BdMO Online Forum

$\textbf{Problem 1:}$
Let $ABC$ be a triangle with circumcenter $O$ and incenter $I$.Also let $R$,$r$ denote the circumradius and inradius of $ABC$ respectively.Prove that \[R^2-OI^2=2rR.\]

$\textbf{Solution :}$
Let $AI \cap \odot(ABC)=A, D$ and $DO \cap \odot(ABC)=D, E$.Let the incircle touch $AB$ at $F$.We have $\triangle AFI\thicksim\triangle EBD$ by easy angle chase.So, $\frac{AI}{DE}=\frac{IF}{BD}$ implying $AI \cdot BD=DE \cdot IF$.As $DE=2R$, $BD=DI$ and $IF=r$,we know that the power of $I$ wrt $\odot(ABC)$ is $AI \cdot DI=AI \cdot BD=2rR$.But the power of $I$ wrt the circumcircle is also $R^2-OI^2$.Therefore,\[R^2-OI^2=2rR\],done.