Vtyjkyuk

Let $O$ be any point on the circumference of a circle circumscribing a regular polygon $A_1,A_2,A_3.., A_2n+1$ such that $O$ lies on the arc $ A_1A_2n+1 $. Show that $OA_1+OA_3+...OA_2n+1=OA_2+OA_4...+OA_2n $.

I tried by find the length of the sides from the angles $OPA_1 , OPA_2 , OPA_3 ... , OPA_2n+1$ Where i considered $P$ to be the circle with ,$OPA_1 = theta$ and the radius of this circle to be $r$ .

Solving further i got struck and got a different answer. Can you pls help me solve this question in this method, If you have some other approach to this question then please share it.

Thank You

The key formula that we need can be found HERE:

$$sum_k=0^nsin(varphi + kalpha)=fracsinfrac(n+1)alpha2sin(varphi+fracnalpha2)sinfracalpha2$$

If you apply the following identity:

$$sinalphasinbeta=frac12(cos(alpha-beta)-cos(alpha+beta))$$

...you get:

$$sum_k=0^nsin(varphi + kalpha)=fraccos(fracalpha2-varphi)-cos(frac2n+12alpha+varphi)2sinfracalpha2tag1$$

Now, denote the center of circumscribed circle with $C$. Introduce:

$$angle OCA_2n+1=-2varphi$$

$$angle A_kCA_2n+1=kalpha$$

$$alpha=frac2pi2n+1tag2$$

You can easily show that:

$$OA_k=2Rsinfracangle A_kCA_2n+1-angle OCA_2n+12=2Rsin(kfracalpha2+varphi)$$

Now calculate two sums:

$$OA_1+OA_3+...OA_2n+1=2Rsin(fracalpha2+varphi)+2Rsin(frac3alpha2+varphi)+...+2Rsin(frac(2n+1)alpha2+varphi)tag3$$

$$OA_2+OA_4+...OA_2n=2Rsin(alpha+varphi)+2Rsin(2alpha+varphi)+...+2Rsin(nalpha+varphi)tag4$$

Now apply formula (1) on (3) and (4) and simplify by taking into account (2). It should not be too complicated to show that (3) and (4) are actually equal.