Difference between revisions of "Circumradius"

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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 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 $d^2=R(R-2r)$

@@ Line 5: / Line 5: @@
 ==Formula for a Triangle==
 Let <math>a, b</math> and <math>c</math> denote the triangle's three sides, and let <math>A</math> denote the area of the triangle. Then, the measure of the of the circumradius of the triangle is simply <math>R=\frac{abc}{4A}</math>. Also, <math>A=\frac{abc}{4R}</math>
+== 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("<math>A</math>", A, W);
+label("<math>B</math>", B, N);
+label("<math>C</math>", C, E);
+label("<math>D</math>", D, S);
+label("<math>O</math>", O, W);
+pair E;
+E=foot(B,A,C);
+draw(B--E);
+dot(E);
+label("<math>E</math>", E, S);
+draw(rightanglemark(B,A,D,20));
+draw(rightanglemark(B,E,C,20));
+[/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
+<cmath>\frac{BD}{BA} = \frac{BC}{BE},</cmath> or <cmath> \frac  {2R} c = \frac  ah.</cmath>
+However, remember that area <math>\triangle ABC = \frac {bh} 2</math>, so <math>h=\frac{2 \times \text{Area}}b</math>. Substituting this in gives us
+<cmath> \frac  {2R} c = \frac  a{\frac{2 \times \text{Area}}b},</cmath> and then bash through algebra.
 ==Formula for Circumradius==

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