+0

# Polygon

-1
7
1
+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.)

Mar 10, 2024

#1
+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
+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 $$15-5\sqrt{5}$$.

hairyberry Mar 10, 2024