[ Homothetic Circles]

Given a triangle $ABC$. Four circles of equal radius $\rho$ are constructed such that one of them is tangent to the other three, and each of these three is tangent to two sides of the triangle. Find $\rho$, if the radii of the inscribed and circumscribed circles of the triangle are $r$ and $R$ respectively.
Let $A_{1}, B_{1}$ and $C_{1}$ be the centers of the circles touching the sides of the triangle, $O$ be the center of the circle touching these circles, $O_{1}$ and $O_{2}$ be the centers of the inscribed and circumscribed circles of triangle $A B C$. The lines $A A_{1}, B B_{1}$ and $C C_{1}$ are the angle bisectors of triangle $A B C$, so they intersect at point $O_{1}$. Therefore, triangle $A_{1} B_{1} C_{1}$ is transformed into triangle $A B C$ by a homothety with center $O_{1}$, and the coefficient of homothety is the ratio of the distances from point $O_{1}$ to the sides of triangles $A B C$ and $A_{1} B_{1} C_{1}$, i.e., it is equal to $(r-\rho) / r$. Under this homothety, the circumscribed circle of triangle $A B C$ is transformed into the circumscribed circle of triangle $A_{1} B_{1} C_{1}$. Since $O A_{1}=O B_{1}=O C_{1}=2 \rho$, the radius of the circumscribed circle of triangle $A_{1} B_{1} C_{1}$ is $2 \rho$. Therefore, $R(r-\rho) / r=2 \rho$, i.e., $\rho=r R /(2 r+R)$. 

Submit a comment