The formula for the circumradius $r$ of a triangle $ABC$ tells me that $r={abc\over{}4\triangle}$, where the lengths of the sides are $a$, $b$, $c$.
I'm not sure, but I occasionaly got wrong values. They might have been calculation mistakes, but then I got fixated on deriving the formula for myself.
So, I took a triangle $ABC$, circumcenter $R$.
I know that:
- $RA=RB=RC$
- Altitudes from $R$ to sides bisect those sides.
Side $AB$ was divided into two parts each of length $c$, $BC$ into two of $a$, and $AC$ into two of $b$. Let the altitudes to $BC$ be of length $h_1$, to $AC$ be of length $h_2$, and to $AB$ of length $h_3$.
So, using the Pythagorean Theorem,
$\begin{align}r^2&=a^2+h_1^2\\r^2&=b^2+h_2^2\\r^2&=c^2+h_3^2\end{align}$
Since I know $a,b,c$, I have four variables, namely $r,h_1,h_2,h_3$. But since I have only three equations, I am unable to solve. I have tried many times, yet I cannot find any other relations. So my primary problem is to find the fourth equation.
Please help.
Best Answer
We can get this formula by the following way.
Let $\angle C$ be an acute angle, $BD$ be an altitude of $\Delta ABC$ and $BE$ be a diameter of the circumcircle.
Thus, $\Delta ABE\sim\Delta DBC$ and $$\frac{c}{h_b}=\frac{2r}{a},$$ which gives $$r=\frac{ac}{2h_b}=\frac{ac}{2\cdot\frac{2\Delta}{b}}=\frac{abc}{4\Delta}.$$