Chứng minh $\Delta \text{ABC}$ là tam giác vuông nếu:
$$\left\{\begin{matrix} \sin^{2}\,\text{A}+ \sin^{2}\,\text{B}= \sin \text{C}\\ \max\left \{ \measuredangle \text{A}- 3\,\measuredangle \text{B},\,\measuredangle \text{B}- 3\,\measuredangle \text{A} \right \}\leqq \frac{\pi }{2} \end{matrix}\right.$$
(Đề xuất! ) Bài toán tổng quát: $\Delta \text{ABC}$ là tam giác vuông nếu:
$$\left\{\begin{matrix} \sin^{2}\,\text{A}+ \sin^{2}\,\text{B}= \sin \text{C}\\ \max\left \{ \measuredangle \text{A}- \text{k}\,\measuredangle \text{B},\,\measuredangle \text{B}- \text{k}\,\measuredangle \text{A} \right \}\leqq \frac{\pi }{2}\\ \left | \text{k} \right |\leqq 3 \end{matrix}\right.$$
[Trường hợp $\text{k}\,=\,0$]:
Đặt: $\sin \text{A}= \frac{2\,a\left ( a+ 1 \right )}{2\,a\left ( a+ 1 \right )+ 1},\,\sin \text{B}= \frac{2\,b+ 1}{2\,b^{2}+ 2\,b+ 1}\,\,\left ( a,\,b\,\geqq \,0 \right )$
$$0= \sin^{2}\text{A}+ \sin^{2}\text{B}- \sin\text{C}= \frac{4\,\left ( a- b \right )\,\left ( a+ b+ 1 \right )\,\left ( 2\,ab+ a+ b \right )\,\left ( 2\,ab+ a+ b+ 1 \right )}{\left ( \left ( 2\,a^{2}+ 2\,a+ 1 \right )\,\left ( 2\,b^{2}+ 2\,b+ 1 \right ) \right )^{2}}$$
$$\Leftrightarrow a\,=\, b\,\Leftrightarrow \, \sin^{2}\text{A}+ \sin^{2}\text{B}= 1\,\Leftrightarrow \measuredangle \text{C}= \frac{\pi}{2}$$