a regular polygon inscribed in a circle with radius $1.$ Compute
P_1 P_2 + P_2 P_3 + P_3 P_4 + \dots + P_9 P_{10} + P_{10} P_1
+9664 Suppose O is the centre of the circle. Then \(\Delta OP_1 P_2\) is an isosceles triangle with \(OP_1 = OP_2 = 1\) and \(\angle P_1OP_2 = \dfrac{360^\circ}{10} = 36^\circ\).
Then \(P_1P_2 ^2 = 1^2 + 1^2 - 2(1)(1) \cos 36^\circ = 2 - \dfrac{1 + \sqrt 5}2 = \dfrac{3 - \sqrt 5}2\\ \). Note that \( \left(\dfrac{\sqrt 5 - 1}2\right)^2 = \dfrac{3 - \sqrt 5}2\). Then \(P_1P_2 = \dfrac{\sqrt 5 - 1}2\).
Repeating the same process 10 times gives \(P_1 P_2 + P_2 P_3 + P_3 P_4 + \cdots + P_9 P_{10} + P_{10} P_1 = 10\cdot \dfrac{\sqrt 5 -1}2 = 5(\sqrt 5 - 1)\).
