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Difference between revisions of "Circumradius"

(→‎Right triangles)
(gdr)
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<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 <math>[ABC] = \frac {bh} 2\implies h=\frac{2 \times [ABC]}b</math>. Substituting this in gives us
 
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 [ABC]}b},</cmath> and then bash through the algebra to get
+
<cmath> \frac  {2R} c = \frac  a{\frac{2 \times [ABC]}b},</cmath> and then simplifying to get
 
<cmath> R=\frac{abc}{4\times [ABC]}\text{ or }[ABC]=\frac{abc}{4R}</cmath>
 
<cmath> R=\frac{abc}{4\times [ABC]}\text{ or }[ABC]=\frac{abc}{4R}</cmath>
 
and we are done.
 
and we are done.
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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>.
 
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:
+
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>
 
<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==
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==Proof==
 
==Proof==
 +
 +
See https://www.cut-the-knot.org/triangle/EulerIO.shtml
  
 
== Right triangles ==
 
== Right triangles ==
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This results in a well-known theorem:
 
This results in a well-known theorem:
 
===Theorem===
 
===Theorem===
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The midpoint of the hypotenuse is equidistant from the vertices of the right triangle.
+
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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<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>
  
−
And this formula comes from the area of Heron and <math>R=\frac{abc}{4A}</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> which is just extension of law of sines
+
<math>2R=\frac{a}{\sin{A}}</math> by the [[Law of Sines]].
  
 
(Extended Law of Sines)
 
(Extended Law of Sines)

Revision as of 15:52, 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.

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 simplifying to get \[R=\frac{abc}{4\times [ABC]}\text{ or }[ABC]=\frac{abc}{4R}\] and we are done.

Formula for Circumradius

$R = \frac{abc}{4rs}$ 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:

$A=\sqrt{s(s-a)(s-b)(s-c)}$

Circumradius, bisector and altitude

R H L angles.png

Circumradius and altitude are isogonals with respect bisector and vertex of triangle.

Euler's Theorem for a Triangle

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

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]

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

$R=\frac{s}{\sqrt3}$

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

$R=\frac{abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}$


Which follows from the Heron's Formula and $R=\frac{abc}{4A}$.

If you know just one side and its opposite angle

$2R=\frac{a}{\sin{A}}$ by the Law of Sines.

(Extended Law of Sines)

See also

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