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|Heron’s Formula]]: | ||
+ | |||
+ | <math>A=\sqrt{s(s-a)(s-b)(s-c)}</math> | ||
+ | ==Circumradius, bisector and altitude== | ||
+ | [[File:R H L angles.png|400px|right]] | ||
+ | Circumradius and altitude are isogonals with respect bisector and vertex of triangle. | ||
==Euler's Theorem for a Triangle== | ==Euler's Theorem for a Triangle== | ||
Line 49: | Line 53: | ||
==Proof== | ==Proof== | ||
+ | |||
+ | See https://www.cut-the-knot.org/triangle/EulerIO.shtml | ||
== Right triangles == | == Right triangles == | ||
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draw(A--B--C--cycle); | draw(A--B--C--cycle); | ||
I=circumcenter(A,B,C); | I=circumcenter(A,B,C); | ||
+ | draw(I--A,gray); | ||
+ | label("$r$",(I+A)/2,NW,gray); | ||
draw(circumcircle(A,B,C)); | draw(circumcircle(A,B,C)); | ||
− | label("$ | + | label("$C$",I,N); |
dot(I); | dot(I); | ||
+ | draw(rightanglemark(B,A,C,10)); | ||
</asy> | </asy> | ||
+ | |||
+ | This results in a well-known theorem: | ||
+ | ===Theorem=== | ||
+ | The midpoint of the hypotenuse is equidistant from the vertices of the right [https://artofproblemsolving.com/wiki/index.php/TOTO_SLOT_:_SITUS_TOTO_SLOT_MAXWIN_TERBAIK_DAN_TERPERCAYA TOTO SLOT] triangle. | ||
== Equilateral triangles == | == Equilateral triangles == | ||
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C=intersectionpoint(arc(A,1,0,90),arc(B,1,90,180)); | C=intersectionpoint(arc(A,1,0,90),arc(B,1,90,180)); | ||
draw(A--B--C--cycle); | draw(A--B--C--cycle); | ||
− | I= | + | I=circumcenter(A,B,C); |
− | draw( | + | draw(circumcircle(A,B,C)); |
− | label("$ | + | label("$C$",I,E); |
+ | dot(I); | ||
+ | label("$s$",A--B,S); | ||
+ | label("$s$",A--C,N); | ||
+ | label("$s$",B--C,N); | ||
</asy> | </asy> | ||
Line 86: | Line 103: | ||
<math>R=\frac{abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}</math> | <math>R=\frac{abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}</math> | ||
− | + | ||
+ | Which follows from the Heron's Formula and <math>R=\frac{abc}{4A}</math>. | ||
== If you know just one side and its opposite angle == | == If you know just one side and its opposite angle == | ||
− | <math>2R=\frac{a}{\sin{A}}</math> | + | <math>2R=\frac{a}{\sin{A}}</math> by the [[Law of Sines]]. |
+ | |||
+ | (Extended Law of Sines) | ||
==See also== | ==See also== |
Latest revision as of 16:52, 19 February 2024
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:
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
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.
This results in a well-known theorem:
Theorem
The midpoint of the hypotenuse is equidistant from the vertices of the right TOTO SLOT triangle.
Equilateral triangles
where is the length of a side of the triangle.
If all three sides are known
Which follows from the Heron's Formula and .
If you know just one side and its opposite angle
by the Law of Sines.
(Extended Law of Sines)