Radius of circle, restricted to Golden Rectangle, with area of half the Golden Ratio

Question

• Let $R$ be the Golden Rectangle (sides of length 1 and $\phi =\frac{1+\sqrt5}{2}$)
• Centered at one of the vertices of $R$, inscribe a circle of radius $\lambda$
• What is the value of $\lambda$ that partitions $R$ into 2 equal halves by area?

• I was able to determine a function for the convex region, parameterized by $\lambda$; $C(\lambda)$ and from there, use software to find a meaningful root of $C(\lambda)-\frac{\phi}{2}$ ($\approx1.016144)$. However, I am disappointed I couldn't find an analytical solution.
• I am looking for an expression for $\lambda$

$Aside:$ I was somewhat expecting $\lambda$ to be more "interesting"

Answer 1

Parameterizing by $\theta$ as in the figure ...

... the goal is to solve ...

$$\frac12\cdot 1 \cdot \phi = X + Y = \frac12 \cdot 1 \cdot \tan\theta + \frac12 \cdot \sec^2\theta \cdot \left(\frac{\pi}{2}-\theta\right) $$ $$\to\qquad 2\phi = 2\tan\theta + (\pi-2\theta)\sec^2\theta$$ $$\to\qquad(\pi-2\theta)\tan^2\theta+2\tan\theta + \pi-2\phi-2\theta = 0 \tag{$\star$}$$ Because $(\star)$ is a non-trivial transcendental equation, it cannot be solved algebraically. Mathematica gives this numerical approximation $\theta = 0.1784924382\ldots$ so that the radius of the circle is $$\lambda = \sec\theta = 1.0161440156726\ldots$$

Answer 2

Let $f(x)=1, g(x)=\sqrt{\lambda^2-x^2}$

My approach was to find the area under curves using integration.

Let it be known that $\lambda>1$ because if $\lambda=1$, then the area of the divided regions of the rectangle would be: $$A_1=\frac\pi4;A_2=\phi-\frac\pi4$$ $$A_1\ne A_2$$

The area of the quarter circle can be written as: $$\frac{A_\circ}4=\int_0^\lambda g(x)dx-\int_0^{\sqrt{\lambda^2-1}}g(x)-f(x)dx$$ Or as I was told, it could be written as: $$\frac{A_\circ}4=\int_0^\lambda g(x)dx-\int_1^{\lambda}\sqrt{\lambda^2-y^2}dy$$

So far, this is the my best representation for finding $\lambda$:

$$\frac\phi2=\int_0^\lambda\sqrt{\lambda^2-x^2}dx-\int_1^{\lambda}\sqrt{\lambda^2-y^2}dy \tag{1}$$

Now, evaluating $(1)$: $$\frac\phi2=\frac\pi4\lambda^2-\frac12(\lambda^2\sec^{-1}\lambda-\sqrt{\lambda^2-1})$$

Wolfram tells me $\lambda=1.01614$, which agrees with OP's answer, but I believe $\lambda$ above can solved algebraically.

Answer 3

As Blue answered, there is no analytical solution to the equation $$(\pi-2\theta)\tan^2(\theta)+2\tan(\theta) + \pi-2\phi-2\theta = 0$$ and numerical methods are required.

However, we can make quite nice approximation of $$f(\theta)=(\pi-2\theta)\tan^2(\theta)+2\tan(\theta) + \pi-2\phi-2\theta$$ using Padé approximants built around $\theta=0$ since the derivatives at this point are quite simple $\{0,2 \pi ,-8,16 \pi ,-128,272 \pi \}$.

For example, the $[2,2]$ Padé approximant would be $$\frac{(\pi -2 \phi )+ \left(\frac{4}{3}-\frac{8 \phi }{3 \pi }\right)\theta +\frac{\left(12 \pi ^2 \phi -32 \phi +3 \pi ^3+16 \pi \right)}{9 \pi ^2}\theta ^2 }{1+\frac{4 }{3 \pi }\theta +\left(\frac{16}{9 \pi ^2}-\frac{2}{3}\right) \theta ^2}$$ and the solution of the quadratic equation in numerator evaluates as $\approx 0.178564$.

Notice that a simple Taylor expansion around $\theta=0$ is $$f(\theta)=(\pi -2 \phi )+\pi \theta ^2+O\left(\theta ^3\right)$$ leading to $$\theta \approx \sqrt{\frac{2 \phi }{\pi }-1}\approx 0.173414$$ form which Newton method converges extremely fast $$\left( \begin{array}{cc} n & \theta_n \\ 0 & 0.17341404095680294715 \\ 1 & 0.17856358886001367599 \\ 2 & 0.17849245180567710065 \\ 3 & 0.17849243822881146264 \\ 4 & 0.17849243822881096809 \end{array} \right)$$