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A right angle triangle with vertices $A,B,C$ ($C$ is the right angle), and the sides opposite to the vertices are $a,b,c$, respectively.

We know that this triangle (and any right angle triangle) has the following properties:

  • $a^2+b^2=c^2$

  • $a+b>c$

  • $a+c>b$

  • $b+c>a$

  • $A+B+C=\pi$

Can we add the property that $A^2+B^2=C^2$ such that this triangle can be formed? If yes, how to find an example for such triangle, finding $A,B,C,a,b,c$?

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Because it's a right triangle, $C=\frac{\pi}{2}=A+B$. That means that $$C^2=(A+B)^2=A^2+B^2+2AB > A^2+B^2$$ So no, it isn't possible. We can come close as we approach angles of $0,90^\circ,90^\circ$, but that's degenerate - it's not a triangle.

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We have $C=\pi/2,B=\pi/2-A$, so we need to solve the quadratic equation$$A^2+(\pi/2-A)^2=\pi^2/4\\\implies2A^2=\pi A$$giving $A=0,\pi/2$ which is not possible.

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