One Triangle from Another

Proof 1

By the Law of Sines,

\begin{align} u&=2R\alpha = a^2+b^2+c^2,\\ v&=2R\beta = ab+bc+ca,\\ w&=2R\gamma = ac+ba+cb. \end{align}

Since obviously $u+v\gt w\;$ and $u+w\gt v,\;$ suffice it to show that $v+w\gt u,\;$ i.e.,

$2(ab+bc+ca)-(a^2+b^2+c^2)\gt 0.$

But the latter is equivalent to

$(\sqrt{a}+\sqrt{b}+\sqrt{c})(-\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c})\gt 0,$

which is true because, since $a,b,c\;$ form a triangle, so do $\sqrt{a},\sqrt{b},\sqrt{c}.\;$

Proof 2

From the Law of Sines,

$\displaystyle \alpha=a\sin A+\frac{b^2}{a}\sin A+\frac{c^2}{a}\sin A,$

i.e.,

$\displaystyle \frac{\alpha}{\sin A}=\frac{a^2+b^2+c^2}{a}.$

Similarly, $\displaystyle\frac{\beta}{\sin B}=\frac{a^2+b^2+c^2}{b}$ and $\displaystyle\frac{\gamma}{\sin C}=\frac{a^2+b^2+c^2}{c}.\;$ Written differently,

$\displaystyle a^2+b^2+c^2=\frac{a\alpha}{\sin A}=\frac{b\beta}{\sin B}=\frac{c\gamma}{\sin C},$

implying $\alpha=\beta=\gamma,\;$ as the sides of an equilateral triangle.

Proof 3

Lemma (a "triangularity" property)

Let $m,n,p\ge 0\;$ (not all zero). If $a,\;$ $b,\;$ $c\;$ are the side lengths of a triangle, then

\begin{align} \alpha &= ma+nb+pc,\\ \beta &= na+pb+mc,\\ \gamma &= pa+mb+nc \end{align}

are aslo the side lengths of a triangle.

Proof of Lemma

We prove that $\alpha+\beta\gt\gamma,\;$ i.e.,