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The polygons in the figure below are all regular polygons(regular heptagon), share a vertex and the orange line crosses the three vertices of the two regular polygons, the area of the small regular polygon and the large regular polygon is denoted as $S_1$, $S_2$, what is $\frac{S_1}{S_2}$?

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Additional question (regular nine-sided polygon)

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  • $\begingroup$ S2/S1 = 3 for nine sided polygons $\endgroup$ – Raffaele Aug 18 '20 at 15:51
  • $\begingroup$ It's easy to see the leftmost line of the bigger 9-gon perpendicular to the bottom line but proving it is pretty damn hard. $\endgroup$ – cr001 Aug 18 '20 at 17:40
  • $\begingroup$ I have found a proof to part $2$ and posted as a separate answer. $\endgroup$ – cr001 Aug 19 '20 at 6:27
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Won't go through the calculation, but this is the idea.

First since $\triangle ADE$ and $\triangle BDF$ are similar, we know $AE$ pass through $G$.

Now we can calculate $DG$,$GC$,$AG$ based on the left heptagon and since $AD\parallel CE$ we can calculate $GE=GC\cdot {AD\over DG}$. Also we know $\angle DGE=180^{\circ}-\angle AGD={5\over 7}180^{\circ}$.

Therefore $DE^2=DG^2+GE^2-2\cos({5\over 7}180^{\circ})DG\cdot GE$.

If you let $a=DG,b=DA,c=DB$, there are some identity here

Using the identity, $\cos({5\over 7}180^{\circ})=-{a^2+c^2-b^2\over 2ac}=-{a+b\over 2c}$

New edit: Actually just realized $\angle GEB=\angle GAD=\angle GBE$ so $GE$ is actually just $b$.

Now the calculation is really simple:

$$ED^2=a^2+b^2+ab\cdot{(a+b)\over c}$$ $$=a^2+b^2+{bc(c-b)+c(c+a)(c-b)\over c}$$ $$=a^2+b^2+bc-b^2+c^2+ac-bc-ab$$ $$=a^2+c^2+ac-ab$$ $$=a^2+c^2+b^2-a^2-c^2+b^2$$ $$=2b^2$$

So the area is exactly twice.

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  • $\begingroup$ Not related to the answer, but since the side ratio is actually $\sqrt{2}$ I would suspect a $45$ degree right triangle solution also exists, albeit I am not able to come up with. $\endgroup$ – cr001 Aug 18 '20 at 14:52
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Solution to part $2$ (additional problem):

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Let $I$ be the point where $AD$ intersect the circumcircle $O$ of $\triangle ABC$. Connect $IO$. Since $AI$ is an angle bisector $BI=CI$.

It is easy to see the trapezoid $BDEC$ is symmetric with respect to $IO$. Furthermore $\angle IBC=\angle ICB=10^{\circ}$ so $\angle IBD=50^{\circ}$.

Now let $\angle IDB=x$. With angle tracing using above information we find $$\angle BID=130^{\circ}-x$$ $$\angle IDE=140^{\circ}-x$$ $$\angle DIE=2x-100^{\circ}$$.

If $ID>DB=DE$, then we have $50^{\circ}>130^{\circ}-x$ and $140^{\circ}-x>2x-100^{\circ}$ so $80^{\circ}>x>80^{\circ}$ which is impossible.

If $ID<DB=DE$, then we have $50^{\circ}<130^{\circ}-x$ and $140^{\circ}-x<2x-100^{\circ}$ so $80^{\circ}<x<80^{\circ}$ which is impossible.

Therefore $ID=DB=DE$ and $\triangle IDE$ is equilateral, hence $\angle IDE=60^{\circ}$ and $\angle ADH=180^{\circ}-40^{\circ}-60^{\circ}=80^{\circ}$. Therefore $BD \perp AC$.

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($N$ is just $C$ re-labelled)

The remaining is simple once $BD\perp AC$. We can find $\angle MDN=360^{\circ}-60^{\circ}--90^{\circ}-120^{\circ}=90^{\circ}$.

Since $\angle DMN=60^{\circ}$, $DN=\sqrt{3} DM$ and the area ratio is exactly $3$.

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  • $\begingroup$ I'm searching the question you posted to prove this. $\endgroup$ – SarGe Aug 19 '20 at 6:27
  • $\begingroup$ I deleted the question because I found the solution myself soon after posting that so the question became unnessesary. $\endgroup$ – cr001 Aug 19 '20 at 6:28
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    $\begingroup$ Alright, it happens. :-) $\endgroup$ – SarGe Aug 19 '20 at 6:29

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