# The ratio of the area of two regular polygons

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 $$\frac{S_1}{S_2}$$?

• S2/S1 = 3 for nine sided polygons – Raffaele Aug 18 '20 at 15:51
• 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. – cr001 Aug 18 '20 at 17:40
• I have found a proof to part $2$ and posted as a separate answer. – cr001 Aug 19 '20 at 6:27

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.

• 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. – cr001 Aug 18 '20 at 14:52

Solution to part $$2$$ (additional problem):

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 $$ID, then we have $$50^{\circ}<130^{\circ}-x$$ and $$140^{\circ}-x<2x-100^{\circ}$$ so $$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 $$BD \perp AC$$.

($$N$$ 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 $$3$$.

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