Difference between revisions of "Circumradius"
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| + | == Equilateral triangles == | ||
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| + | <math>R=\frac{s}{\sqrt3}</math> | ||
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| + | where <math>s</math> is the length of a side of the triangle. | ||
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| + | == If all three sides are known == | ||
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| + | <math>R=\frac{abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}</math> | ||
| + | |||
| + | And this formula comes from the area of Heron and <math>R=\frac{abc}{4A}</math>. | ||
==See also== | ==See also== | ||
Revision as of 18:48, 1 August 2017
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.
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 
. Also, 
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 area 
, so 
. Substituting this in gives us
![\[\frac {2R} c = \frac a{\frac{2 \times \text{Area}}b},\]](http://latex.artofproblemsolving.com/a/8/1/a81bf57ae56e5402c1f795781c7bd7db5e069c86.png)
and then bash through algebra to get
![\[R=\frac{abc}{4\times \text{Area}},\]](http://latex.artofproblemsolving.com/a/5/2/a52b9d32b1179fdcbd9cf95a599740f468ad20cf.png)
and we are done.
--Nosaj 19:39, 7 December 2014 (EST)
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 
.
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
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(circumcircle(A,B,C)); label("$I$",I,N); dot(I); [/asy]](http://latex.artofproblemsolving.com/4/2/9/4291500c2d43f6c03d70d1fe39675b49215155e3.png)
Equilateral triangles
where 
is the length of a side of the triangle.
If all three sides are known
And this formula comes from the area of Heron and .
