1st ibmo - colombia 1985./q6

carlosbr

Given an acute triangle $ABC$ , let $D$ , $E$ and $F$ be
points in the lines $BC$ , $AC$ and $AB$ respectively. If the lines
$AD$ , $BE$ and $CF$ Pass by $O$ the center of the circumcircle
of the triangle $ABC$ , whose radio is $R$ , show that
$\frac{1}{AD}+\frac{1}{BE}+\frac{1}{CF}=\frac{2}{R}$ .

**Posted:** Thu Apr 06, 2006 2:49 am

_________________
Comunidades en Español!!

hucht

**Posted:** Fri Apr 07, 2006 10:34 am

_________________
$\bigstar$ It's good that life is unfair because I can get what I don't deserve $\bigstar$

math10

I think this issue really simple, I resolved by methods similar to hucht.
On some mathematics journals and youth magajine have mentioned this problem.(it is only a lemma to sloved problem that)

**Posted:** Sat Jun 13, 2009 3:15 pm

stsamster

By $\textbf{\textit{Sine}}$ law to $\textbf{\textit{ABD}}\Delta$ , $\frac{\textbf{\textit{AB}}}{\textit{\textbf{Sin(B+90-C)}}}=\frac{\textbf{\textit{AD}}}{\textbf{\textit{Sin(B)}}}$ . So $\frac{\textbf{\textit{1}}}{\textbf{\textit{AD}}}=\frac{\textbf{\textit{Cos(C-B)}}}{\textbf{\textit{2RSinBSinC}}}=\frac{\textb...$ . Thus
$\frac{\textbf{\textit{1}}}{\textbf{\textit{AD}}} + \frac{\textbf{\textit{1}}}{\textbf{\textit{BE}}} + \frac{\textbf{\textit{1...$ $=\frac{\textbf{\textit{1}}}{\textbf{\textit{2R}}} \cdot\textbf{\textit{[CotACotB+CotBCotC+CotCCotA+3]}}$ . But we can show $\textbf{\textit{CotACotB+CotBCotC+CotCCotA}}=\textbf{\textit{1}}$ . Hence the result.

$\textbf{\textit{Q.E.D.}}$ Smile

**Posted:** Sat Jun 13, 2009 11:55 pm