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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.

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 of the circumradius of the triangle is simply $R=\frac{abc}{4A}$. Also, $A=\frac{abc}{4R}$


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 area $\triangle ABC = \frac {bh} 2$, so $h=\frac{2 \times \text{Area}}b$. Substituting this in gives us \[\frac  {2R} c = \frac  a{\frac{2 \times \text{Area}}b},\] and then bash through algebra.

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. Note that this is similar to the previously mentioned formula; the reason being that $A = rs$.

Euler's Theorem for a Triangle

Let $\triangle ABC$ have circumradius $R$ and inradius $r$. Let $d$ be the distance between the circumcenter and the incenter. Then we have \[d^2=R(R-2r)\]

See also

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