<s>[INST] Regular polygon areas in ratio 3:2 Two regular polygons are inscribed in the same circle of radius $r$. First one has $k$ sides and second has $p$ sides. We are given that their areas have a ratio of $1.5$.
Calculate the area of a regular polygon inscribed in the same circle, having number of sides the sum of the other two numbers.
Area of the first polygon: 
$$A_k = \frac{1}{2}\cdot k \cdot r^2 \cdot \sin\bigg(\dfrac{2\pi}{k}\bigg)$$
and area of the second:
$$A_p = \frac{1}{2} \cdot p \cdot r^2 \cdot \sin\bigg(\dfrac{2\pi}{p}\bigg)$$
Also $\frac{A_k}{A_p} = 1.5$ (assuming $k>p$ WLOG)
So
$$\frac{A_k}{A_p} = \frac{k}{p} \cdot \frac{\sin\bigg(\dfrac{2\pi}{k}\bigg)}{\sin\bigg(\dfrac{2\pi}{p}\bigg)} = 1.5$$
Now obviously:
$$A_{k+p} = \frac{1}{2} \cdot (k+p) \cdot r^2 \cdot \sin\bigg(\dfrac{2\pi}{k+p}\bigg)$$
But I don't know how to continue, i.e. how to find a relation between the sins.
 [/INST] There are only limited possibilities for $n$ because we require $n\ge3$ to get any area and also $\frac32\frac n2\sin\frac{2\pi}n<\pi$ because the bigger polygon has to be smaller than the circumcircle. So we enumerate the possibilities:
$$\begin{array}{r|l}n&\text{Area}\\\hline
3&1.299\\
4&2.000\\
5&2.378\\
11&2.974\\
12&3.000\\
13&3.021\\
\end{array}$$
So $n=5$ is too big because $1.5\times2.378>\pi$. $3$ won't work because $1.5\times1.299$ doesn't match anything. Thus we are left with $n=4$ for the smaller polygon, so the area of the bigger polygon is $1.5\times2.000=3.000$, and this matches $n=12$. The final polygon must have $n=4+12=16$ sides with area
$$A=\frac{16}2\sin\frac{2\pi}{16}=8\sin\frac{\pi}8=4\sqrt{2-\sqrt2}$$
So actually @Toby Mak's answer turned out to be correct in that the areas were in fact rational.  
EDIT: Given that only one possibility is feasible it is easy to check that for $n=4$
$$A=\frac42\sin\frac{2\pi}4=2\sin\frac{\pi}2=2$$
While for $n=12$
$$A=\frac{12}2\sin\frac{2\pi}{12}=6\sin\frac{\pi}6=3$$
So the identity is exactly satisfied.
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