246. R1-R2=OI (nice).

by Virgil Nicula, Mar 6, 2011, 5:03 PM

PP. Let $\triangle ABC$ and $d$ for which $A\in d$ , $d\perp IO$ and meet $BC$ at $K$ . Denote the lengths $R_1$ , $R_2$ of the circumradii for $\triangle AKB$ , $\triangle AKC$ . Prove that $OI=R_1-R_2$ .

A preliminary study. Suppose $K\in (BC)\ ,\ AK\perp OI$ and $b\ne c\ .$ Denote $\alpha =m(\widehat {BAK})$ and $\beta =m(\widehat {CAK})\ .$ Using an elementary metrics we obtain :

$1.\blacktriangleright \sum a(a-b)(a-c)=4pr(R-2r)\ \ \ ;\ \ \ 2.\blacktriangleright\frac{KB}{c(b-a)}=$ $\frac{KC}{b(a-c)}=\frac{1}{b-c}\ ;$

$3.\blacktriangleright \boxed {\ AK^2=\frac{bc}{a(b-c)^2}\cdot \sum a(a-b)(a-c)\ }\ \ \ ;\ \ \ 4.\blacktriangleright\tan \alpha=$ $\frac{\sin A-\sin B}{\cos A+\cos B-1}\ ,$ $\tan \beta =\frac{\sin A-\sin C}{\cos A+\cos C-1}\ .$

Proof of the proposed problem. Denote the circumradius $R_1$ of the triangle $BAK$ and the circumradius $R_2$ of the triangle $CAK\ .$

Using the relations $(1)$ and $(3)$ we obtain $:\ AK=2R_1\sin B=2R_2\sin C\Longrightarrow (R_1-R_2)^2=AK^2\cdot \frac{\sin B-\sin C)^2}{4\sin^2B\sin^2C}=$

$\frac{bc}{a(b-c)^2}\sum a(a-b)(a-c)\cdot \frac{R^2(b-c)^2}{b^2c^2}=$ $\frac{R^2}{abc}\sum a(a-b)(a-c)=$ $\frac{R^2}{4Rpr}\cdot 4pr(R-2r)=R(R-2r)=OI^2\ .$
This post has been edited 3 times. Last edited by Virgil Nicula, Nov 22, 2015, 12:22 PM

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