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, <cmath>R-2r \ge 0\\R \ge 2r</cmath> as desired
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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 00:44, 25 June 2015

Euler's Inequality states that \[R \ge 2r\] where R is the circumradius and r is the inradius of a non-degenerate triangle

Proof

Let the circumradius be $R$ and inradius $r$. Let $d$ be the distance between the circumcenter and the incenter. Then \[d=\sqrt{R(R-2r)}\] From this formula, Euler's Inequality follows as \[d^2=R(R-2r)\] By the Trivial Inequality, $R(R-2r)$ is positive. Since $R$ has to be positive as it is the circumradius, $R-2r \ge 0$ $R \ge 2r$ as desired.

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