105. A problem with three circles.

by Virgil Nicula, Sep 9, 2010, 2:00 PM

Quote:
Consider three circles $w_k\ ,\ k\in\overline {1,3}$ which have only one common point $D$ , i.e. $\{D\}=w_1\cap w_2\cap w_3$ . Prove that

$\left\|\begin{array}{ccc} \{D,A\}=w_2\cap w_3 & , & A_1\in (AD\cap w_1\\\\ \{D,B\}=w_3\cap w_1 & , & B_1\in (BD\cap w_2\\\\ \{D,C\}=w_1\cap w_2 & , & C_1\in (CD\cap w_3\end{array}\ \right\|\ \ \Longrightarrow\ \ \frac {AD}{AA_1}+\frac {BD}{BB_1}+\frac {CD}{CC_1}=1$ .

Proof. Denote $\{B,C_2\}=A_1B\cap w_3$ and $\{C,B_2\}=A_1C\cap w_2$ . Prove easily that $A\in B_2C_2\ ,\ B_1B_2\parallel A_1C_2\ ,$

$C_1C_2\parallel A_1B_2$ . Define $B_0\in DB_2\cap A_1C_2$ and $C_0\in DC_2\cap A_1B_2$ . Thus, $\left\|\begin{array}{c} B_1B_2\parallel A_1C_2\ \Longrightarrow\ \frac {BD}{BB_1}=\frac {B_0D} {B_0B_2}\\\\ C_1C_2\parallel A_1B_2\ \Longrightarrow\ \frac {CD}{CC_1}=\frac {C_0D}{C_0C_2}\end{array}\right\|$ $\Longrightarrow$

$\sum \frac {AD}{AA_1}=\frac {AD}{AA_1}+\frac {B_0D}{B_0B_2}+\frac {C_0D}{C_0C_2}=1$ because $D\in AA_1\cap B_0B_2\cap C_0C_2$ in $\triangle A_1B_2C_2$ .
This post has been edited 7 times. Last edited by Virgil Nicula, Nov 23, 2015, 8:17 AM

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