Difference between revisions of "1985 OIM Problems/Problem 6"

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== Problem ==
 
== Problem ==
 
Given triangle <math>ABC</math>, we consider the points <math>D</math>, <math>E</math>, and <math>F</math> of lines <math>BC</math>, <math>AC</math>, and <math>AB</math> respectively.  If lines <math>AD</math>, <math>BE</math>, and <math>CF</math> all pass through the center <math>O</math> of the circumference of triangle <math>ABC</math>, which radius is <math>r</math>, proof that:
 
Given triangle <math>ABC</math>, we consider the points <math>D</math>, <math>E</math>, and <math>F</math> of lines <math>BC</math>, <math>AC</math>, and <math>AB</math> respectively.  If lines <math>AD</math>, <math>BE</math>, and <math>CF</math> all pass through the center <math>O</math> of the circumference of triangle <math>ABC</math>, which radius is <math>r</math>, proof that:
<math></math>\frac{1}{AD}+\frac{1}{BE}+\frac{1}{CE}=\frac{2}{r}
+
<cmath>\frac{1}{AD}+\frac{1}{BE}+\frac{1}{CE}=\frac{2}{r}</cmath>
  
 
== Solution ==
 
== Solution ==
 
{{solution}}
 
{{solution}}

Revision as of 12:42, 13 December 2023

Problem

Given triangle $ABC$, we consider the points $D$, $E$, and $F$ of lines $BC$, $AC$, and $AB$ respectively. If lines $AD$, $BE$, and $CF$ all pass through the center $O$ of the circumference of triangle $ABC$, which radius is $r$, proof that: \[\frac{1}{AD}+\frac{1}{BE}+\frac{1}{CE}=\frac{2}{r}\]

Solution

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