The circumradius of a cyclic polygon is the radius of the circumscribed 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 $R=\frac{abc}{4A}$. This can be rewritten as $A=\frac{abc}{4R}$.

## 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 $[ABC] = \frac {bh} 2\implies h=\frac{2 \times [ABC]}b$. Substituting this in gives us $$\frac {2R} c = \frac a{\frac{2 \times [ABC]}b},$$ and then bash through the algebra to get $$R=\frac{abc}{4\times [ABC]}\text{ or }[ABC]=\frac{abc}{4R}$$ and we are done. $R = \frac{abc}{4rs}$ 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 $A = rs$.

But, if you don't know the inradius, you can find the area of the triangle by Heron's Formula: $A=\sqrt{s(s-a)(s-b)(s-c)}$

## Euler's Theorem for a Triangle

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

## Right triangles

The hypotenuse of the triangle is the diameter of its circumcircle, and the circumcenter is its midpoint, so the circumradius is equal to half of the hypotenuse of the right triangle. $[asy] pair A,B,C,I; A=(0,0); B=(0,3); C=(4,0); draw(A--B--C--cycle); I=circumcenter(A,B,C); draw(I--A,gray); label("r",(I+A)/2,NW,gray); draw(circumcircle(A,B,C)); label("C",I,N); dot(I); draw(rightanglemark(B,A,C,10)); [/asy]$

This results in a well-known theorem:

### Theorem

The midpoint of the hypotenuse is equidistant from the vertices of the right triangle.

## Equilateral triangles $R=\frac{s}{\sqrt3}$

where $s$ is the length of a side of the triangle. $[asy] pair A,B,C,I; A=(0,0); B=(1,0); C=intersectionpoint(arc(A,1,0,90),arc(B,1,90,180)); draw(A--B--C--cycle); I=circumcenter(A,B,C); draw(circumcircle(A,B,C)); label("C",I,E); dot(I); label("s",A--B,S); label("s",A--C,N); label("s",B--C,N); [/asy]$

## If all three sides are known $R=\frac{abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}$

And this formula comes from the area of Heron and $R=\frac{abc}{4A}$.

## If you know just one side and its opposite angle $2R=\frac{a}{\sin{A}}$ which is just extension of the law of sines.

(Extended Law of Sines)