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 through $O$ the centre of the circumcircle of the triangle $ABC$, whose radius is $R$, show that: \[\frac{1}{AD}+\frac{1}{BE}+\frac{1}{CF}=\frac{2}{R}\]
Problem
Source: Spanish Communities
Tags: Columbia, geometry, circumcircle, trigonometry, geometry unsolved
hucht
07.04.2006 21:34
carlosbr wrote: $1^{st}$ Iberoamerican Olympiad Villa de Leyva, COLOMBIA. [1985] Problem 6 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 $\frac{1}{AD} + \frac{1}{BE} + \frac{1}{CF} = \frac{2}{R}$. $\text{\LaTeX}{}ed$ by Carlos Bravo - [carlosbr] Lema. (Gergonne) Given $\triangle ABC$ and a point $P$ in its plane, let $\triangle P_aP_bP_c$ be the cevian triangle of $P$ respect to $\triangle ABC$, then $\frac{PP_a}{AP_a} + \frac{PP_b}{BP_b} + \frac{PP_c}{CP_c} =1$ and $\frac{AP}{AP_a} + \frac{BP}{BP_b} + \frac{CP}{CP_c} =2$. Proof. By areas it follows $[BPC]/[ABC]=PP_a/AP_a$, $[APC]/[ABC] = PP_b/BP_b$, $[APB]/[ABC]=PP_c/CP_c$, which, when added up, we have $PP_a/AP_a+PP_b/BP_b+PP_c/CP_c=\frac{[BPC]+[APC]+[APB]}{[ABC]}=1$. if we pay attention at $PP_a=AP_a-AP$, $PP_b=BP_b-BP$, $PP_c=CP_c-CP$ we get $\frac{AP}{AP_a} + \frac{BP}{BP_b} + \frac{CP}{CP_c} =2$. $\square$ So, using this we have: $\frac{R}{AD} + \frac{R}{BE} + \frac{R}{CF} = 2$. and thus $\frac{1}{AD} + \frac{1}{BE} + \frac{1}{CF} = \frac{2}{R}$. $\square$
math10
14.06.2009 02:15
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)
stsamster
14.06.2009 10:55
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{\textbf{\textit{1}}}{\textbf{\textit{2R}}}\cdot[\textbf{\textit{CotCCotB+1}}]$. Thus $ \frac{\textbf{\textit{1}}}{\textbf{\textit{AD}}} + \frac{\textbf{\textit{1}}}{\textbf{\textit{BE}}} + \frac{\textbf{\textit{1}}}{\textbf{\textit{CF}}}$ $ =\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.}}$