Exact Answer: Cut a circular pizza into 16 Pieces, 12 of them with equal area. Find where the cuts should be made to satisfy these conditions.

Question

EDIT:

It's almost been a year since I attempted this problem, so I decided to come back to it.

One thing I was trying to do before was find a "closed form" of this problem, but really I wanted a way to infinitely approximate $w$ while isolating $w$ to one side of the equation.

Going back to a previously found equation:

$4w^2 = arcsin(w) + w\sqrt{1 - w^2}$

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TAYLOR EXPANSIONS
I kept the Taylor expansion of $arcsin(x)$ and $\sqrt{1-x^2}$ in the back of my mind decided to finally use them. (Note, I will also prove the following using integration later.)

Where $arcsin(w) = \sum_{n=0}^{\infty}\frac{(2n)!w^{2n+1}}{2^{2n} (n!)^2(2n+1)}$

And $w\sqrt{1 - w^2} = \sum_{n=0}^{\infty}\frac{(2n)!w^{2n+1}}{2^{2n} (n!)^2(-2n+1)}$

$\implies 4w^2 =\sum_{n=0}^{\infty}\frac{(2n)!w^{2n+1}}{2^{2n} (n!)^2(2n+1)} + \sum_{n=0}^{\infty}\frac{(2n)!w^{2n+1}}{2^{2n} (n!)^2(-2n+1)}$

$\implies 4w^2 =\sum_{n=0}^{\infty}\frac{(2n)!w^{2n+1}}{2^{2n} (n!)^2(2n+1)} + \frac{(2n)!w^{2n+1}}{2^{2n} (n!)^2(-2n+1)}$

$\implies 4w^2 =\sum_{n=0}^{\infty}\frac{(2n)!w^{2n+1}}{2^{2n} (n!)^2} (\frac{1}{(2n+1)} + \frac{1}{(-2n+1)}) $

$\implies 4w^2 =\sum_{n=0}^{\infty}\frac{(2n)!w^{2n+1}}{2^{2n} (n!)^2} (\frac{2}{(2n+1)(-2n+1)}) $

$\implies 4w^2 =\sum_{n=0}^{\infty}\frac{(2n)!w^{2n+1}}{2^{2n-1} (n!)^2(2n+1)(-2n+1)} $

$\implies w^2 =\sum_{n=0}^{\infty}\frac{(2n)!w^{2n+1}}{2^{2n+1} (n!)^2(2n+1)(-2n+1)} $

$\implies w =\sum_{n=0}^{\infty}\frac{(2n)!w^{2n}}{2^{2n+1} (n!)^2(2n+1)(-2n+1)} $

let $a_{n} = \frac{(2n)!}{2^{2n+1} (n!)^2(2n+1)(-2n+1)}$

$\implies w =\sum_{n=0}^{\infty}a_{n}w^{2n} $

INTEGRAL PROOF

$2w^2 = \int_{0}^{w}{\sqrt{1 - x^2}}dx$

$\implies \int_{0}^{w}{\sum_{n=0}^{\infty}\frac{(2n)!x^{2n}}{2^{2n} (n!)^2(-2n+1)}}dx$

$\implies [\sum_{n=0}^{\infty}\frac{(2n)!x^{2n+1}}{2^{2n} (n!)^2(-2n+1)(2n+1)}]_{0}^{w}$

$\implies \sum_{n=0}^{\infty}\frac{(2n)!w^{2n+1}}{2^{2n} (n!)^2(-2n+1)(2n+1)}$

$2w^2 = \sum_{n=0}^{\infty}\frac{(2n)!w^{2n+1}}{2^{2n} (n!)^2(-2n+1)(2n+1)}$

$w^2 = \sum_{n=0}^{\infty}\frac{(2n)!w^{2n+1}}{2^{2n+1} (n!)^2(-2n+1)(2n+1)}$

$w = \sum_{n=0}^{\infty}\frac{(2n)!w^{2n}}{2^{2n+1} (n!)^2(-2n+1)(2n+1)}$

$\implies w =\sum_{n=0}^{\infty}a_{n}w^{2n} $

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RECURSIVE APPROXIMATION
let $g_{k}$ be a recursive function for insertion purposes.
Where $g_0 = \sum_{n=0}^{\infty}{a_{n}}$

Similarly $g_k = \sum_{n_k=0}^{\infty}{a_{n_k}*(g_{k-1})^{2*n_k}}$

This substitution method is by reinserting $w$ into the right side an infinite amount of times recursively until it disappears from the right side entirely (as it will never reach the base case).

Therefore, $w=\lim{_{k\rightarrow\infty}}{g_k}$

This is also proven to work computationally in both python and Mathematica.

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WORK PREVIOUS I've recently been stumped on a simple pizza cutting problem that was brought to me by one of my friends at work.

Cut a pizza into 16 pieces where 12 of the pieces have equal area. More specifically A=B=C=D=E=F=G=H=I=J=K=L in Find where the pizza should be cut to satisfy this problem.

My first step was to make a visualization of the pizza in Desmos. From this visualization (and later in algebra) you can see that the size of the pizza doesn't matter, and the length of the inner square $w$ remains constant in relation to r.

My next step was to isolate a symmetric quarter of the pizza and find a relation between the known formulas. The area shaded in blue at the bottom was the key to me finding the formula that I need to be solved.

From this quarter, we know that:
$3(wr)^2 + s = \frac14 \pi r^2 $

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From the triangle, we know:
To $\theta,$
$wr$ is opposite
$\sqrt {r^2 - (wr)^2}$ is adjacent and
$r $ is the hypotenuse.

$sin(\theta) = \frac{wr}{r} = w $
$\implies \theta = \arcsin(w) $

(Triangle Area):
$\frac12b* h$
$b=w*r$
$h=\sqrt{r^2 - (wr)^2}$
$\implies \frac{1}{2}*w\sqrt{1 - w^2}*r^2$

(Arced Area):
$\frac{\theta}{2\pi} * \pi r^2 = \frac{\theta}{2} * r^2 = \frac{\arcsin(w)}{2}*r^2$

The combined area must equal $2(wr)^2$ because A=B=C=D=E=F=G=H=I=J=K=L ...

$[\frac{1}{2}*w\sqrt{1 - w^2}*r^2]$ + [$\frac{\arcsin(w)}{2}*r^2] = 2(wr)^2$
...
Multiply both sides by $\frac{2}{r^2}$
...

The equation in question:
$w\sqrt{1 - w^2} + arcsin(w) = 4w^2$

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Eureka! So plug and chug into Wolfram Alpha right?
Sadly that only works for approximations of the answer. $(w ≈ 0.480068840868035244581566735...)$
I was wondering if it's even possible to have an exact answer for this problem, or if it's impossible due to the approximate nature of $\arcsin(x)$'s Taylor expansion.

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Note: Approaching this problem by using the accumulation function of a circle up to wr will yield the same result.
Because $\int\sqrt{r^2 - x^2} dx = \frac{1}{2} * x\sqrt{r^2 - x^2} + \frac{1}{2}\arcsin(\frac{x}{r})*r^2 + C$

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Alternate Forms of the equation:
Trig Approach:
$sin(\theta)*cos(\theta) + \theta = 4sin^2(\theta)$
Hyperbolic Trig Approach:
$(4+i)e^{-2i\theta} + (4-i)e^{2i\theta} = 8 - 4\theta$
$(θ≈0.5007331859243733442276693702257483572649...)$

Answer 1

You are correct, the problem boils down to finding the $z\in(0,1)$ such that $$\int_{0}^{z}\sqrt{1-x^2}\,dx = 2z^2, $$ i.e. to solving $$ z\sqrt{1-z^2}+\arcsin(z)=4z^2 \tag{A}$$ or, by setting $z=\sin\theta$ and $\varphi=2\theta$, $$ 4\cos\varphi+\sin\varphi=4-\varphi. $$ Solving $(A)$ is essentially equivalent to solving Kepler's equation (see also this recent answer).
An approximate solution can be derived from the Shafer-Fink approximation $$\arcsin(x)\approx\frac{3x}{2+\sqrt{1-x^2}}\tag{B} $$ leading to $z\approx 0.48$. Such initial approximation can be refined through Newton's method, till reaching $z\approx 0.480068840868$, but there is no hope of getting a nice closed form for the solution of the trascendental equation appearing in (A).