A golden incircle

Let the apex angle be $\theta =2x$ , $0<x<\dfrac{\pi }{2}$ . Referring to the figure, triangles $\triangle AIN$ and $\triangle ACM$ are similar, hence $\frac{IN}{MC}=\frac{AI}{AC}\Rightarrow \frac{IN}{MC}=\frac{AM-IM}{AC}\Rightarrow \frac{r}{a\cdot \sin x}=\frac{a\cdot \sin \left( \frac{\pi }{2}-x \right)-r}{a} \ \ \ \ \ (1)$ Without loss of generality, we take $a=1$ , so from $(1)$ we get: $\begin{aligned} \frac{r}{\sin x}=\sin \left( \frac{\pi }{2}-x \right)-r & \Leftrightarrow r=\sin x\cdot \cos x-r\sin x \\ & \Leftrightarrow r=\frac{1}{2}\cdot \sin \left( 2x \right)-r\sin x \\ & \Leftrightarrow r=\frac{1}{2}\cdot \frac{\sin \left( 2x \right)}{1+\sin x} \\ \end{aligned}$ To maximise $r$ we set $\begin{aligned} \dfrac{dr}{dx}=0 & \Leftrightarrow {{\left[ \dfrac{1}{2}\cdot \dfrac{\sin \left( 2x \right)}{1+\sin x} \right]}^{\prime }}=0 \\ & \Leftrightarrow \dfrac{1}{2}\cdot \dfrac{2\cos \left( 2x \right)\cdot \left( 1+\sin x \right)-\sin \left( 2x \right)\cos x}{{{\left( 1+\sin x \right)}^{2}}}=0 \\ & \Leftrightarrow 2\cos \left( 2x \right)\cdot \left( 1+\sin x \right)-\sin \left( 2x \right)\cos x=0 \\ & \Leftrightarrow 2\left( 1-2{{\sin }^{2}}x \right)\cdot \left( 1+\sin x \right)-2\sin x{{\cos }^{2}}x=0 \\ & \Leftrightarrow {{\sin }^{3}}x+2\cdot {{\sin }^{2}}x-1=0 \\ & \Leftrightarrow \left( \sin x+1 \right)\cdot \left( {{\sin }^{2}}x+\sin x-1 \right)=0 \\ & \overset{0<x<\frac{\pi }{2}}{\mathop{\Leftrightarrow }}\,{{\sin }^{2}}x+\sin x-1=0 \\ & \Leftrightarrow \sin x=\dfrac{\sqrt{5}-1}{2} \\ \end{aligned}$ The function $r\left( x \right)=\dfrac{1}{2}\cdot \dfrac{\sin \left( 2x \right)}{1+\sin x}$ is concave down, since ${r}''\left( x \right)=-\dfrac{\cos x\cdot \left( 3{{\sin }^{2}}x+2 \right)}{{{\left( 1+\sin x \right)}^{3}}}<0$ , for $0<x<\dfrac{\pi }{2}$ , hence, at the zero of $\dfrac{dr}{dx}$ , $r$ gets maximised. This means that the incircle is the largest, when the apex angle is $\theta =2x=2\cdot {{\sin }^{-1}}\left( \dfrac{\sqrt{5}-1}{2} \right).$ For the answer, $p=2$ , $q=5$ , $r=-1$ , $s=2$ , hence $p+q+r+s=\boxed{8}$ .

Image from Fletcher Mattox

Let the apex angle and radius of the incircle be $\theta$ and $r$ respectively. Then the area of the isosceles triangle is given by $A_\triangle = sr$ , where $s = \dfrac {2a+2a\sin \frac \theta 2}2$ is the semiperimeter of the triangle. And the area of triangle is also given by $A_\triangle = \dfrac {a^2\sin \theta}2$ . Therefore $\dfrac {2a\left(1+\sin \frac \theta 2\right)}2r = \dfrac {a^2\sin \theta}2 \implies r = \dfrac {a\sin \theta}{2\left(1+\sin \frac \theta 2\right)}$ . Since the larger the $r$ the larger the incircle, the largest incircle occurs when $\dfrac {\sin \theta}{1+\sin \frac \theta 2}$ is maximum. Then $\dfrac d{dx}\left(\dfrac {2r}a\right) = \dfrac {\cos \theta \left(1+\sin \frac \theta 2\right) - \sin \theta \left(\frac 12 \cos \frac \theta 2\right)}{\left(1+\sin \frac \theta 2\right)^2}$ . Equating $\dfrac d{dx}\left(\dfrac {2r}a\right)$ to zero, we have:

$\begin{aligned} \cos \theta \left(1+\sin \frac \theta 2\right) & = \sin \theta \left(\frac 12 \cos \frac \theta 2\right) \\ \left(1-2\sin^2 \frac \theta 2\right)\left(1+\sin \frac \theta 2\right) & = \sin \frac \theta 2 \cos^2 \frac \theta 2 \\ 1 + \sin \frac \theta 2 -2\sin^2 \frac \theta 2 - 2\sin^3 \frac \theta 2 & = \sin \frac \theta 2 \left(1 - \sin^2 \frac \theta 2 \right) \\ 1 -2\sin^2 \frac \theta 2 - \sin^3 \frac \theta 2 & = 0 \\ \sin^3 \frac \theta 2 + 2 \sin^2 \frac \theta 2 - 1 & = 0 \\ \left(\sin \frac \theta 2 + 1 \right) \left(\sin^2 \frac \theta 2 + \sin \frac \theta 2 - 1\right) & = 0 \end{aligned}$

$\implies \begin{cases} \begin{array} {lll} \sin \dfrac \theta 2 = - 1 & \implies \dfrac \theta 2 = \dfrac 32 \pi & \red{\text{Not valid, since }\theta < \pi} \\ \sin^2 \dfrac \theta 2 + \sin \dfrac \theta 2 - 1 = 0 & \implies \sin \dfrac \theta 2 = \dfrac {\sqrt 5 -1}2 & \blue{\implies \theta = 2\sin^{-1}\left(\dfrac {\sqrt 5 -1}2\right)} \end{array} \end{cases}$

Therefore, $p+q+r+s = 2+5-1+2 = \boxed 8$ .

We want to express the inradius, $r$ , as a function of the apex angle, $\theta$ . First find the area of the triangle.

$A = \dfrac{{b}\cdot{h}}{2} = \dfrac{1}{2}\cdot{2a} \cdot \sin\dfrac{\theta}{2} \cdot{a}\cos{\dfrac{\theta}{2}} = \dfrac{1}{2}{a}^2\sin\theta$

And the perimeter, $P$ of the triangle. $P = 2\cdot{a}+{b}$

The inradius can be expressed as $r = \dfrac{2\cdot{A}}{P} =\dfrac{a^2\sin{\theta}}{2a+2a\cdot\sin{\frac{\theta}{2}}} = \dfrac{a\cdot\sin{\theta}}{2(1+\sin{\frac{\theta}{2} }) }$

Now find extrema if any exist: $\displaystyle \hspace{.5cm}\frac{dr}{d\theta} = \frac{2\cdot(\sin{\frac{\theta}{2}}+1)\cdot\cos{\theta} - \sin{\theta}\cdot\cos{\frac{\theta}{2}} } {4\cdot(\sin{\frac{\theta}{2}}+1)^2} = 0$

Since the denominator is always positive, $\hspace{.2cm} 2\cos{\theta} \left(\sin{\frac{\theta}{2}} + 1\right) - \sin(\theta)\cos(\frac{\theta}{2}) = 0$

Let $\theta = 2x \implies \hspace{.5cm} 2\cos{2x}\cdot(\sin{x}+1) - \sin{2x}\cdot\cos{x} = 0$

With some trigonometry, this can be reduced to a cubic in $\sin x$ :

$\sin^3x + 2\sin^2x - 1 = (\sin{x}+1) \cdot (\sin^2{x} - \sin{x}+1) = 0$

Taking each factor: $\sin{x} + 1 = 0 \implies \theta = \pi$ , which can be ignored.

The quadratic factor, however, can be solved: $\sin{x} = \dfrac{\sqrt{5} \pm 1}{2}$

Only the negative surd leads to a real root: $\theta = 2x = 2\cdot\sin^{-1} (\dfrac{\sqrt{5}-1}{2}) \approx 76.34^{\circ}$

So the required response $p+q+r+s$ , is $2+5-1+2=\boxed{8}$