Difference between revisions of "Euler's inequality"
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==Proof== | ==Proof== | ||
| − | Let the circumradius be <math>R</math> and inradius <math>r</math>. Let <math>d</math> be the distance between the circumcenter and the incenter. Then <cmath>d=\sqrt{R(R-2r)}</cmath> From this formula, Euler's Inequality follows as <cmath>d^2=R(R-2r)</cmath> By the [[Trivial Inequality]], <math>R(R-2r)</math> is positive. Since <math>R</math> has to be positive as it is the circumradius, < | + | Let the circumradius be <math>R</math> and inradius <math>r</math>. Let <math>d</math> be the distance between the circumcenter and the incenter. Then <cmath>d=\sqrt{R(R-2r)}</cmath> From this formula, Euler's Inequality follows as <cmath>d^2=R(R-2r)</cmath> By the [[Trivial Inequality]], <math>R(R-2r)</math> is positive. Since <math>R</math> has to be positive as it is the circumradius, <math> R-2r \ge 0 </math> <math> R \ge 2r </math> as desired. |
Latest revision as of 23:44, 24 June 2015
Euler's Inequality states that ![\[R \ge 2r\]](http://latex.artofproblemsolving.com/b/6/7/b67a21d28c721bc8c4932576a99292686f277554.png)
where R is the circumradius and r is the inradius of a non-degenerate triangle
Proof
Let the circumradius be 
and inradius 
. Let 
be the distance between the circumcenter and the incenter. Then ![\[d=\sqrt{R(R-2r)}\]](http://latex.artofproblemsolving.com/3/5/0/35000a85b7898dabbc4a2f8ccbb8e845d8aff7b5.png)
From this formula, Euler's Inequality follows as ![\[d^2=R(R-2r)\]](http://latex.artofproblemsolving.com/7/e/2/7e2fcd01749b32bd1072381de2ac6589dd730703.png)
By the Trivial Inequality, 
is positive. Since has to be positive as it is the circumradius, 
as desired.
