Circumradius
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.
Contents
Formula for a Triangle
Let 
and 
denote the triangle's three sides and let 
denote the area of the triangle. Then, the measure of the circumradius of the triangle is simply 
. This can be rewritten as 
.
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]](http://latex.artofproblemsolving.com/3/a/d/3ad58f915ab8e62f7ffd12a4d01aa92205a280f7.png)
We let 
, 
, 
, 
, and 
. We know that 
is a right angle because 
is the diameter. Also, 
because they both subtend arc 
. Therefore, 
by AA similarity, so we have
![\[\frac{BD}{BA} = \frac{BC}{BE},\]](http://latex.artofproblemsolving.com/c/e/e/ceef2bbb1ea4d0e9e833a5e37cf786c21b9d465f.png)
or ![\[\frac {2R} c = \frac ah.\]](http://latex.artofproblemsolving.com/d/6/c/d6cbf0ed51fd35d263b5f200b1fde96b578ae5e7.png)
However, remember that ![$[ABC] = \frac {bh} 2\implies h=\frac{2 \times [ABC]}b$](http://latex.artofproblemsolving.com/b/8/1/b81625da59243767b9673379f8f483c99c5dbaf8.png)
. Substituting this in gives us
![\[\frac {2R} c = \frac a{\frac{2 \times [ABC]}b},\]](http://latex.artofproblemsolving.com/a/1/8/a182e4c2fdc633771a79bdcde0256d738ce7496b.png)
and then simplifying to get
![\[R=\frac{abc}{4\times [ABC]}\text{ or }[ABC]=\frac{abc}{4R}\]](http://latex.artofproblemsolving.com/b/d/0/bd016117c7013997e7379b59234c4176f9ddf383.png)
and we are done.
Formula for Circumradius
Where 
is the circumradius, 
is the inradius, and 
, 
, and are the respective sides of the triangle and 
is the semiperimeter. Note that this is similar to the previously mentioned formula; the reason being that 
.
But, if you don't know the inradius, you can find the area of the triangle by Heron’s Formula:
Circumradius, bisector and altitude
Circumradius and altitude are isogonals with respect bisector and vertex of triangle.
Euler's Theorem for a Triangle
Let 
have circumcenter 
and incenter 
.Then ![\[OI^2=R(R-2r) \implies R \geq 2r\]](http://latex.artofproblemsolving.com/d/3/0/d30d2cc82daeaaf90572fb9069456896a1310b78.png)
Proof
See https://www.cut-the-knot.org/triangle/EulerIO.shtml
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]](http://latex.artofproblemsolving.com/7/f/a/7fa53fd535608055e8b16af17f60b76d0766a768.png)
This results in a well-known theorem:
Theorem
The midpoint of the hypotenuse is equidistant from the vertices of the right triangle.
Equilateral triangles
where 
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]](http://latex.artofproblemsolving.com/c/a/1/ca10ddc9a1840bd3a175658125dfc16d353a3b39.png)
If all three sides are known
And this formula comes from the area of Heron and .
If you know just one side and its opposite angle
which is just extension of the law of sines.
(Extended Law of Sines)
