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Difference between revisions of "Circumradius"

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==Euler's Theorem for a Triangle==
 
==Euler's Theorem for a Triangle==
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Let <math>\triangle ABC</math> have circumcenter <math>O</math> and incenter <math>I</math>.Then <cmath>OI=R(R-2r) \implies R \geq 2r</cmath>
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Let <math>\triangle ABC</math> have circumcenter <math>O</math> and incenter <math>I</math>.Then <cmath>OI^2=R(R-2r) \implies R \geq 2r</cmath>
  
 
==Proof==
 
==Proof==

Revision as of 15:48, 8 June 2017

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The circumradius of a cyclic polygon is the radius of the cirumscribed circle of that polygon. For a triangle, it is the measure of the radius of the circle that circumscribes the triangle. Since every triangle is cyclic, every triangle has a circumscribed circle, or a circumcircle.

Formula for a Triangle

Let $a, b$ and $c$ denote the triangle's three sides, and let $A$ denote the area of the triangle. Then, the measure of the circumradius of the triangle is simply $R=\frac{abc}{4A}$. Also, $A=\frac{abc}{4R}$

Proof

[asy] pair O, A, B, C, D; O=(0,0); A=(-5,1); B=(1,5); C=(5,1); dot(O); dot (A); dot (B); dot (C); draw(circle(O, sqrt(26))); draw(A--B--C--cycle); D=-B; dot (D); draw(B--D--A); label("$A$", A, W); label("$B$", B, N); label("$C$", C, E); label("$D$", D, S); label("$O$", O, W); pair E; E=foot(B,A,C); draw(B--E); dot(E); label("$E$", E, S); draw(rightanglemark(B,A,D,20)); draw(rightanglemark(B,E,C,20)); [/asy]

We let $AB=c$, $BC=a$, $AC=b$, $BE=h$, and $BO=R$. We know that $\angle BAD$ is a right angle because $BD$ is the diameter. Also, $\angle ADB = \angle BCA$ because they both subtend arc $AB$. Therefore, $\triangle BAD \sim \triangle BEC$ by AA similarity, so we have \[\frac{BD}{BA} = \frac{BC}{BE},\] or \[\frac {2R} c = \frac ah.\] However, remember that area $\triangle ABC = \frac {bh} 2$, so $h=\frac{2 \times \text{Area}}b$. Substituting this in gives us \[\frac {2R} c = \frac a{\frac{2 \times \text{Area}}b},\] and then bash through algebra to get \[R=\frac{abc}{4\times \text{Area}},\] and we are done.

--Nosaj 19:39, 7 December 2014 (EST)

Formula for Circumradius

$R = \frac{abc}{4rs}$ Where $R$ is the Circumradius, $r$ is the inradius, and $a$, $b$, and $c$ are the respective sides of the triangle and $s = (a+b+c)/2$ is the semiperimeter. Note that this is similar to the previously mentioned formula; the reason being that $A = rs$.

Euler's Theorem for a Triangle

Let $\triangle ABC$ have circumcenter $O$ and incenter $I$.Then \[OI^2=R(R-2r) \implies R \geq 2r\]

Proof

See also

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