Strange point in a strange polygon

We can express all the vertices using complex numbers: $A_k=Re^\dfrac{2\pi k i}{n}$ , where $R$ is the radius of the circumscribed circumference, but also the following condition holds using simple trigonometry: $L=2R \sin \dfrac{\pi}{n}$ , so $A_k=\dfrac{L}{2\sin\frac{\pi}{n}} \times e^\dfrac{2\pi k i}{n}$ .

We can also express $P$ in this way, where $r$ is the radius of the inscribed circumference: $P=re^{\theta i}$ , but $L=2r\tan \dfrac{\pi}{n}$ , so $P=\dfrac{L}{2\tan\frac{\pi}{n}} \times e^{\theta i}$ .

Next, we calculate the required distances:

$\overline{A_kP}^2=|A_k-P|^2$

If we simplify that, we obtain $\overline{A_kP}^2=\dfrac{L^2}{4}\times\left(\dfrac{1+\cos^2 \frac{\pi}{n}}{\sin^2 \frac{\pi}{n}}-\dfrac{2\cos(\frac{2\pi k}{n}-\theta)}{ \sin \frac{\pi}{n} \tan \frac{\pi}{n} }\right)$

Finally, let's take the sum:

$\displaystyle \sum_{k=0}^{n-1}\overline{A_kP}^2=\dfrac{L^2}{4} \times \left(\displaystyle \sum_{k=0}^{n-1}\dfrac{1+\cos^2 \frac{\pi}{n}}{\sin^2 \frac{\pi}{n}}-2\displaystyle \sum_{k=0}^{n-1}\dfrac{\cos(\frac{2\pi k}{n}-\theta)}{ \sin \frac{\pi}{n} \tan \frac{\pi}{n} }\right)$

The second sumation is $0$ , using the formula for the summation of cosines in arithmetic progression. And we are left with:

$\displaystyle \sum_{k=0}^{n-1}\overline{A_kP}^2=\dfrac{L^2n}{4}\times \dfrac{1+\cos^2 \frac{\pi}{n}}{\sin^2 \frac{\pi}{n}}$

So, we have to solve $\dfrac{L^2n}{4}\times \dfrac{1+\cos^2 \frac{\pi}{n}}{\sin^2 \frac{\pi}{n}}=(45+24\sqrt{3})L^2 \implies n(1+\cos^2 \frac{\pi}{n})=(180+96\sqrt{3})\sin^2 \frac{\pi}{n}$

We can use trial and error method to obtain that $n=\boxed{12}$ or plot the equation for $n \geq 3$ . If you know a better method to solve that, I'd like to know.

I present a slightly more geometric method to obtain an expression for the $LHS$ so that it's a factor of $L^2$ . The rest of the problem, finding $n$ , follows from the other solutions.

Let $B_0,...,B_{n-1}$ denote the point of tangency of the inscribed circumference. Since they are midpoints of each side of the polygon, namely $A_0A_1,...,A_{n-1}A_0$ respectively. By the median formula we have $A_0P^2+A_1P^2=2(B_0P^2+\frac {L^2}{4})$ . Adding all the equations we obtain for each side $LHS=\sum_{i=0}^{n-1} (B_iP^2+\frac {L^2}{4})$ . It is well known that $\sum_{i=0}^{n-1} B_iP^2=2nr^2$ where $r$ is the radius of the inscribed circle. We know $\frac {L}{2r}=\tan(\frac {\pi}{n})$ , thus $LHS=2n(\frac {L^2}{4\tan^2(\frac {\pi}{n})})+n\frac {L^2}{4}=L^2(\frac {n}{2\tan^2(\frac {\pi}{n})}+\frac {n}{4})=\frac {L^2n}{4}(\frac {\cos^2(\frac {\pi}{n})+1}{\sin^2 (\frac {\pi}{n})})$

$\displaystyle \sum_{r=0}^{n-1} \overline{A_{r}P}^{2}$ = $\frac{nL^{2}}{2\sin^{2}\frac{\pi}{n}}=(45+24\sqrt{3})L^{2}$

$\frac{nL^{2}}{2\sin^{2}\frac{\pi}{n}}=(45+24\sqrt{3})L^{2}$

Here, n can be large enough to have $\sin\theta \approx \theta$ . $\therefore \sin^{2}\frac{\pi}{n} \approx (\frac{\pi}{n})^{2}$

$\therefore \frac{nL^{2}}{2\sin^{2}\frac{\pi}{n}}=(45+24\sqrt{3})L^{2} \Rightarrow n^{3} \approx (90+48\sqrt{3})\pi^{2} \Rightarrow n \approx 12$