Circles and enneagon

Question

Using that in a triangle ABC, $\tan\frac A2=\frac{r}{p-a}$ where $p=\frac{a+b+c}{2}$, I found that the radius are equal if $\tan^220°=\frac{\frac{1+2\cos40°}{\cos20°-1}}{1-\frac{1}{2\cos40°}+\frac{\sin60°}{\sin80°}}$ Can someone help me to prove this? Thanks for antention!

Answer 1

Refer to the annotated figure.

Claim 1. The intersecton of $A_1 A_6$ and $A_3 A_7$ at $O_2$ is also the incenter of $\triangle A_5 A_8 D$.

Proof. Let $O_2$ be the intersection of $A_1 A_6$ and $A_3 A_7$. Note $\triangle A_6 A_7 O_2$ is equilateral. Thus, $\triangle A_8 A_7 O_2$ is isosceles with $A_8 A_7 = O_2 A_7$. Since $\angle A_8 A_7 O_2 = \frac{4\pi}{9}$, it follows that $$\angle A_7 A_8 O_2 = \angle A_8 O_2 A_7 = \frac{1}{2}\left(\pi - \frac{4\pi}{9}\right) = \frac{5\pi}{18}.$$ But since $\angle A_7 A_8 A_5 = \frac{2\pi}{9}$ and $\angle A_5 A_8 A_4 = \frac{\pi}{9}$, we have $$\angle M_2 A_8 O_2 = \angle A_7 A_8 O_2 - \angle A_7 A_8 A_5 = \frac{\pi}{18} = \frac{1}{2} \angle A_5 A_8 A_4.$$ Therefore, $A_8 O_2$ bisects $\angle D A_8 A_5$, and by symmetry, $A_5 O_2$ bisects $\angle D A_5 A_8$. Consequently $O_2$ is also the incenter of $\triangle A_5 A_8 D$ as claimed.

As a result of this claim, we can easily calculate the inradius $r_2 = O_2 M_2$. Without loss of generality let the side length of the $9$-gon be $1$, so in particular $A_6 A_7 = A_7 O_2 = A_7 A_8 = 1$. Since $A_5 A_8 || A_6 A_7$ and $\angle A_7 A_8 A_5 = \frac{2\pi}{9}$, we have $PM_2 = \sin \frac{2\pi}{9}$, hence $$r_2 = \frac{\sqrt{3}}{2} - \sin \frac{2\pi}{9}.$$

Next, we proceed with calculating the inradius $r_1 = O_1 M_1$. Note $\angle A_8 AB = \frac{\pi}{3}$ and $\angle A_1 A_8 M_1 = \frac{\pi}{9}$, hence $$AM_1 = r_1 \sqrt{3}, \quad M_1 A_8 = r_1 \cot \frac{\pi}{9},$$ so again assuming unit side length, we obtain $$r_1 = \frac{1}{\sqrt{3} + \cot \frac{\pi}{9}}.$$ All that is left is to show $r_1 = r_2$; to this end, we consider the ratio $$\begin{align} \frac{r_2}{r_1} &= \left(\frac{\sqrt{3}}{2} - \sin \frac{2\pi}{9}\right)\left(\sqrt{3} + \cot \frac{\pi}{9}\right) \\ &= \left(\sin \frac{3\pi}{9} - \sin \frac{2\pi}{9}\right)\left(\tan \frac{3\pi}{9} + \cot \frac{\pi}{9}\right) \\ &= 2 \cos \frac{5\pi}{18} \sin \frac{\pi}{18} \left(\frac{\sin \frac{3\pi}{9} \sin \frac{\pi}{9} + \cos \frac{\pi}{9} \cos \frac{3\pi}{9}}{\cos \frac{3\pi}{9} \sin \frac{\pi}{9}} \right) \\ &= \sin \frac{2\pi}{9} \frac{\sin \frac{\pi}{9}}{\cos \frac{\pi}{18}} \frac{\cos \frac{2\pi}{9}}{\cos \frac{3\pi}{9} \sin \frac{\pi}{9}} \\ &= \frac{\sin \frac{2\pi}{9} \cos \frac{2\pi}{9}}{\frac{1}{2} \sin \frac{4\pi}{9}} \\ &= 1. \end{align}$$

I have no doubt there must be a more elegant way, possibly even a purely geometric one, but this was the most immediate approach that was sufficiently elementary that I could find.

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Edit.

I have found an alternative solution. Refer to the following figure.

Claim 2. The intersection $O_2$ of $A_1 A_6$ and $A_3 A_7$ is also the incenter of $\triangle A_1 A_5 A_7$.

Proof. $\angle A_1 A_7 A_3 = \angle A_3 A_7 A_5 = \frac{2\pi}{9}$, and $\angle A_7 A_1 A_6 = \angle A_6 A_1 A_5 = \frac{\pi}{9}$.

As a result of this claim, we conclude that the inradius $$r_3 = O_2 S = \sin \frac{2\pi}{9},$$ since $S$ is the midpoint of $O_2 A_8$ of isosceles $\triangle O_2 A_7 A_8$ and also the point of tangency of the corresponding incircle.

Next, note that $\triangle A_1 A_5 A_7 \sim \triangle A_8 A B$, simply by comparing their angle measures. So $$r_1 = r_3 \frac{AA_8}{A_1 A_5} = 2\sin \frac{2\pi}{9} \sin \frac{\pi}{18}.$$ But we already established that $$r_2 = \sin \frac{3\pi}{9} - \sin \frac{2\pi}{9} = 2\cos \frac{5\pi}{18} \sin \frac{\pi}{18},$$ and since $\cos \frac{5\pi}{18} = \sin \frac{2\pi}{9}$, we are done.