Can You Bake The Radish Pie?

Riddler Express

Start with any positive amount of flour. Replace it by the average of itself and $3$ divided by itself, and repeat forever. How many cups does the recipe converge to?

Solution

Call the running amount $f$. Each step sends $f$ to $\tfrac12\!\left(f + \tfrac3f\right)$. If this settles to a limit, the limit is a fixed point, unchanged by the step: $$f = \frac12\left(f + \frac3f\right) \;\Longrightarrow\; 2f = f + \frac3f \;\Longrightarrow\; f^2 = 3 \;\Longrightarrow\; f = \boxed{\sqrt3}.$$ The recipe is Newton’s method for $\sqrt a$ in disguise: the iteration $x \mapsto \tfrac12(x + a/x)$ converges to $\sqrt a$ from any positive start, here with $a = 3$. So whatever amount of flour you begin with, you end at $\sqrt3 \approx 1.732$ cups.

The computation

Iterate the recipe from an arbitrary start and watch it converge.

def num_cups(start=2.0):
    s = start
    while True:
        c = 0.5 * (s + 3 / s)
        if abs(s - c) < 1e-12: return c
        s = c
print(num_cups())                              # 1.7320508... = sqrt(3)

Riddler Classic

In a regular $1000$-gon with side length $2$, take a longest diagonal (joining opposite vertices). The diagonals perpendicular to it slice it into $500$ pieces. What is the product of the piece lengths? Extra credit: in a regular $1001$-gon with side $2$, slice the altitude from a vertex the same way into $500$ pieces, and find that product.

Solution

Put the long diagonal on the $x$-axis. Going up the right half of the polygon, each successive side turns by $\tfrac{2\pi}{n}$, and the perpendicular diagonals meet the axis at the vertices’ $x$-coordinates. The piece between two consecutive cuts is the projection of one side onto the axis, which for the $k$-th side works out to $$V_k = 2\sin\!\Big((2k-1)\frac{\pi}{n}\Big), \qquad k = 1, \dots, \tfrac n2,$$ so the product is $P = \prod_{k=1}^{n/2} 2\sin\!\big((2k-1)\tfrac{\pi}{n}\big)$. To evaluate it, recall that the non-trivial $n$-th roots of unity give $\prod_{k=1}^{n-1}\big(1 - e^{-2\pi i k/n}\big) = n$. Writing each sine through complex exponentials and separating the odd-indexed factors from the even-indexed ones turns the product into $n$ divided by $\tfrac n2$: $$P = \frac{\prod_{k=1}^{n-1}\big(1 - e^{-2\pi i k/n}\big)}{\prod_{k=1}^{n/2-1}\big(1 - e^{-2\pi i k/(n/2)}\big)} = \frac{n}{\,n/2\,} = \boxed{2}.$$ Remarkably the side length and the value $n = 1000$ drop out entirely. For the odd case the same sine-product identity $\prod_{k=1}^{n-1}2\sin\tfrac{k\pi}{n} = n$ splits into odd and even terms, each equal to $P$, giving $P^2 = n$: $$P = \sqrt{n} = \sqrt{1001} \approx \boxed{31.64}.$$

The computation

For $n = 1000$, build the polygon from coordinates, find where the perpendicular (vertical) diagonals cut the horizontal long diagonal, and multiply the gaps. For the odd case multiply the piece lengths $V_k$ directly.

import numpy as np
def diagonal_product(n):                       # even n: direct construction
    R = 1 / np.sin(np.pi / n)                   # circumradius, side length 2
    cuts = sorted({R * np.cos(2 * np.pi * j / n) for j in range(n // 2 + 1)})
    return np.prod(np.diff(cuts))
print(diagonal_product(1000))                  # 2.0
k = np.arange(1, 501)
print(np.prod(2 * np.sin((2 * k - 1) * np.pi / 1001)))  # 31.638... = sqrt(1001)