# Difference between revisions of "Circumradius"

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− | + | 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 [[circumscribe]]s the triangle. Since every triangle is [[cyclic]], every triangle has a circumscribed circle, or a [[circumcircle]]. | |

− | |||

− | The '''circumradius''' of a [[cyclic]] [[polygon]] is the radius of the | ||

==Formula for a Triangle== | ==Formula for a Triangle== | ||

− | Let <math>a, b</math> and <math>c</math> denote the triangle's three sides | + | 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 circumradius of the triangle is simply <math>R=\frac{abc}{4A}</math>. This can be rewritten as <math>A=\frac{abc}{4R}</math>. |

== Proof == | == Proof == | ||

Line 31: | Line 29: | ||

draw(rightanglemark(B,E,C,20)); | draw(rightanglemark(B,E,C,20)); | ||

</asy> | </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 | 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> | <cmath>\frac{BD}{BA} = \frac{BC}{BE},</cmath> or <cmath> \frac {2R} c = \frac ah.</cmath> | ||

− | However, remember that | + | However, remember that <math>[ABC] = \frac {bh} 2\implies h=\frac{2 \times [ABC]}b</math>. Substituting this in gives us |

− | <cmath> \frac {2R} c = \frac a{\frac{2 \times | + | <cmath> \frac {2R} c = \frac a{\frac{2 \times [ABC]}b},</cmath> and then simplifying to get |

− | <cmath> R=\frac{abc}{4\times \text{ | + | <cmath> R=\frac{abc}{4\times [ABC]}\text{ or }[ABC]=\frac{abc}{4R}</cmath> |

and we are done. | and we are done. | ||

− | |||

− | |||

==Formula for Circumradius== | ==Formula for Circumradius== | ||

<math>R = \frac{abc}{4rs}</math> | <math>R = \frac{abc}{4rs}</math> | ||

− | Where <math>R</math> is the | + | Where <math>R</math> is the circumradius, <math>r</math> is the inradius, and <math>a</math>, <math>b</math>, and <math>c</math> are the respective sides of the triangle and <math>s = (a+b+c)/2</math> is the semiperimeter. Note that this is similar to the previously mentioned formula; the reason being that <math>A = rs</math>. |

+ | |||

+ | But, if you don't know the inradius, you can find the area of the triangle by Heron's Formula: | ||

+ | |||

+ | <math>A=\sqrt{s(s-a)(s-b)(s-c)}</math> | ||

==Euler's Theorem for a Triangle== | ==Euler's Theorem for a Triangle== | ||

− | Let <math>\triangle ABC</math> have | + | Let <math>\triangle ABC</math> have circumcenter <math>O</math> and incenter <math>I</math>.Then <cmath>OI^2=R(R-2r) \implies R \geq 2r</cmath> |

+ | |||

+ | ==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(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 == | ||

+ | |||

+ | <math>R=\frac{s}{\sqrt3}</math> | ||

+ | |||

+ | where <math>s</math> 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 == | ||

+ | |||

+ | <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>. | ||

+ | |||

+ | == If you know just one side and its opposite angle == | ||

+ | |||

+ | <math>2R=\frac{a}{\sin{A}}</math> which is just extension of the [[law of sines]]. | ||

+ | |||

+ | (Extended Law of Sines) | ||

==See also== | ==See also== |

## Latest revision as of 17:55, 20 July 2020

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

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
or
However, remember that . Substituting this in gives us
and then simplifying to get
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:

## Euler's Theorem for a Triangle

Let have circumcenter and incenter .Then

## 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.

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.

## 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)