Cubic Equation for the Sides of a Triangle
Here's a popular fact that is hard to locate on the web:
Let, $a,\,b,\,c,\,p,\,r,\,R$ be the sides, the semiperimeter, the inradius, and the circumradius of a triangle. Then $a,\,b,\,c$ are the roots of the cubic equation
$x^{3}-2px^{2}+(p^{2}+r^{2}+4rR)x-4prR=0.$
Proof
The basic idea is to employ the relations between the roots and the coefficients of an equation, following Viète's theorem.
By definition, $2p=a+b+c$ which verifies the quadratic coefficient.
For the free coefficient, note that $S=pr,$ where $S$ is the area of the triangle, which also satisfies $abc=4SR.$
Thus we only have to show that $ab+bc+ca=p^{2}+r^{2}+4rR.$ I shall reduce that to a trigonometric identity. Let $\alpha,\,\beta,\,\gamma$ denote the angles of the triangle.
From the Law of Sines,
$ab+bc+ca=4R^{2}(\sin\alpha\sin\beta+\sin\beta\sin\gamma+\sin\gamma\sin\alpha ).$
Also,
$p=\displaystyle 4R\cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\gamma}{2}$
$r=\displaystyle 4R\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}$
It remains to show that $X=Y$ where
$\displaystyle X=\sin\alpha\sin\beta+\sin\beta\sin\gamma+\sin\gamma\sin\alpha$
and
$\displaystyle\begin{align} Y&=4(\cos^{2}\frac{\alpha}{2}\cos^{2}\frac{\beta}{2}\cos^{2}\frac{\gamma}{2}\\ &+\sin^{2}\frac{\alpha}{2}\sin^{2}\frac{\beta}{2}\sin^{2}\frac{\gamma}{2}\\ &+\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}). \end{align}$
We'll need two identities:
$\displaystyle\begin{align} \cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\gamma}{2}&=\sin\alpha+\sin\beta+\sin\gamma,\\ \sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}&=\cos\alpha+\cos\beta+\cos\gamma-1. \end{align}$
The substitution reduces $X=Y$ to
$\begin{align} \cos (\alpha +\beta )&+\cos (\beta +\gamma )+\cos (\gamma +\alpha )\\ &+\cos\alpha+\cos\beta+\cos\gamma=0 \end{align}$
which is obvious.
|Contact| |Front page| |Contents| |Algebra|
Copyright © 1996-2018 Alexander Bogomolny71536515