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.
[Math] The golden ratio and a right triangle
algebra-precalculusgeometrytrigonometry
Related Solutions
Here is a diagram of the situation.
Let us calculate the relevant lengths, assuming that the radius of the circle is $1$ so that we don't have to deal with scaling. One should note that $\angle CAB$ is $\pi/6$, equal to $\angle CDE$. Thus, the length of $AC$ is $\frac{1}{\cos(\pi/6)}$. The length of $CE$ and $EF$ separately is $\tan(\pi/6)$. Noting that $\cos(\pi/6)$ is $\sqrt{3}/2$ and $\tan(\pi/6)=\frac{1}{\sqrt{3}}$, we find that the length from $A$ to $F$ is $\frac{4}{\sqrt{3}}$. The length from $F$ to $G$ is $2-\frac{1}{\sqrt{3}}$. The ratio of these $$\frac{\frac{4}{\sqrt{3}}}{2-\frac{1}{\sqrt{3}}}=\frac{4}{2\sqrt{3}-1}$$ which is not the golden ratio, but is $1.623$ instead.
By the "chord-chord" aspect of the Power of a Point Theorem, we have $$|\overline{AX}||\overline{XB}| = |\overline{CX}||\overline{XD}|\quad\to\quad p\cdot p = (p+q)\cdot q \quad\to\quad\frac{p}{q} = \frac{p+q}{p} = \frac{|\overline{CX}|}{|\overline{XE}|}$$
The relation between $p$ and $q$ says exactly that $\frac{p}{q}$ is the golden ratio, $1.618\dots$. $\square$
I can't say that I've seen this construction before, but I don't exactly have a catalog of such things in my head. The power-of-a-point connection seems (almost) "obvious" in retrospect, so it likely has been observed before. Still, it's pretty neat. (It's my favorite of your constructions so far.)
Best Answer
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.