Solution

Answer: 3.2370342194

Let $E(x)$ denote the expected number of repetitions when the stopping threshold is

$$F=\frac1x.$$

We begin with the whole pie remaining, i.e. remaining fraction $=1$.

1. Mathematical analysis

Step 1: Distribution of the remaining piece

Suppose a fraction $r$ of the pie remains.

Jeff chooses two points independently and uniformly on the remaining boundary arc.

Parameterize the remaining arc by $[0,r]$, measured counterclockwise from the initial cut.

Let the two random cut positions be $U,V\sim \text{Uniform}(0,r)$.

The pie is divided into three sectors. Jeff eats the first two sectors counterclockwise from the initial cut, so the only part left is the sector from

$$\max(U,V)$$

to $r$.

Hence the new remaining fraction is

$$r' = r-\max(U,V).$$

Define

$$M=\frac{\max(U,V)}{r}.$$

Then $M\in[0,1]$ and

$$r' = r(1-M).$$

Step 2: Density of $M$

For two iid uniform variables on $[0,1]$,

$$\Pr(M\le m)=m^2.$$

Therefore,

$$f_M(m)=2m.$$

Let

$$T=1-M.$$

Then the remaining fraction multiplier is $T$, with density

$$f_T(t)=2(1-t), \qquad 0\le t\le1.$$

Thus after one step,

$$r' = rT.$$

Step 3: Integral equation for $E(x)$

We stop once the remaining fraction becomes $<1/x$.

Equivalently, starting from remaining fraction $1$, we continue while

$$r\ge \frac1x.$$

By scale invariance, the expected number of future steps depends only on $x$.

Conditioning on the first cut:

$$E(x) = 1+\int_0^1 E(xt),2(1-t),dt, \qquad x\ge1.$$

Also,

$$E(x)=0 \qquad (x<1).$$

2. Convert to a differential equation

Rewrite:

$$E(x)=1+2\int_0^1 E(xt)(1-t),dt.$$

Substitute $u=xt$:

$$E(x) = 1+\frac{2}{x}\int_0^x E(u),du -\frac{2}{x^2}\int_0^x uE(u),du.$$

Define

$$A(x)=\int_0^x E(u),du, \qquad B(x)=\int_0^x uE(u),du.$$

Then

$$E(x)=1+\frac{2A}{x}-\frac{2B}{x^2}.$$

Differentiate:

$$E'(x)=-\frac{2A}{x^2}+\frac{4B}{x^3}.$$

Using

$$A'(x)=E(x), \qquad B'(x)=xE(x),$$

one obtains after elimination:

$$x^2E''(x)+4xE'(x)-2=0.$$

3. Solve the ODE

Let

$$Y(x)=E'(x).$$

Then

$$x^2Y'+4xY-2=0,$$

$$Y'+\frac4xY=\frac2{x^2}.$$

Using integrating factor $x^4$:

$$(x^4Y)'=2x^2.$$

Integrate:

$$x^4Y=\frac23x^3+C,$$

$$E'(x)=\frac{2}{3x}+\frac{C}{x^4}.$$

Integrating again:

$$E(x)=\frac23\ln x-\frac{C}{3x^3}+D.$$

Step 4: Determine constants

We know

$$E(1)=1.$$

Also, from the integral equation, since $E(u)=0$ for $u<1$,

$$E'(1)=0.$$

Compute:

$$E'(1)=\frac23+C=0 \quad\Rightarrow\quad C=-\frac23.$$

Hence

$$E(x) = \frac23\ln x+\frac{2}{9x^3}+D.$$

Using $E(1)=1$:

$$1=0+\frac29+D$$

$$D=\frac79.$$

Therefore the exact formula is

$$\boxed{ E(x)=\frac23\ln x+\frac79+\frac{2}{9x^3} }.$$

4. Compute $E(40)$

$$E(40) = \frac23\ln(40) +\frac79 +\frac{2}{9\cdot40^3}.$$

Now evaluate numerically:

$$\ln(40)\approx 3.6888794541139363.$$

Thus

$$E(40) \approx \frac23(3.6888794541139363) +0.7777777777777778 +\frac{2}{576000}.$$

$$E(40) \approx 2.459252969409291 +0.7777777777777778 +0.000003472222222222.$$

$$E(40)\approx 3.237034219409291.$$

Rounded to 10 decimal places:

$$\boxed{3.2370342194}$$

5. Python implementation

import math

def E(x):
    return (2/3) * math.log(x) + 7/9 + 2/(9*x**3)

print(f"{E(1):.10f}")
print(f"{E(2):.10f}")
print(f"{E(7.5):.10f}")
print(f"{E(40):.10f}")

6. Code walkthrough

import math

Imports logarithm support.

def E(x):
    return (2/3) * math.log(x) + 7/9 + 2/(9*x**3)

Implements the closed-form formula derived above.

print(f"{E(1):.10f}")

Checks:

$$E(1)=1.$$

print(f"{E(2):.10f}")

Produces

$$1.2676536759\ldots$$

matching the statement.

print(f"{E(7.5):.10f}")

Produces

$$2.1215732071\ldots$$

again matching the problem statement.

print(f"{E(40):.10f}")

Computes the required value.

7. Final verification

Growth is logarithmic, which makes sense because each step removes a random positive fraction of the remaining pie.
The value $E(40)\approx 3.24$ is reasonable:
$E(7.5)\approx2.12$,
increasing the threshold to $40$ should require somewhat more than one extra expected step.

Everything is consistent.

Answer: 3.2370342194

Project Euler Problem 394