Difference between revisions of "Circumradius"

(Formula for a Triangle)
(Proof)
Line 8: Line 8:
 
== Proof ==
 
== Proof ==
 
Proof:
 
Proof:
[asy]
+
<asy>
 
pair O, A, B, C, D;
 
pair O, A, B, C, D;
 
O=(0,0);
 
O=(0,0);
Line 19: Line 19:
 
D=-B; dot (D);
 
D=-B; dot (D);
 
draw(B--D--A);
 
draw(B--D--A);
label("<math>A</math>", A, W);
+
label("$A$", A, W);
label("<math>B</math>", B, N);
+
label("$B$", B, N);
label("<math>C</math>", C, E);
+
label("$C$", C, E);
label("<math>D</math>", D, S);
+
label("$D$", D, S);
label("<math>O</math>", O, W);
+
label("$O$", O, W);
 
pair E;
 
pair E;
 
E=foot(B,A,C);
 
E=foot(B,A,C);
 
draw(B--E);
 
draw(B--E);
 
dot(E);
 
dot(E);
label("<math>E</math>", E, S);
+
label("$E$", E, S);
 
draw(rightanglemark(B,A,D,20));
 
draw(rightanglemark(B,A,D,20));
 
draw(rightanglemark(B,E,C,20));
 
draw(rightanglemark(B,E,C,20));
[/asy]
+
</asy>
  
 
We let <math>AB=c</math>, <math>BC=a</math>, <math>AC=b</math>, <math>BE=h</math>, and <math>BO=R</math>. We know that <math>\angle BAD</math> is a right angle because <math>BD</math> is the diameter. Also, <math>\angle ADB = \angle BCA</math> because they both subtend arc <math>AB</math>. Therefore, <math>\triangle BAD \sim \triangle BEC</math> by AA similarity, so we have
 
We let <math>AB=c</math>, <math>BC=a</math>, <math>AC=b</math>, <math>BE=h</math>, and <math>BO=R</math>. We know that <math>\angle BAD</math> is a right angle because <math>BD</math> is the diameter. Also, <math>\angle ADB = \angle BCA</math> because they both subtend arc <math>AB</math>. Therefore, <math>\triangle BAD \sim \triangle BEC</math> by AA similarity, so we have

Revision as of 19:36, 7 December 2014

This article is a stub. Help us out by expanding it.

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 of the circumradius of the triangle is simply $R=\frac{abc}{4A}$. Also, $A=\frac{abc}{4R}$

Proof

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.

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. 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 circumradius $R$ and inradius $r$. Let $d$ be the distance between the circumcenter and the incenter. Then we have \[d^2=R(R-2r)\]

See also

Invalid username
Login to AoPS