Calculate the area of a triangle with four circles inside

Let us construct a similar configuration by choosing the radius $r$ of the involved circles and the "critical distance" $d$:

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The area of the innermost triangle is $4r^2$ and the innermost triangle is similar to $ABC$. The angle bisectors of $ABC$ are also the angle bisectors of the innermost triangle, so the ratio between a side length of $ABC$ and the length of the parallel side in the innermost triangle is $\frac{i+r}{i}$, with $i$ being the inradius of the innermost triangle. It follows that the area of $ABC$ is $$ 4r^2\left(\frac{i+r}{i}\right)^2 $$ and $\frac{i+r}{i}=\frac{AC}{4r}$. But... wait! This gives that the area of $ABC$ is just $\color{red}{\frac{1}{4}AC^2}$, we do not need to know neither $r$, or $i$, or $d$!!!

In particular, the distance of $B$ from $AC$ is exactly $\frac{AC}{2}$.

Tags:

Geometry

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