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Let $S_{\mathcal T}$ be the surface of triangle $\mathcal T$. Let $3$ isometrical circles meet at a certain point, and name $A$, $B$, $C$ their $3$ pairwise intersections. Let $XYZ$ be the triangle containing the $3$ circles such that each side is tangent to $2$ of them. Show that $S_{XYZ}\ge 9S_{ABC}$

After thinking about this problem for a bit, and looking at a drawing of it, the circumradius of $XYZ$ seems to be $r_1+r_2$, where $r_1$ is the common radius of the $3$ isometrical circles and $r_2$ is the inradius of the triangle formed by the centers of the circles (which I believe is also the inradius of $ABC$, since both triangles seem isometrical). If this is proven, then we can easily conclude with Euler's relation...

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Let us assume that the radius of each original circle is $1$, and let us denote with $D,E,F$ the centers of such circles and with $P$ their common point. enter image description here

A homotethy with center $P$ and factor $\frac{1}{2}$ maps $ABC$ into the medial triangle of $DEF$, hence $[ABC]=[DEF]$. $XYZ$ and $DEF$ have parallel sides and the inradius of $XYZ$ is just one plus the inradius of $DEF$, so

$$ \frac{[XYZ]}{[ABC]}=\frac{[XYZ]}{[DEF]}=\left(\frac{r_{DEF}+1}{r_{DEF}}\right)^2 $$ and it is enough to show that $r_{DEF}\leq \frac{1}{2}$. On the other hand the circumradius of $DEF$ is $1=PD=PE=PF$ and by Euler's inequality $r_{DEF}\leq \frac{1}{2}R_{DEF}$: we are done.

Since $DEF$ and $ABC$ are isometrical, we have also proved Johnson's theorem $R_{ABC}=1$.

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