Circle in a quarter circle

Let the triangle inscribed in the unit quarter circle be $ABC$ , $\angle A = \theta$ , and the radius of the incircle be $r$ . Then

$\begin{aligned} r & = \frac As & \small \blue{\text{where }A \text{ and }s \text{ are the area and}} \\ & = \frac {\frac 12\cdot 1^2 \cdot \sin \theta}{\frac 12 \left(1+1+2\sin \frac \theta 2\right)} & \small \blue{\text{semiperimeter of }\triangle ABC.} \\ & = \frac {\sin \theta}{2+2\sin \frac \theta 2} & \small \blue{\text{To find }\max (r)} \\ \frac {dr}{d\theta} & = \frac {\cos \theta(2+2\sin \frac \theta 2)- \sin \theta \cos \frac \theta 2}{(2+2\sin \frac \theta 2)^2} & \small \blue{\text{Putting }\frac {dr}{d\theta} = 0} \\ 2\cos \theta \left(1 + \sin \frac \theta 2 \right) & = \sin \theta \cos \frac \theta 2 \\ 2\left(1-2\sin^2 \frac \theta 2 \right) \left(1 + \sin \frac \theta 2 \right) & = 2 \sin \frac \theta 2 \left(1 - \sin^2 \frac \theta 2 \right) \\ 1-2\sin^2 \frac \theta 2 & = \sin \frac \theta 2 \left(1 - \sin \frac \theta 2 \right) \\ \sin^2 \frac \theta 2 + \sin \frac \theta 2 - 1 & = 0 \\ \implies \sin \frac \theta 2 & = \frac {\sqrt 5 -1}2 = \varphi - 1 = \frac 1\varphi & \small \blue{\text{where }\varphi \text{ denotes the golden ratio.}} \\ \cos \frac \theta 2 & = \sqrt{1-\frac 1{\varphi^2}} = \sqrt{1-\frac {\varphi -1}\varphi} = \frac 1{\sqrt \varphi} \end{aligned}$

Therefore

$\begin{aligned} \max(r) & = \frac {\sin \theta}{2+2\sin \frac \theta 2} = \frac {\frac 2{\varphi\sqrt \varphi}}{2+2(\varphi -1)} = \left(\frac 1\varphi \right)^{5/2} = \boxed{\left(\frac {\sqrt 5-1}2\right)^{5/2}} \end{aligned}$

A triangle 'petal' of a square is the one with the largest possible area that can be inscribed in a larger quarter circle, but its incircle can't be the biggest because the petal itself is not optimising the convex area available in the quarter (see why traditionalists are called squares, they're mostly thinking in one direction of two dimensions for the petal area but not the incircle itself). A triangle 'petal' of a hexagon on the other hand is "too optimised" that we can fill in another circle the size of its incircle if we have the full 6 petals in all 4 quarters (in the middle of all 6 incircles), such that we can safely assume that this arrangement is more communal / social-centric and less individual. With a pentagonal petal as the middle ground, seeing the √5 is golden enough to choose the above option of ((√5 – 1)/2)^(5/2).

Since two sides of the triangle are the same as the quarter circle's radius of $1$ , it is an isosceles triangle.

If the triangle's vertex angle is $\theta$ , then its base is $2 \sin \frac{1}{2}\theta$ , so that its perimeter is $P = 2 + 2 \sin \frac{1}{2}\theta$ and its area is $A = \frac{1}{2} \cdot 1^2 \cdot \sin \theta = \sin \frac{1}{2}\theta \cos \frac{1}{2}\theta$ .

That means its inradius is $r = \cfrac{2A}{P} = \cfrac{2\sin \frac{1}{2}\theta \cos \frac{1}{2}\theta}{2 + 2 \sin \frac{1}{2}\theta} = \cfrac{\sin \frac{1}{2}\theta \cos \frac{1}{2}\theta}{1 + \sin \frac{1}{2}\theta}$ .

Letting $x = \sin \frac{1}{2} \theta$ , $\cos \frac{1}{2}\theta = \sqrt{1 - \sin^2 \frac{1}{2} \theta} = \sqrt{1 - x^2}$ and $r = \cfrac{x \sqrt{1 - x^2}}{1 + x}$ .

The largest radius $r$ will occur when $r' = -\cfrac{x^2 + x - 1}{(x + 1)\sqrt{1 - x^2}} = 0$ , which solves to $x = \cfrac{\sqrt{5} - 1}{2} = \phi - 1 = \cfrac{1}{\phi}$ for $x > 0$ .

This makes the largest radius $r = \cfrac{x \sqrt{1 - x^2}}{1 + x} = \sqrt{\cfrac{x^2(1 - x)}{1 + x}} = \sqrt{\cfrac{\frac{1}{\phi^2} \cdot \frac{1}{\phi^2}}{\phi}} = \sqrt{\cfrac{1}{\phi^5}} = \boxed{\bigg(\cfrac{\sqrt{5} - 1}{2}\bigg)^{5/2}}$ .