# Difference between revisions of "Circumradius"

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. $R = \frac{abc}{4rs}$ 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 $A = rs$.
Let $\triangle ABC$ have circumradius $R$ and inradius $r$. Let $d$ be the distance between the circumcenter and the incenter. Then we have $$d^2=R(R-2r)$$