Triangles with a common incircle

Question

I am given two triangles that intersect at 6 points, which share an incircle of radius $r$.

There are also 6 circles each tangent to two sides of one triangle from the inside, and tangent to a side of the other triangle from the outside. I will number these circles 1 through 6, $R_i$ is the radius of circle $i$.

I need to prove $R_1 R_3 R_5 = R_2 R_4 R_6$.

I know the inner circle with radius $r$ is an excircle of each of the 6 small triangles, so $$r=\frac{R_i \cdot s_i}{s_i - a_i}$$ for all $i$ between 1 and 6, where $s_i$ is the semiperimeter of each small circle, and $a_i$ is the side of the triangle tangent to the circle with radius $r$.

How can I continue from here?

Answer 1

Let us use the following notation of points:

We will use index notation modulo six, with representatives $1,2,3,4,5,6$ for the indices, and denote by

• $R_i$ the radius of the incircle of $\Delta A_iA_{i,i+1}A_{i-1,i}$,
• $A_{ij}$ also the angle in $A_{ij}$ in $\Delta A_iA_{i,i+1}A_{i-1,i}$, for each of the vertices labeled with two indices,

and we observe that the notations are compatible. For instance, the angles in $A_{12}$ in the two triangles $\Delta A_{12}A_{61}A_1$ and
$\Delta A_{12}A_{23}A_2$ are equal.

The proof and the generalization are now immediate, given the following lemma:

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Lemma: Let $\Delta ABC$ be a triangle, let $r$ be the radius of the incircle, and let $r_a$ be the radius of the ex-circle tangent to $BC$. Then we have the relation: $$\frac r{r_a}=\tan \frac B2\tan\frac C2\ .$$

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Proof: We can use the hint from the OP, $$ \begin{aligned} \frac r{r_a} &= \frac{s-a}s= \frac{b+c-a}{b+c+a}= \frac {2R(\sin B+\sin C-\sin A)} {2R(\sin B+\sin C+\sin A)} \\ &= \frac {8R\;\sin \frac B2\;\sin \frac C2\;\cos \frac A2} {8R\;\cos \frac B2\;\cos \frac C2\;\cos \frac A2} = \tan\frac B2\;\tan\frac C2\ . \end{aligned} $$ Alternatively, and for a "shorter" proof, let $A'$ be the projection of the incenter $I$ on $BC$, and $A''$ the projection of the $A$-ex-center $I_a$ on $BC$. Then $A'$ and $A''$ are symmetric to each other w.r.t. the mid point of $BC$, and considering the triangles $\Delta BA'I$ and $\Delta CA''I_a$ we have $BA'=s-b=CA''$, and thus $$ \begin{aligned} \tan\frac B2 &=\frac{IA'}{BA'}=\frac r{s-b}\ ,\\ \tan\frac {\pi-C}2 &=\frac{I_aA''}{CA''}=\frac {r_a}{s-b}\ ,\\[3mm] \tan\frac B2\; \tan\frac C2 &=\frac{\tan\frac B2}{\tan\frac {\pi-C}2} =\frac{r/(s-b)}{r_a/(s-b)}=\frac r{r_a}\ . \end{aligned} $$ $\square$

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Now let us pass to the relation in the OP, $$ R_1R_3R_5=R_2R_4R_6\ , $$ which becomes equivalent to $$ \color{brown} { \left(\tan\frac{A_{61}}2\tan\frac{A_{12}}2\right) \left(\tan\frac{A_{23}}2\tan\frac{A_{34}}2\right) \left(\tan\frac{A_{45}}2\tan\frac{A_{56}}2\right) } = \color{blue} { \left(\tan\frac{A_{12}}2\tan\frac{A_{23}}2\right) \left(\tan\frac{A_{34}}2\tan\frac{A_{45}}2\right) \left(\tan\frac{A_{56}}2\tan\frac{A_{61}}2\right) } \ . $$

$\square$