# Area of a Polygon in a Polygon

I'm dealing with a regular polygon with 7 corners. In this polygon is another polygon defined by connecting one point with the two opposite points of the same polygon.

I made a small sketch of the polygons

I need to calculate the ratio of the area of the big and of small polygon. I know how to calculate the area of the big polygon but I don't know how to get area of the small polygon.

Easy to see that $$P_1Q_2P_3P_2$$ and $$P_1P_7Q_1P_2$$ are parallelograms, which says that $$P_1P_7=Q_1P_2=P_1P_2=P_1Q_2=P_2P_3.$$

Now, let $$P_1P_4\cap P_6P_2=\{A\}$$.

Thus, from $$\Delta Q_1Q_2A$$ we obtain: $$Q_1Q_2=2AQ_2\sin\frac{\measuredangle Q_1AQ_2}{2}=2AQ_2\sin\frac{3\pi}{14}.$$ Also, from $$\Delta P_1P_2A$$ we obtain: $$P_1P_2=2AP_1\sin\sin\frac{\measuredangle P_1AP_2}{2}=2AP_1\sin\frac{3\pi}{14}.$$ Since $$AQ_2+AP_1=P_1Q_2=P_2P_3=P_1P_2,$$ we obtain: $$\frac{Q_1Q_2}{2\sin\frac{3\pi}{14}}+\frac{P_1P_2}{2\sin\frac{3\pi}{14}}=P_1P_2,$$ which gives $$\frac{P_1P_2}{Q_1Q_2}=\frac{1}{2\sin\frac{3\pi}{14}-1},$$ which gives the answer: $$\frac{1}{\left(2\sin\frac{3\pi}{14}-1\right)^2}.$$ Another way.

Let $$R$$ be a radius of the circumcircle of $$P_1P_2...P_7.$$

Thus, $$Q_1Q_2=P_7P_3-2P_1P_2=2R\sin\frac{3\pi}{7}-4R\sin\frac{\pi}{7}$$ and $$P_1P_2=2R\sin\frac{\pi}{7}.$$ Id est, $$\frac{P_1P_2}{Q_1Q_2}=\frac{\sin\frac{\pi}{7}}{\sin\frac{3\pi}{7}-2\sin\frac{\pi}{7}}=\frac{1}{3-4\sin^2\frac{\pi}{7}-2}=$$ $$=\frac{1}{1-2\left(1-\cos\frac{2\pi}{7}\right)}=\frac{1}{2\cos\frac{2\pi}{7}-1},$$ which gives a ratio of areas again.

Let $$O$$ be the centre of the circumscribed circle of our heptagons, and the outer heptagon $$P_1P_2P_3P_4P_5P_6P_7$$ has circumradius $$R$$.

Then the distance from $$O$$ to side $$P_1P_2$$ is $$R\cos(\pi/7)$$ (as seen by drawing the isosceles triangle $$OP_1P_2$$.

Similarly, can you find the distance from $$O$$ to side $$Q_1Q_2=P_3P_7$$ and finish it from here?