n-Polygon

If $R$ is the radius of the circumscribed circle, then $A_1A_2 = 2R\sin\left(\frac{360^\circ}{2n}\right),\ \ \ A_1A_3 = 2R\sin\left(2\cdot\frac{360^\circ}{2n}\right),\ \ \ A_1A_4 = 2R\sin\left(3\cdot\frac{360^\circ}{2n}\right).$ We simplify this by writing $\theta = 360^\circ/2n = 180^\circ$ .

$\frac{1}{\sin\theta} = \frac{1}{\sin2\theta} + \frac{1}{\sin3\theta};$ $\frac1s = \frac1{2sc} + \frac1{4sc^2-s};\ \ \ \ \ s=\sin\theta,\ c = \cos\theta$ $1 = \frac1{2c} + \frac1{4c^2-1} = \frac{2c(4c^2-1)}{2c+4c^2-1};$ $8c^3 - 4c^2 - 4c + 1 = 0.$ I solved this numerically and found $c = \cos\theta = 0.90096887;\ \ \ \ \theta = 25{\tfrac5 7}^\circ;$ Therefore $n = \frac{180^\circ}{25{\tfrac5 7}^\circ} = \boxed{7}.$

$WLOG\ let\ n\!-\!gon\ be\ inscribe\ in\ an\ unit\ circle.\\ Let\ \dfrac{360^o}n=2X^o, \ where\ n\ is\ the\ number\ of \ sides.\\ \therefore\ A_1A_2=SinX, \ \ \ A_1A_3=Sin2X=2sinX*CosX,\ \ \ A_1A_4=Sin3X= - 4Sin^3X+3SinX=1 - 4Cos^3X\\ \dfrac{1}{A_1A_2} =\dfrac{1}{A_1A_3} +\dfrac{1}{A_1A_4}\\ \implies\ \dfrac{1}{SinX} =\dfrac{1}{2SinXCosX} +\dfrac{1}{SinX(1 - 4Cos^2X)}\\ SinX\neq 0,\ \ \therefore\ multiplying\ by\ SinX,\ and \ simplifying \ we\ get\ \ 8*Cos^3X-4Cos^2X - 4CosX+1=0.\\ Solving\ this\ cubic\ and \ selecting\ the\ relevant\ root\ \ \ CosX=0.90096886.\\ \ \implies\ \ X=25.71428676. \ \ \therefore\ n=\dfrac{360}{2*25.71428676}= 6.99999971.\\ So\ \ n=\ \ \color{#D61F06}{7}$ )