This is my answer to the puzzler presented here: http://fivethirtyeight.com/features/can-you-solve-the-puzzle-of-the-picky-eater/

The answer is:

The total proportion of the sandwich eaten by the picky eater is 0.2189514.

Work

We first draw a representation of the “sandwich”, in light blue below.

The four parabolas represent the collection of points that are equidistant from the center and the four sides. Their equations are \[ \begin{eqnarray*} y &=& x^2 - \frac{1}{4}\\ x &=& y^2 - \frac{1}{4}\\ y &=& \frac{1}{4} - x^2\\ x &=& \frac{1}{4} - y^2 \end{eqnarray*} \] The total area in the center bounded by the four parabolas is the region of “acceptable” points: the part of the sandwich that will be eaten by the picky eater.

We first work out length \(a\). The diagonal of the triangle from \((0,0)\) to \((1/2,1/2)\) has length \(\sqrt{2}/2\). Thus \[ a\sqrt{2} + 2a = \frac{\sqrt{2}}{2} \] which implies that \(a = 1/(2\sqrt{2} + 2)\). This means that area within the central square bounded by the red lines at the \(y=\pm a\) and \(x=\pm a\) is \(4a^2 = 1/(3+2\sqrt{2})\).

Using the knowledge that the parabola represented by the dashed black line has equation \(y = x^2 - 1/4\), the area of the blue hatched region is \[ \int_0^a x^2\,dx = \frac{1}{3}a^3. \] This means that the area of the pink shaded region is \[ \frac{a}{4}-a^2 - \frac{1}{3}a^3. \] In the whole figure, there are eight such regions.

The total area in the center bounded by the parabolas is thus: \[ 4a^2 + 8\left(\frac{a}{4}-a^2 - \frac{1}{3}a^3\right) \] which is 0.2189514. The total light blue area is 1, meaning that the area bounded by the parabolas is 21.9% of the “sandwich”.

Monte Carlo check

We can check the answer via Monte Carlo by randomly sampling points within the “sandwich” and working out the proportion of points that are acceptable to the picky eater:

## This is R code.
## Number of simulations to perform
M = 1000000
x = replicate(M,{
  ## Sample a point in the "sandwich"
  coord = runif(2,-.5,.5)
  ## Find the distance to the center
  dcen = sqrt(sum(coord^2))
  ## Is it closer to an edge than the center?
  any((.5-abs(coord))<dcen)
})

## Proportion of "good" points
p_good = mean(!x)
p_good
## [1] 0.219046

This agrees well with our exact answer.