The golden ratio and a right triangle Assume the square of the hypotenuse of a right triangle is equal to its perimeter and one of its legs  is  $1$ plus its inradius(the radius inside the circle inscribed inside the triangle.) Find an expression for the hypotenuse  $c$  in terms of the golden ratio. 

Draw a picture. Let $r$ be the radius of the incircle, and let the legs be $r+x$ and $r+y$. We are told that one of the legs, say $r+x$, is equal to $r+1$, so $x=1$. Thus the hypotenuse is $1+y$. 
The condition that the square of the hypotenuse is equal to the perimeter says that 
$(1+y)^2=2r+2y+2$, which simplifes to $y^2=2r+1$.
The Pythagorean Theorem says that 
$$(1+y)^2=(1+r)^2+(r+y)^2,$$
which simplifies to 
$$y(1-r)=r^2+r.$$
Substitute $\sqrt{2r+1}$ for $y$, and square both sides. We get
$$(1-r)^2(2r+1)=(r^2+r)^2,$$
which miraculously simplifies to 
$$r^4+4r^2-1=0.$$
Solve. We get $r^2=\sqrt{5}-2$, and it's over.
