Let $(a,b)$ be any point in the disk $D = \{(x,y): x^2 + y^2 < 1\}$. Put $r=\sqrt{a^2 + b^2}$. Let $R_{(a,b)}$ be the open rectangle with vertices at the points $\left(a\pm\frac{1-r}{8}, b \pm\frac{1-r}{8} \right)$. Verify that $R_{(a,b)}\subset D$.
I drew the following example on Mathematica - the border of the circle has to be dotted, I know. Let's just pretend it is :P
Anyways, so my reasoning is as follows:
We must show that the furthest possible distance from the origin to a vertex on the rectangle, is always less than $r$, i.e, it is still within the disk $D$. I am having trouble finding an equation for this furthest distance though. Can anybody please help me?
I know that it will be the length of the hypotenuse from the right angled triangle drawn from the origin to the furthest "corner vertex" of the rectangle, but I'm having trouble finding this formula.
Can anyone please help me?
Just check the four points of the rectangle are within the circle. That's enough since both the circle and the rectangle are convex sets.
For example for the $\;(x,y):=\left(a+\frac{1-r}8,\,\,b-\frac{1-r}8\right)$ :
$$x^2+y^2=a^2+\frac{1-r}4+\frac{(1-r)^2}{64}+b^2-\frac{1-r}4+\frac{(1-r)^2}{64}=$$
$$=a^2+b^2+\frac{(1-r)^2}{32}<r^2+\frac{r^2-2r+1}{32}=\frac{33}{32}r^2-\frac r{16}+\frac1{32}$$
We must prove that last quantity is less than one, so:
$$\frac{33}{32}r^2-\frac r{16}+\frac1{32}<1\iff33r^2-2r-31<0\;\;(**)$$
The last quadratic's discriminant is
$$\Delta=4+4092=(64)^2\implies r_{1,2}=\frac{2\pm64}{66}=\begin{cases}1\\{}\\-\frac{31}{33}\end{cases} \implies$$
$$(**)\;\;33\left(r-1\right)\left(r+\frac{31}{33}\right)<0\iff -\frac{31}{33}<r<1$$
and this is true since $\;a^2+b^2=r^2<1\;$