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. 
 A: 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.
A: For any triangle,
$$
2\times\text{area} = \text{inradius}\times\text{perimeter}
$$
For a right triangle,
$$
2\times\text{area} = ab
$$
We are given
$$
\text{inradius}=a-1\quad\text{and}\quad\text{perimeter}=c^2
$$
Therefore,
$$
\begin{align}
\overbrace{(a-1)}^{\text{inradius}}\overbrace{(a+b+c)}^{\text{perimeter}}&=ab\tag{1}\\
(a-1)(a+c)&=b\tag{2}
\end{align}
$$
and
$$
\overbrace{a+b+c}^{\text{perimeter}}=c^2\tag{3}
$$
Combining $(2)$ and $(3)$ to eliminate $b$ yields
$$
\begin{align}
(a-1)(a+c)&=c^2-c-a\tag{4}\\
a(a+c)&=c^2\tag{5}\\
\left(\frac ac\right)^2+\frac ac&=1\tag{6}\\
\frac ac&=\frac1\phi\tag{7}
\end{align}
$$
Equation $(5)$ and $a^2+b^2=c^2$ gives
$$
\begin{align}
b^2&=ac\tag{8}\\
\left(\frac bc\right)^2&=\frac ac\tag{9}\\
\frac bc&=\frac1{\sqrt\phi}\tag{10}
\end{align}
$$
Dividing equation $(3)$ by $c$ yields
$$
\begin{align}
c
&=\frac ac+\frac bc+1\tag{11}\\
&=\frac1\phi+\frac1{\sqrt\phi}+1\tag{12}
\end{align}
$$
