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.

$ \begingroup $ EDIT:
It ‘s about 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 kind ” of this problem, but very I wanted a way to infinitely approximate $ tungsten $ while isolating $ west $ to one english of the equation.

Going rear to a previously found equation :


$ 4w^2 = arc sine ( west ) + w\sqrt { 1 – w^2 } $

TAYLOR EXPANSIONS
I kept the Taylor expansion of $ arc sine ( x ) $ and $ \sqrt { 1-x^2 } $ in the back of my thinker decided to ultimately use them. ( Note, I will besides prove the following using integration late. )
Where $ arc sine ( w ) = \sum_ { n=0 } ^ { \infty } \frac { ( 2n ) ! w^ { 2n+1 } } { 2^ { 2n } ( normality ! ) ^2 ( 2n+1 ) } $

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

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

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

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

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

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

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

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

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

INTEGRAL PROOF

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

$ \implies \int_ { 0 } ^ { watt } { \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 } ( nitrogen ! ) ^2 ( -2n+1 ) ( 2n+1 ) } ] _ { 0 } ^ { west } $

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

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

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

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

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

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

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

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

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

This is besides proven to work computationally in both python and Mathematica .
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 peer sphere. More specifically A=B=C=D=E=F=G=H=I=J=K=L in enter image description here Find where the pizza should be cut to satisfy this problem.

My first step was to make a visual image of the pizza in Desmos. From this visual image ( and belated in algebra ) you can see that the size of the pizza does n’t matter, and the length of the inner squarely $ west $ remains changeless in sexual intercourse to r .
My adjacent step was to isolate a symmetrical quarter of the pizza enter image description here and find a relation between the known formula. The area shaded in blue at the penetrate was the key to me finding the formula that I need to be solved .

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

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

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

( Triangle Area ) :
$ \frac12b* planck’s constant $
$ 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 ( west ) } { 2 } *r^2 $

The blend 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 ( tungsten ) } { 2 } *r^2 ] = 2 ( wr ) ^2 $

Multiply both sides by $ \frac { 2 } { r^2 } $

The equation in interrogate :
$ w\sqrt { 1 – w^2 } + arc sine ( w ) = 4w^2 $

eureka ! indeed chew 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 potential to have an demand answer for this problem, or if it ‘s impossible due to the approximate nature of $ \arcsin ( x ) $ ‘s Taylor expansion .
eminence : approach this trouble by using the accumulation officiate of a circle up to wr will yield the same leave.
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 $
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 … ) $