How Do Triangle Sides Relate to Acute Angles?

[anemone]

Here is the POTW for the week 52, 2019:

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The sides $a,\,b,\,c$ and $u,\,v,\,w$ of two triangles $ABC$ and $UVW$ are related by the equations

$u(v+w-u)=a^2\\v(w+u-v)=b^2\\w(u+v-w)=c^2\\$

Prove that $ABC$ are acute, and express the angles $U,\,V$ and $W$ in terms of $A,\,B$ and $C$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!

 

[anemone]

Congratulations to MegaMoh for his correct solution!(Cool)

Below is a suggested solution:

Spoiler

Note that $a^2+b^2-c^2=w^2-u^2-v^2+2uv=(w+u-v)(w-u+v)>0$

By the triangle inequality, $\cos C >0$.

By this reasoning, all of the angles of triangle $ABC$ are acute. Moreover,

$\begin{align*} \cos C &=\dfrac{a^2+b^2-c^2}{2ab}\\&=\sqrt{\frac{(w+u-v)(w-u+v)}{4uv}}\\&=\sqrt{\dfrac{w^2-u^2-v^2+2uv}{4uv}}\\&=\dfrac{1}{\sqrt{2}}\sqrt{1-\cos U}\end{align*} $

from which we deduce

$\cos U = 1-2\cos^2 A = \cos (\pi -2A)$

Therefore,

$U=\pi-2A$

Similarly,

$V=\pi-2B$ and $W=\pi-2C$