# Is there a right triangle with angles $A$, $B$, $C$ such that $A^2+B^2=C^2$?

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 $$A,B,C,a,b,c$$?

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.

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.