consider a square of side length $x$, find the area of the region which contains the points which are closer to its centre than the sides.

Question

Any ideas how to start.

I am having trouble figuring out the region itself

All ideas are appreciated

thanks

Answer 1

We consider the square $Q = [0,x]^2$, because of symmetry, we will look only at the lower left subsquare $Q_1 := [0,\frac x2]^2$. $Q$s center is $(\frac x2, \frac x2)$. For a point $y \in Q_1$, with $y_1 \le y_2$, its distance to the sides is $y_1$, its distance to the center is $\sqrt{(y_1-\frac x2)^2 + (y_2 - \frac x2)^2}$. We have \begin{align*} && y_1 &\le \sqrt{\left(y_1-\frac x2\right)^2 + \left(y_2 - \frac x2\right)^2}\\ & \iff &y_1^2 &\le y_1^2 - x(y_1+y_2) + y_2^2 + \frac 14x^2\\ &\iff & xy_1 &\le y_2^2 -xy_2 + \frac14x^2\\ & \iff & y_1 &\le \frac 1xy_2^2 - y_2 + \frac 14x \end{align*} That gives the part of the region in $Q_1$ above the line $y_1 = y_2$, the region below (where $y_2$ is the distance to the sides) is handled along the same lines. Symmetry gives the whole square then.