areaeuclidean-geometrygeometrypolygons
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.
We calculate explicitly that the area of the limiting polygon is $\dfrac{4}{7}A$ where $A$ is the area of the original triangle.
Note that on the $(n+1)$-th iteration we cut off twice as many triangles as we did in the $n$-th iteration. Consider the triangles we cut off in the $(n+1)$-th iteration. Each of these has $\dfrac{1}{3}$ the height and $\dfrac{1}{3}$ the base of a triangle cut off in the $n$-th iteration. Hence the ratio of the areas of these triangles is $\dfrac{1}{9}$. Since the total area of the first triangles we cut off is $\dfrac{1}{3}A$, we have that the area of the limiting polygon is
$$A-\sum\limits_{n=0}^\infty2^n\left(\frac{1}{3}A\right)\left(\frac{1}{9}\right)^n=A-\frac{1}{3}A\frac{1}{1-\frac{2}{9}}=\frac{4}{7}A$$
Whether this actually helps us determine the shape of the limiting figure I am not so sure.
This was motivated by Hagen von Eitzen's calculation and Michael's subsequent observation.
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EDIT: Explanation of $\dfrac{1}{3}$ base and $\dfrac{1}{3}$ height statement.
Consider the polygon at a particular vertex $V$ just before making the $n$-th iteration cuts. Choose a particular edge bordering $V$ and let it have length $9x$.
Now take the $n$-th and $(n+1)$-th iteration cuts. In the figure above, the green triangle is one that is cut off in the $n$-th iteration cuts, and the red triangle is one cut off in the $(n+1)$-th iteration cuts.
Because we trisect each time, one side of the green triangle has length $3x$ and one side of the red triangle has length $x$. Call these sides the base of each triangle. Consider now the length of the altitudes corresponding to these bases, letting the altitude of the green triangle be of length $3y$. By similar triangles we find that the altitude of the red triangle has length $y$ (the similar triangles are outlined in black).
Hence the ratio of area of the red triangle to the area of the green triangle is $$\frac{\frac{1}{2}xy}{\frac{1}{2}(3x)(3y)}=\frac{1}{9}$$ as claimed.
Fun note: we did not use the fact that the triangles in question are isosceles.
When you scale a polygon (or any subset of the plane whose area you can define) by $\alpha$, its area scales by $\alpha^2$. Proving this rigorously takes some work and requires you to have a rigorous definition of "area", but you can see that this makes sense by considering a rectangle. If you scale a rectangle by $\alpha$, you scale each of its sides by $\alpha$, so since the area is the product of the two side lengths, you scale the area by $\alpha^2$.
So in your case, you want $\alpha^2A=A'$, or $\alpha=\sqrt{A'/A}$.
Best Answer
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.