Suppose $a$, $b$, $c$ are sides of a triangle and $$\begin{align} x^2-ax+b^2+ac-y&=0 \\ ax+bc-y&=0 \end{align}$$ has one solution. Classify the triangle.
Then $x^2-2ax+b^2+ac-bc=0$ has one solution. $$\Delta_x=4a^2-4b^2-4ac+4bc=0$$ $$a^2=b^2+(a-b)c$$ Since $b+c>a$, $$a^2<b^2+c^2$$ I think this means the triangle is acute. However, we still need to prove that $a$ is the longest side. How would you prove that?
Can anyone help, please?