Prove $\Delta \text{ABC}$ is a right triangle with $\left | \text{k} \right |\leqq 3$

Question

Prove $\Delta \text{ABC}$ is a right triangle if: $$\left\{\begin{matrix} \sin^{2}\,\text{A}+ \sin^{2}\,\text{B}= \sin \text{C}\\ \max\left \{ \measuredangle \text{A}- \text{k}\,\measuredangle \text{B},\,\measuredangle \text{B}- \text{k}\,\measuredangle \text{A} \right \}\leqq \frac{\pi }{2}\\ \left | \text{k} \right |\leqq 3 \end{matrix}\right.$$ I have a proof for my problem with $\text{k}= 0$. See here:

$\lceil$ https://diendantoanhoc.net/topic/185093-measuredangle-textc-pi/#entry716757 $\rfloor$

I also have another solution but ugly, I try to use similar method with $\left | \text{k} \right |\leqq 3$ but without success !

Answer 1

One idea: Using the formulas $$A=\frac{1}{2}ab\sin(\gamma)$$ etc we get

$$\frac{4A^2}{b^2c^2}+\frac{4A^2}{a^2c^2}=\frac{2A}{ab}$$ and using $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ where $$s=\frac{a+b+c}{2}$$ plugging this in our formula we get after squaring and simplifying

$$- \left( {a}^{6}-{a}^{4}{b}^{2}-{a}^{4}{c}^{2}-{a}^{2}{b}^{4}-6\,{a}^{ 2}{b}^{2}{c}^{2}+{b}^{6}-{b}^{4}{c}^{2} \right) \left( {a}^{2}+{b}^{2 }-{c}^{2} \right) =0$$ it remaines to show that the first factor can not be zero.

Answer 2

With ${\rm k}= 0$ and $t= constant$, for $\sin {\rm A}= \dfrac{2\,ta(\!ta+ 1\!)}{2\,ta(\!ta+ 1\!)+ 1},\,\sin {\rm B}= \dfrac{2\,tb+ 1}{2\,t^{2}b^{2}+ (\!2\,tb+ 1\!)}$, we have : $$\sin^{\,2}\,{\rm A}+ \sin^{\,2}\,{\rm B}- 1= \frac{4\,t^{\,2}(\,a- b\,)(\,ta+ tb+ 1\,)(\,2\,tab+ a+ b\,)(\,2\,t^{2}ab+ ta+ tb+ 1\,)}{(\,2\,t^{\,2}a^{\,2}+ 2\,ta+ 1\,)(\,2\,t^{\,2}b^{\,2}+ 2\,tb+ 1\,)}= 0\,\,\,\,\,.$$ WLOG, suppose that $a> 0,\,b> 0,\,t> 0\,\therefore\,a= b\,\therefore\,\sin^{2} {\rm A}+ \sin^{2} {\rm B}= 1\,\therefore\,\measuredangle {\rm C}= \dfrac{\pi}{2}$ / $\lceil$ q.e.d! $\rfloor$