Discover the Area of Points in a Unit Square | POTW #471

[anemone]

Here is this week's POTW:

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Find the area of the set of all points in the unit square, which are closer to the center of the square than to its sides.

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[anemone]

Congratulations to lfdahl for his correct solution(Cool), which you can find below:

Spoiler

By symmetry, it suffices to solve the problem in one half quadrant of the unit square.

The boundary of the solution set comprises of the points in the plane for which $|OA| = |AB| = \frac{1}{2}-y$.

From the figure, one immediately has

(a). $\cos \theta = \frac{y}{\frac{1}{2}-y}$

(b). $z^2 = \left ( \frac{1}{2} \right )^2+ x^2$

Since $\angle OAB = \pi - \theta$ the acute angles in the isosceles triangle $OAB$ are $\frac{\theta}{2}$, and it follows, that

(c). $z = 2 (\frac{1}{2}-y) (\cos \left ( \frac{\theta}{2} \right )$.

With the help of the half-angle formula: $\cos \theta = 2\cos^2\left ( \frac{\theta }{2} \right ) -1$ and combining (a), (b) and (c), we get the result:

\[y = \frac{1}{4}-x^2\]. The graph has endpoints in $\left ( 0,\frac{1}{4} \right )$ and $(r,r)$.

The right end point is determined by the condition: $r = \frac{1}{4}-r^2$, which has the (positive) solution:

$\frac{1}{2}\left ( \sqrt{2} -1\right )$.

The solution set for the 1st quadrant thus comprises the area under the red y-curve minus the area of the coloured triangle:

\[A = \int_{0}^{r}\left ( \frac{1}{4}-x^2 \right )dx - \frac{1}{2}r^2 = \left ( \frac{1}{4}-\frac{1}{2}r-\frac{1}{3}r^2\right )r\]

Thus, by symmetry, the solution is: $A_{sol} = 8 A = \frac{1}{3}\left ( 4\sqrt{2}-5 \right ) \approx 0.218951$