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 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
From this formula, Euler's Inequality follows as
By the Trivial Inequality,
is positive. Since
has to be positive as it is the circumradius,
as desired.